Hello, welcome to module 49 of NPTEL NOC, an introductory course on point set topology particle. So, we continue our study of ordinal topology. In a totally order set, every sequence has a monotone subsequence. This is an elementary result which goes under the name peak valley lemma. Maybe you have not seen it. On the other hand, this is actually important for us now. So, I would like to recall it completely. So, what is the meaning? See, you can imitate the real numbers. Real numbers, maybe you know this one, but do not use the full properties of real number. You have to use only that is totally order. No addition, no subtraction, no multiplication and so on. That is the whole idea. You can still prove this one just by using totally order. Every subsequence has a monotonically increasing sequence or a monotonically decreasing sequence. Some monotone subsequence should be there. That is the thing. So, fixing a sequence, consider the following two properties. Given any n inside n, you will have m bigger than n such that this s of m is you know less than or equal to s n follows s n. So, this is not to be confused with the order of real numbers. That is all. So, I have carefully written this one. s m is s n is s n. There is a larger m for which s m is smaller. Second one is given n inside n, there exists larger m such that s n is greater than s n. Rather, rather wrong. If s satisfies d, it follows that as such a decreasing subsequence. Similarly, if f satisfies i, it will have the increasing subsequence. That is why I have put d and i. So, decreasing sequence, increasing sequence will come forward by this property. So, there is nothing to prove. Therefore, let us assume that neither d nor i is true. Under this, let us see what happens to the sequence. So, if possible, let s be such that neither d nor i is true. Starting with s 1, for definiteness sake, we may assume that s 2 is bigger than 1. It is less than or equal to s 1. One of them has to be there. So, this is a symmetrical operation. So, we assume s 1 is less than or equal to s 2. Since s does not satisfy i, that means what? If you took s 2, I must be getting s 3 bigger than that. Some s and s k bigger than that. That is not possible. So, that is a positive. One, we always do not find. But is it true? It is not true indefinitely. It follows that there will be one n 1 belonging to n such that for all m bigger than n 1, s m will be smaller than s n 1. You cannot go further. So, pick up that n 1. What you may assume is that I have started the sequence at s n 1 itself. Just forget about the earlier part.  It has to be less than or equal to this one. m is bigger, h n plus 1 is bigger than m. So, it must be less than that. So, that is why I could have assumed. See in the beginning, I assume s 1 is s 2 is bigger than n 1. Now, I have another thing. Namely, this s n plus 1 is smaller than s n plus 2. The role is interchangeable. But now, can I keep going down and down and so on? d says it is not possible because I have assumed that d is not true. So, d is not possible. So, that means what? You see, this is what the picture says. I started with s n s 1 here. I keep going up. Maybe s 2 is there. But maybe s 5 is there. I have come somewhere here s n 1. After that, if you look at the whole thing, everything is smaller. There is nothing bigger than this. What is the meaning of this one? This is a peak. So, that is why p 1. So, now the next one is definitely smaller than that because everything is smaller now. So, you keep going down, down, down. If you have what? Subsequence which is decreasing. You keep coming down. In between, there may be something more. Don't worry.  So, this is a valley. What is the meaning of this? Everything after that is bigger than p 1. So, they are above. But they are all smaller than p 1. See, because the p 1 has a property, this is bigger. So, repeat the process whatever you have done from here. The next one will be bigger. Keep going up. How far you can go somewhere wherein the same thing what happened to p 1 will happen to p 2. Namely, everything after that is smaller than p 2. Keep repeating. You get p 1, v 1, p 2, v 2 and so on. An interlaced, what is interlaced subsequence. What is the property of p 1? It is bigger than everything after that. What is the property of p 2? It is bigger than everything after that. Therefore, p 1, p 2, p 3 is a monotonically decreasing sequence. Indeed, you are getting two of them. v 1, v 2, v 3 will be monotonically decreasing sequence. So, either of them contradicts the assumption. Therefore, there must be a subsequence which is monotone. So, that is the proof. So, I have written down it completely. Started with n 1, it pick up n 2, etcetera. So, it is that n i is now bigger than s of n i plus 1, but the assumption does not allow you to continue. So, you have to stop somewhere. So, you have to keep changing the role each time. So, finally, you get two monotones of sequences like this. One is increasing, one is decreasing. The next thing about now I come back to the ordinal 0 omega. Both 0 omega closed and 0 omega open, they are sequentially compact. Remember one sequentially compact, every sequence has subsequence which is cannot be changed. So, we are discussing this sequential compactness. So, this 0.9 here is done in that background. So, how do you prove that something is sequentially compact? Start with a n, any sequence in 0 omega. As seen above, just using total order. First of all, it is bounded. Any sequence is bounded that we have seen before and before that. In a well-ordered stack, we have seen that every countable set is bounded. It is bounded inside 0 omega itself. Therefore, a monotone subsequence will be also bounded above. Bounded above is what is important. Going down is always true here. But then it will convert to either supremum or supremum. According as the sequence is decreasing or increasing. Depending upon decreasing sequence, it will convert to the infimum. Increasing sequence will convert to supremum. Therefore, we have proved that every sequence has a convergent subsequence. So, 0.9 is used only to get a monotone subsequence. This monotone subsequence, I do not know whether it is increasing or decreasing. So, both of them you have to take the care. Both of them will converge in any case because it is bounded. This is also standard result in analysis. Namely, in real line, if we have a bounded monotone sequence, it is converted. There it is the property of the existence of least upper bound and the latest lower bound. So, we have more or less proved that theorem also here in this approach because all that you have to assume, all that you have to do is here is just a total order. That is all. And of course, assumption that least upper bound and latest lower bound exist, which we have proved separately for this 0 omega or well order sets. The next thing is 0 omega open is first countable and T1, actually T2 we have seen. And then the above result implies that it is limit point compact and countably compact also. So, I will try to recall these things. So, let me go through this one. Sequentially compactness implies limit point compactness under the axiom T1 and first countability. So, the converse is also true. Always sequential compactness implies these two, but under the T1-ness first countability then T1-ness, so they will imply limit point compactness implies sequential compactness. This is what we have seen. You can have a look at this picture. Remember, so this picture will tell you at this point. So, this is limit point compactness here. This is countable compactness here. Plus T1, limit point compact plus T1 will imply is countable compactness. T1 plus first countability limit point compact implies sequential compactness. From here to here you can come back. So, you can recall. I have just recalled that one for your ready reference. So, let us go back here now. So, 0 omega is first countable. 0 omega closure is not first countable. That is what we have seen. And of course, it is T1 because it is T2 also. Hence, the above result that it is sequential compactness implies that it is limit point compact and countably compact also. Of course, 0 omega, since it is not compact, this implies that 0 omega open is not Lindelof. Alright? Because once it is Lindelof and countably compact, it will be compact. Remember, Lindelof means what every open cover has a countable sub cover. Countably compact means what? Every countable sub cover has a countable cover has a finite sub cover. So, combining these two would have got compactness. So, important thing we have derived is 0 omega is not Lindelof. Okay? Next, if A and B are any two nonempty disjoint closed subsets in 0 omega, then at least one of them is countable and its closed ring 0 omega. So, we are slowly going towards normality here. So, take two disjoint closed subsets of 0 omega. Okay? Of course, nonempty are starting. At least one of them is countable. It is a very strong condition and that countable set is closed in 0 omega, omega closed. A closed subset here may not be closed here, right? How? Because this capital omega may be a limit point. So, this says that capital omega is not a limit point. This we know already because any countable subset of 0 omega is already bounded inside 0 omega. So, this part we know already, right? So, let us see how this one works. Part of it we have already seen. Let y bar denote the closure of y in 0 omega, omega closed in the larger space. Just for in this part. Otherwise, closure closure where 0 omega or 0 omega, I will have too many notations here. I do not want to have that. Okay? So, for any subset y of 0 omega, let the y bar, you know, the closure is 0 omega closure. Now, look at the case wherein a bar union b bar is contained inside 0 omega. It need not be, this is a case I am talking about. It need not happen this. Suppose this is true, then both are bounded in 0 omega, right? Because they are closed subset of now inside 0 omega. They are closed subset. Content in here does not matter. So, both are inside this one. Therefore, from our larger result, which we have just seen last time, both of them are commutable.  Okay? It is a stronger thing we should have. So, we are through. So, now come to otherwise means what now? Suppose a bar in b bar will contain the this element capital omega. Otherwise, by symmetry, you may assume that capital omega is inside a bar. Okay? Interchange a and b if you want that is all. Suppose this capital omega is in b bar also. That can also happen, right? Then given b naught inside 0 omega open, we have this a intersection b naught omega closed is non-empty because all these open subsets must intersect a because omega is inside, capital omega is inside a bar. It is non-empty and then we get an element b naught.  Okay? Now, apply the same thing to a1 to omega. This is neighborhood of capital omega, intersection b that is non-empty. So, you get an element b1 inside b bigger than a1. Repeat this process. What you get? You will get an internalized sequence a1 less than b1 less than a2 less than b2 less than and so on, right? So, where are they going? Each time they are inside a smaller thing, the increasing sequence that is the whole idea. Okay? So, we observe that this internalized sequence, okay? They must have a same supremum. They have same limit. This is what we have seen earlier and that limit must be inside 0 omega. Okay? But then because it is any sequence which converges has to converge inside 0 omega only. The sequence is inside 0 omega. So, it does not converge to capital omega. No sequence converges to capital omega. But then s itself will be in a bar intersection 0 omega which is a and s is also in b bar intersection b. This means s is inside a intersection b and that is a contradiction because the intersection b is empty to begin with. So, they are disjoint closed subsets of 0 omega to begin with. So, therefore, if we assume a is inside a bar instead of omega is inside a bar, then omega is not inside b bar. But that means b itself is closed inside, b itself is closed inside 0 omega. Okay? Anything is closed inside 0 omega that is a compact. But now this compact subset is inside 0 omega open. Therefore, b is countable. This we have seen already. So, we wanted to prove one of them is countable. Right? Automatically, it will be closed. That part will be there. Countable and closed here. That is all. Now we will use this one in a meaningful way. Namely, to prove that 0 omega open is normal. We already know that 0 omega open is Hausdorff. Use this fact that 0 omega closed is normal and T2 actually is a compact and T2 space which is normal. Right? Therefore, starting with the two disjoint closed subsets here. Okay? You go to the closures of them inside 0 omega closure. Just now what we have shown is they are themselves closed. The closures are closed and disjoint. Because only omega bar is the question that belongs to only one of them. What does that mean? You can separate them inside 0 omega bar, 0 omega closure by two open subsets, disjoint open subsets. Intersect these disjoint open subsets with 0 omega open. So, that will give you disjoint closed subsets of these two closed subsets. Disjoint open subsets containing respectively these two open subsets within 0 omega open. Okay? That this space is normal. It is already T2. So, it is a T4. Okay? The next thing is the aim of all this. If you take the product of 0 omega open cross with 0 omega closed. Okay? Let us denote this one by T. This is going to be a wonderful example now. So, I am not taking 0 omega cross 0 omega. That will be compact cross compact. It will be compact. Okay? So, this is going to give you a number of counter examples. Okay? So, now we have some notational problem here. See, we have been using the ordered pair kind of A, B to denote intervals. Both inside real number systems as well as any ordered set. Right? But now the ordered pair we have to write here. So, I will not use that notation for ordered pairs. I will restrict it to open intervals only. So, ordered pairs will be denoted by X cross Y now. So, this is an element in the standard ordered pair notation. It would have been X comma Y brackets. Right? So, that notation will not be used for some time in this part now. Okay? So, with that convention, we shall show that T, this product is not normal. This is normal. This is actually compact order set. Yet the product is not normal. So, that is a strong example for all. You know, product is not normal was promised in part one but we did not prove any examples there. So, here is an example. Beautiful example. So, let us have this standard notation. Delta denotes the diagonal X cross X where X varies over 0 omega closed. But I do not want to go to 0 omega cross 0 omega but I want to be inside T. Take A to be delta intersection T. Okay? So, last point omega cross omega is not there. That is the problem. And B equal to 0 omega open cross singleton omega. So, second factor has omega here. Right? I can take that. So, this A and B are subsets of T. Obviously, they are disjoint. Alright? This part omega cross omega is not there but here there is no omega cross omega either. So, anyway, these are all. So, this is like a top line and this is a diagonal course all the way near but the intersection point is not there. That would have been omega cross omega. So, these two are disjoint closed subsets. Since 0 omega bar with this closure is half door, this delta is closed. Therefore, A intersection that delta intersection T is A. A is closed in T. Okay? Similarly, singleton is closed. So, this opens up that is closed in 0, T. So, this is closed in T. Alright? Also, A intersection B is empty. Is that clear? So, now we claim that there are no disjoint open subsets U and B inside T such that A is inside U and B is inside V. So, that will complete the proof that T is not normal. So, you have to do a little bit of work here. Assume on the contrary. That means what?  You will arrive at a contrary. That is the whole idea. Okay? Assume the contrary. Fix each, for each x belonging to 0 omega, you just fix an element 0 omega. Look at the subspace, the vertical subspace mx equal to singleton x cross 0 omega. That is an element of T, that is a subset of T, right? And it is homomorphic 0 omega because I first fact prime singleton set. Then if you take V, U and V are open subsets, right? V intersection mx, that will be a neighborhood of x cross omega. X cross omega is one of the points inside B, right? So, V contain B. So, x cross omega is inside V. So, it is also inside mx. So, this is a neighborhood of x cross omega inside mx. Okay? Therefore, V intersection mx minus singleton x cross omega is a punctured neighborhood. That is non-empty. Okay? It is an open subset and this is a singleton. Okay? This is non-empty subset. If you put rx equal to all of second coordinate y such that this x is less than y. Okay? This is our old notation, sorry. This is just the right ray. It follows that this is this non-empty set. What is this? V intersection mx minus this one. This is non-empty set. But this is actually contained inside x cross rx minus U. Okay? So, let me show all these things in a picture just to see that you are not completely lost. So, this is my T. This is 0 cross, you know, this is just 0 cross. This whole thing is omega open and this way I put 0 comma omega closed. So, that is a cross. This is the second factor. This is the first factor. Okay? So, I have taken U here, this A here. This is the diagonal minus this point. This point omega cross omega is not there. Okay? Now, B is here. This is just this singleton x, singleton 0 cross or this B here except this point is not there. So, these two disjoints are set and I have taken this dot dot dot here U is an open neighborhood of this A and V is an open neighborhood of V and I am pretending they are disjoint. What is happening when you keep coming closer to closer? I do not know. Okay? So, that is where the mystery is there. Finally, we will get a contradiction there. So, what I have done? Fixy x here on this axis, similarly x cross 0. Look at this vertical line. Okay? That is my Mx. This V intersection Mx, this portion is the neighborhood of this x cross omega. If you throw away this point, this part is non-empty. Okay? This U is disjoint from that whether I throw it away or not, this will still be non-empty. That is all I am telling you. Okay? So, Rx is what? Take any x here, x has come here somewhere. Okay? x is here, x comma x. So, take everything above that, that is Rx. So, this is from here to here, it is x cross Rx. Throw away U, this is U part. Okay? So, this part will be still non-empty all the time. And it contains V minus, sorry, V intersection Mx, this line minus this point. So, that is all so far we have done. This V intersection Mx minus x cross omega, it contains the x cross Rx minus 0. Okay? So, that is non-empty set. So, we can define a function here. Namely, take the minimum of all elements y such that x cross y is inside this non-empty set, x cross Rx minus 0. So, what is this in this picture? This is the minimum alpha x. Okay?  So, that is the definition, x cross Rx minus 0. Take the minimum. So, this is the function now because these are non-empty subsets for each x. Okay? Alpha x is some point of 0 omega for each x. Okay? Starting with any x one times at 0 omega, inductively you define xn equal to alpha of xn minus 1. So, you see now the induction process starts. Maybe I started here. Okay? Now alpha of this one will be some point here. Okay? So, that point will be somewhere here. Then you have to take alpha of that will be some point here. That point will be somewhere here. Now alpha of that point will be somewhere here. You take alpha of that and so on. Okay? So, this is the induction process here. Namely, starting with x1 belongs to 0 omega, anything wherever you want to start. Inductively define x2 equal to alpha of x1, x3 equal to alpha of x2 and so on. So, obtain a sequence xn inside 0 omega which is monotonically increasing. Why? Look at this one. This alpha x is bigger than x because x, x is inside u. So, no chance that this alpha x will be the minimum of this one will be less than that. It is actually away from that. It is bigger than that. Okay? So, this is a strictly monotonically increasing sequence, converging to some point in 0 omega. Any sequence in 0 omega which is monotonic has to converge inside 0 omega. That is all. But then the sequence xn cross alpha xn, where does it converge to? xn is going to, why? Alpha xn is bigger than that. So, it must go to bigger than that one.  So, alpha xn is nothing but the same sequence xn, except the indexing is changed. So, both of them converge to same point. So, their same sequence is except indexing is changed. That is all. So, it will converge to y cross y inside a which is contained inside u by definition. Whereas, none of the terms of this sequence is inside u. Right? Look at this point, x comma alpha x is this point. Right? When you take xn plus 1, xn plus 1, alpha xn plus 1 will be in this point like that. Though they are not inside u. So, that is a contradiction. So, this proves that t is not normal. So, finally, why this notation t? Because this is called Ticknoff's Planck. This example is called Ticknoff's Planck. In conclusion, we say normality is not finite productive. As a conclusion, we can say that Ticknoff's Planck is not para-compact. Para-compact, Hoss-Dorff's space is para-compact. Regular space will be normal. That is what we have seen already. It follows that 0 omega is not para-compact because para-compact, cross para-compact would have been para-compact. So, 0 omega itself is not para-compact. So, in one single example, you have so many contra-examples here. There will be more to come. This is another wonderful property of 0 omega. Every continuous function 0 omega to R is eventually a constant. Real value continuous functions are all eventually constant. See, the role is reversed here. Usually, if you have a connected space and a continuous function R into a disconnected space, a connected space into a disconnected space or a discrete space, then it is a constant. That is the way you have been all the time doing. So, this is highly connected. This is a nice connected space. This is the one which has problems in a disconnected space. But this is the other way around now. This is a eventually constant. So, what is the meaning of that? There are some x inside 0 omega such that f of the right ray Rx, x2 omega, that is a single term. By composing with a homeomorphism g from R to 0 1, because any homeomorphism, you can compose with a 0 1 because you can take a tannin or whatever. There are many. You may assume that the function is bounded first of all. It is actually taking value 0 1. If I show this is eventually constant, then f will be also eventually constant. So, let us assume f is bounded. Put Rx equal to all y inside 0 omega. This is the old notation. I am changing here. Now, I am taking a closure here instead of R bar of x. We shall claim that there exists a monotonically increasing sequence xn in 0 omega such that the diameter of f of Rxn, f of Rxn is instead of R. So, diameter of that makes sense because this is a matrix space. That is less than 2 third rays to n. Some number raised to n, that number is less than 1 is important for r. Then, if you take x as a limit of xn, it will follow that f of Rx is a singleton. This kind of term, you know balls of radius smaller and smaller converging to 0, it will be in the intersection f of Rxn. So, it must be singleton. So, that is the way we are going to prove this theorem. Just by proving that, you know a construction of monotonically increasing sequence in 0 omega, which will automatically converge and that converging point will be our x. Then, f of Rx contained inside each of them being the smallest subset of that. Therefore, it is contained in the intersection. Therefore, it is happening right now. So, we will have to construct this sequence. Given, so we are going to prove slightly different statement here, inductive statement here and then apply it. So, this statement is followed. This s given x belong to 0 omega, let g from this right ray to AB be a continuous function. Right now, recall this is I have taken as a closed ray. Suppose you have continuous function, which is bounded. That is the whole A to B. Then, there exists a y bigger than x. That is an element of Rx such that g of R y is contained inside the first two-third or the second two-third. The first two-third means what? A to A plus two-third of B minus A. B minus A is the length of this one. I am adding that.  So, that is again a two-third length. The first two-third or second two-third. They are not disjoint by the way. Remember that. That is fine. I want to say it is either here or here. It may be in the intersection also. That is good enough for me. No problem. It is contained in one of them is important. That is what happens. Then, we construct a sequence x n as follows. See, first I had A B. Then, if this A B is zero one, then what happens? R of g of R y will be the maximum radius of the diameter of that one will be two-third. Apply this g. Apply the same thing to this function. Repeat this process. You will get another y two such that diameter of the new thing will be two-third of that. That is what we are going to do. So, then we construct a sequence x n as follows. Start with x naught belongs to zero omega. Take g equal to s. Take x 1 equal to y as given in s. Having defined x n, repeat the step to the function g which is restricted to f x n. You get x n equal to 1. This s is a general statement. I am starting with this hypothesis. f is from zero one to zero one to r, but I have converted the g from r. This f is zero one to zero omega to zero one. So, begin with radius is one. This is the second stage. The first stage I am getting the diameter two-third. Next, I will get two-third into two-third and so on. So, we have to prove this statement. Once we prove this one, this inductive step is over. Then the proof is over. So, let us prove this statement in s now. So, let us prove.  This means these are one-third one-third now. Like cantor set, okay, in the construction of cantor set. A to one-third length and then one-third length to entire B. They are disjoint. If you take g inverse of that, they will be disjoint subsets of zero comma omega. What do you know about the disjoint rows subsets of zero comma omega? We have done something, you see. So, that can be used now. As seen before, one of them is countable, okay, and so as an upper bound, say the first one is bounded by y. Which one? This one, that one, does not matter symmetrically. So, let us say, assume this is bounded by y. This y is an element of what? Element of g inverse of, you know, sorry, element of the bound of the main leaf AB. g is a function, sorry, in rx. So, y is bigger than x, that is all, okay. So, we get a y. That is all as seen above. One of them is countable and hence we have this. This means that g of r y is contained inside this part. Likewise, if the second one is bounded by y, then you will get g of r y is contained inside this part. So, that is all, okay. So, why I separated out this one, the normality, directly instead of proving a and b are disjoint rows subsets, find open subsets and so on. Disjoint rows subsets have themselves some interesting property. Namely, one of them has to be countable and that is here. So, next thing is a strong conclusion, okay. The first thing is 0 omega closure is the Alexander compact, one point compactification of 0 omega open. This is very easy to see by the very definition of this, there is only one extra point, right. What are the neighborhoods of this omega? Take any neighborhood system, something like alpha and x open to omega capital. Compliment is what? Compliment is a bounded subset, it is countable, right. So, it is compact also. That is the definition of neighborhoods for Alexander of one point compactification. The complements must be compact subsets of the given space, okay. In particular, if you want Alexander's compactification, you should see that this is a horse door space to begin with 0 omega and it is locally compact also, okay. Namely, finally it is a subspace of a compact horse door space. So, it is locally compact also. So, Alexander's one point compactification makes sense. If the extra point here is denoted by infinity, that infinity can send it to this capital omega to get a homeomorphism, that is all. So, whichever way you want to, so it is easy to see that this is the Alexander of one point compactification of the open 0 omega. It is also the stone-checked compactification because of this property that every continuous function 0 omega to r is eventually a constant. Therefore, you can extend it, so you can take 0 omega open, 0 omega, instead of taking the whole of r, take just the real rad functions taking value inside this closed interval, okay. It must be eventually constant, right. Once it is eventually constant, you can extend it continuously to take f of capital omega to that constant. So, every continuous function from 0 omega to 0 1 can be extended to 0 omega closed. That is the characterization of stone-checked compactification. So, that is what we have to remember now, okay. Any space which contains this one and compact, okay, saving this property of extension of continuous functions into any, into 0 1, closed interval 0 1 has the property that it must be a homeon, you know equivalent, equivalent to the stone-checked compactification. That is the meaning of characterization of stone-checked compactification. Therefore, the closed ordinal 0 to omega closed is a stone-checked compactification of 0 to omega also, okay. There is a final thing here. Since 0 omega is a T4 space, okay, T1 and regular that is enough. So, it is T4 actually. Its wallman compactification exists. And being normal, we have seen that it was an exercise to you. The, if you have T4 space, the wallman compactification is house door. So, we have got this one as a house door space. Then we have made a remark that whenever the wallman compactification is a house door space, it is the same as stone-checked compactification, okay. So, that is another remark which we have studied all these things. So, we conclude that 0, omega closed is three different compactifications of 0, omega open. I do not know any other space having this such a property. So, here is some more concluding remark. Though 0 omega is compact house door, it has a subspace Y which is not compactly generated. We wanted to have an example of a space which is, you know, compactly generated. The subspace is not compactly generated. This seems to be a simplest example. I have studied all these. This now is easy example for us. The original space is compact house door. Obviously, it is compactly generated. But the subspace, we have a subspace here which is not compactly generated. So, what is that subspace? Let us see. Take Y to be obtained by 0 omega, you know, as a subspace of this, by deleting all the limit ordinars except the top one, capital omega. Like for the first thing I will omit is little omega. Then I will omit twice omega. Then I will omit three times omega. Then I will omit omega square and all those limit ordinars are omitted. I will deal with the last one I will keep. We first claim that every compact subset of subset K of Y is finite now. Every compact subset of this one which does not contain omega, we have seen that it is countable. Now we are claiming something more stronger. Every compact subset of this subset Y, automatically it is compact in 0 omega but it is subset of Y. Then it is finite. For K will then be compact as a subspace of 0 omega. If K were infinite, first of all it is countable. If it is infinite countable, we can then extract a strictly increasing sequence in K which will converge to 0, converge to some point in 0 omega. And the limit point that will be limit ordinal. The limit being the limit point, it will be inside K but K is inside Y but Y does not contain any limit ordinal if inside 0 omega open, so that is the countable. So K has to be finite.  This will make every compact subset of Y in a closed subset because it is just finite. Yet U is not closed. I am taking U equal to Y minus just omega. Yet U is not closed inside Y obviously because Y is the limit point of this. Remember compactly generated means what?  And that is violated here because take any subset K which is compact inside Y that is called finite obviously. Then U intersection that will be a finite subset and finite subsets are closed inside the host of space. So they are all closed inside K. That means for every K this is true.  But this U contains this capital omega as a limit point. So that is a contradiction. Therefore this space Y is not compactly generated. I think you must have been satisfied with so many properties, wonderful properties, topological properties of this zero omega. So that justifies the efforts in studying these ordinals. So next time we will do one more example using the ordinal but we will construct something more, some more interesting things from the topology point of view. Thank you.


I'm Maurizio Maïs and this is the Service Design Show. Welcome to your two-weekly service design update, where you get to learn what some of the world's best service designers are currently doing. We talk about the current state of the industry, exciting new developments and the challenges up ahead. The Service Design Show is all about helping you to become a better service designer so you can make a bigger impact on the world around us. We bring you a new episode every two weeks on Thursday. So if you don't want to miss anything, be sure to subscribe to the channel. My guest in this episode is Maurizio Manes. Maurizio is currently a professor at SCAT. He's also involved with LiveWork and he's been traveling all around the world doing service design. For the next 30 minutes or so, we'll be talking about topics like who are service designers and what are the three fundamental pillars. We'll talk about why service design makes sense from a business economic cycle perspective. And finally, we'll also talk about service dominant logic and how it has the power to transform the way we see service design. If you want to fast-forward to one of these topics, check out the episode guide down below in the description or just stick around and enjoy the whole episode. So let's jump right in. Welcome to the show Maurizio. Thank you. I'm really glad to be here. Maurizio, you have a long history in service design, you have an interesting background. And I'm really curious, do you actually recall your very first memory that you got in touch with service design? Yes, it was 2008 and I was working on an IT company and at some point the way we were developing software at that time, it wasn't working and I tried to figure out how could I solve this? How could I involve the users in a better way? And actually researching on the web, I found out the service design network and right away I got in contact with Birgit Magher, she was really welcoming. So that's how everything started. Back in 2010, there was also the period that actually the first service design network conference was in Amsterdam. Yes, 2008. Were you there in 2008? No, no, I wasn't, unfortunately, but since then I've been to most of them. So you've handed me three topics that we can discuss in this episode and all three, like we said, we can talk for hours, but we need to keep it short and for the people who haven't seen any of the previous episodes, I have your three topics here on the paper and you also have a few papers containing question starters, right? Yeah. All right. So I'll pick one of your topics and I'll invite you to pick up a question starter and then we'll talk about this topic, right? Yeah. Okay. So let's start with this topic, Maurizio. It's a topic called three pillars. Do you have a question starter and can you hold it up for those along with this one? The three pillars of service design I would choose, who are service designers? Explain. We have here at SCAD a lot of contact from companies wanting to know how they hire service design, how would they assess these candidates, what they would expect from them and actually to be able to respond to their curiosity or need, we end up developing three pillars. So to help them mostly HR people to understand what service designers are and what to expect from them. All right. So one of the pillars is understanding stakeholders. The first one understanding stakeholders. The second one would be developing innovative opportunities and the third one would be understanding institutional transitions. Of course we end up developing a whole way to assess and explain these three pillars but basically is a service designer should be able to understand stakeholders both from a qualitative and quantitative perspective being able to assess these stakeholders in a meaningful way. So they know the now, what's at stake now. The second one would be developing innovative opportunities. Therefore is the whole design, co-creation, creativity, and the third one is to understand how to move from A to B. So A would be the stakeholders now. B would be these innovative opportunities and a service design should be able to know how to get from A to B in the most effective and sensible way. And the last pillar that you described seems like, all three seem like an important one but I can imagine people ask you the question is a third pillar really a part of service design because it sounds like change management and transitioning organizations. Do you believe that this should be a core skill of service designers? Yeah. If we understand design and service design as a way to augment the potential to act of people and stakeholders for instance. Service designers should be able at least to understand the challenges of doing this. It's really at some point it's really easy let's say to create innovative opportunities but making this transition from A to B is not very or it's not as simple as it seems and being able to understand this may set the standard for a good performance service designer. So how do the HR people respond that you've presented these three pillars to? Yeah, actually those three pillars were developed in a partnership with several companies and HR people. So it's kind of a language that they are at least being accustomed to. So yeah it's been useful they really enjoy so we break down those three pillars in how to assess a better definition of what implies each one of them. So it helps them better picture both the candidates and the employee that is supposed to develop service design in an organization. What do you find the most interesting about these three pillars or what are the things that you're thinking about the next steps? I guess the next step is exactly what you just asked is like how to turn these three pillars and the whole explanation about them and how to assess them and turn this into a more easy and accessible way. So deliver to companies a common understanding of how to assess and evaluate even the service designers from this three pillars perspective. Do you see also students using this terminology to explain at home what they do? Yes, yes, yes for sure and then they use a very much a much more simple way to explain this to their parents. It's quite nice to hear them like echoing this to their parents it's interesting. For the people who want to learn more about this is there any documentation literature presentations or should they just go to SCAT and study with you? So far it's an internal document. Our students have access, we have access, our partners in the industry have access to this but it's not yet been released as a public document but it will be so. Let's use the comments in this video to start a discussion about this so we'll see what happens. Yes, yes. Welcome. I like discussions. So Mauricio let's move on to the second topic and it's quite a different one and this second topic it's called the economic business cycle. Yes, yes. And then you have a question started that goes along with this one. Yeah, it's the why service design from an economic business cycle perspective. So the first question Mauricio what do you mean with the economic business cycle? Let's clear that up. Okay, the economic business cycle as defined by Schumpeter in the 20s of the last century is the dynamic that occurs from innovation. So something, a new product or a new way of doing things is created. It creates value profit at some point it becomes routine, it's copied so the laggers come in and then profits dries and it usually ends in a crisis. So this way, this economic way that we've seen all over the world is said or was named by Schumpeter as the economic business cycle. And it has two major, if I can summarize the economic business cycle is one is the what Schumpeter called brain activity phase and I called the design phase and another one is the routine phase. So creating routines that so the whole efficiency process that drives profit and mass produce products. So those two phases we've been seeing from the I don't know the last 60 years a shorten of the routine phase. It's becoming easier and easier to produce mass produce products and figure out routines to make them efficient. Thanks to the engineering and management researchers and practitioners, it became really easy at some point you have companies today that are just focused on the design part, the brain activity and just outsource the production and management to third parties or China, whatever. So but the funny thing is that although the routine part has shortened the design part because it's always the new trying out new things, hasn't. So it seems to I don't know the layman let's say that the design is getting more attention but it's just because the other part is being sort out in a way that it's been easier and easier to routinize things. So what you're saying basically is that design isn't becoming bigger or more important. It's just the other part, the routine part is getting smaller and that's why we all believe as being in the design industry that we are getting more important but it's the other way around. The other industry is getting smaller or efficient in a sense that it's becoming really easy to routinize things. So and also what he proposed and then the service dominant logic reinforces is that service as Bastiat said in 1800 is the begin, the middle and the end of the economic cycle. So this cycle is basically a service life cycle. So being in service design at this moment when the routine part gets shortened is an awesome opportunity. So we started with the question how does service design fit in into the economic business cycle? How would you summarize that? So the economic business economic cycle is basically a service design cycle. You create something, you try to routinize this to implement this and you drive all the value that it can create. At some point it's so used or drawn, the whole values is drawn from it that it dries out and you have to start again. So basically it's what service designers do. The designers do at large but service designers are more focused on the core of the economic business cycle which is service. What is the thing that excites you the most about this topic because you seem to be excited? Yeah, I know it might not seem sexy but from a consistent perspective, if you need to have a strong theoretical background to support what service designers do, it's just perfect. The father of innovation all the way through service domain logic today points clearly towards service design. At some point it's possible to say that even companies that don't know service design, they must be doing service design although not in the better way but there's no way because they are part of the economic cycle. Economic business cycle, yeah and everything is service so they are and nowadays with the prominent design role in the economic business cycle, they cannot help but do service design. They might be doing wrong or badly let's say but they can't do anything else but service design. And then do you have a question around this topic? What is the thing that you're trying to figure out or is this all specifically to this topic right? Yeah, again I guess I have as a professor of service design my main question how it's always how to get this message across. How to transform this in a way in a narrative that augments the potential to act of people like when they hear they say aha oh now I understand what's happening. It's always my main concern is how to get this very dry and theoretical and from my perspective awesome theories to the people and enable them to act better. That's my main question all the time so how to get the message across. That should be a thing that designers and service designers should be good at so it shouldn't take too much time. Mauricio we're moving on because time is flying by and I have the third topic in it's again a cryptic one maybe one of the most cryptic ones we've had on the show so far and it's the SDL XNM number five so service design logic XNM number five And do you have a question starter that goes along with that one? Yeah Mark this is the for me it's a very dear one and I say how can we define service design from the service dominant logic. Service dominant logic XNM number five how can we use that to define service design? Again that's not a very sexy thing but actually it enables us to define service design a very consistent way just let me say something first we all know that service design has many definitions like it's I don't think that one definition would be correct over others but this one especially because it comes from a very good body of knowledge that's been involving throughout the years the service dominant logic and at some point the service dominant logic proposed five axioms so it's kind of not easily question five sentences and the fifth one when I read for the first time was like a very good surprise for me which is the I'll read it is value co-creation is coordinated through actor-generated institutions and institutional arrangements not sexy at all but when you read you say okay value co-creation is a product of a coordination of institution mm-hmm so at some point to create value to co-create value someone something must coordinate institutions so when you think of going to a restaurant there's an institution called restaurant there's an institution called waiter there's an institution called food all the expectations that evolve around those institutions someone has to coordinate them and from a service design perspective that is actually what we do right right yeah you have like okay we have to create a new I don't know insurance service so there's a you have to understand the institution and not the organization the the institution insurance what people expect from it what values can be created from it and then you have to coordinate this and at some point design new arrangements mm-hmm so although again it's not a message that get across easily from academic let's say and mostly a theoretical perspective it defines in a way service design that it's very very helpful to drive further research to drive more friendly discussions and definitions of service side what do you mean with more friendly instead of saying like value corporations coordinated through actor-generated institutions to show arrangement yeah yeah you would say like a service design is a way to create and I like to stress this to create new forms of augmenting people's potential to act so when you create a new service you're concerned or you should be concerned about all the stakeholders not only the service provider or are the customer but all involved stakeholders into augmenting their potential to act which is creating value and and then do you see this domain of service dominant logic as something that we as service designers should know more about and explore more definitely actually it's it's nowadays the first paper that was published on 2004 by professor Varga and Robert Robert Lush it's the most cited academic paper of the century so yeah yeah we have like 16 years only but it's it's nowadays the most cited paper so as I always say you can hate or love service design serves them the logic there's no problem what you can't do by now is not knowing serves them the logic it's it's a very consistent open-ended body of work constantly evolving so it's not it's not gospel it can be reviewed revised and change as it evolves and actually this is the main force behind serves them logic is this not and then the the whole body and the whole researchers involved they they are not professing a faith this is it's it's a work that's evolving it's it's not the truth it's a perspective on service that it's really really interesting so if you would have to point somebody who wants to learn more about this what is the first research that you would say to look up the actual papers that explain the service dominant logic are quite dry and very academic but I've been writing or trying to translate this for a very day-to-day language so I have got a couple of texts about this published but anyway Google it Google service dominant logic and and you find amazing things for service designer I guess at least I tell this to my students you can't not know this and I must be honest so far we've done 20 episodes nobody has mentioned this before so that's quite that's quite interesting yeah and it's it's it's how academia things evolve so it was the first time it was published was in 2004 so in terms of writing about service design there's a very sharp rupture in 2004 papers before 2004 are considered old they are so and and papers that don't cite certain dominant logic either way so even either supporting or criticizing or being against certain logic today they don't I have for instance I have a lot of contacts I receive a lot of contacts from students all over the world to discuss about their research project master and PhD projects and the first thing that I asked them is have you read serves them dot it serves them the logic papers if they didn't I asked them to first read and then we can talk because if you didn't read this there's not much we can discuss about I think you're giving a lot of viewers homework I'm a ratio but that's good that's good so I'm a professor yeah that's good it's it's interesting to know that there's such an amazing pocket body of knowledge that we can tap in tap into Maricio final question for now I'm sure that you have questions that you want to figure out yourself and you already touched upon that you are thinking every day about how to spread this to a broader audience is that the main question you have or is there is something else you'd like to ask the viewers were viewing or listening this episode yeah I guess my main concern now is how to get this message across there are valid consistent and meaningful theories that could support our work there are their mature it's hanging low hanging fruits and I see a lot of discussions going on in the service design world that they are old now by now those theories could jump us ahead like ten years and there's still discussions about those not most but a lot of the issues that we discussed in the service design world were already solved by research we just have to okay this is it okay let's move on let's talk about something else it's like the saying that service design should be human-centered we destroy our planet by being human-centered so we serve design should be holistic not centered at all should be a very 360 degree perspective it's not just the human that's there but should be a lot more holistic and that's why I always teach and try to to to get my my students very comfortable and knowledgeable about ambiguity so how can we jump into a context and be effective by dealing and and and perceiving this context as ambiguous as possible and drive value from this instead of trying to fight ambiguity right exactly the other way around like drive value from this ambiguity so just to serverize this if people want to help you out it's especially about how to spread this message across through the service design community and maybe even in a broader community right yes yes in a lot not even repeating what I said but translating what I said right so okay yeah making accessible okay I got you more this oh this but that should be said or the message should be that way to to be more meaningful to to a broader audience hopefully some people will pick up the the gloves and start working on it chewing on it yeah yeah yeah that's that's one of my main desires by being at your show yeah awesome so Mauricio thanks for your time I have the feeling that we have much more to discuss but we we scratch the surface and again thanks for making the time and sharing your your thoughts yeah Mark was my player and that's completely thank you very much for inviting me and this is the service design show


Hey friends today, I am hanging out at Epcot as you guys mostly know Disney World is closing on the 16th Which is just two days away until the end of March because of the corona virus So I wanted to come out to Epcot see what it was like here See what the crowds are like and kind of enjoy it before I take a nice little vacay from Disney World a Vacay away from Disney World which could be great I'm very excited because I feel like if you spend a couple days away from something. It's More you're gonna appreciate it once you get back. So let's go do this I guess a good start would be to walk on Spaceship Earth through the standby line There's nobody here Really appreciating Spaceship Earth right now because in two months It's gonna be closing for two years and with the Walt Disney World closing for basically the last part of March It just cuts down the time that you can ride it before it's not here anymore Well, it's gonna be here. Just not the same one There's nobody in the carts ahead of me That was a great start to the day, I mean when Spaceship Earth is a five-minute wait Well a walk on you just do it you just you just do it But now I think we're gonna head on over to the land look at that beautiful Spaceship Earth I love it when the nighttime lights and effects come on just when it's still daylight out. It's majestic, isn't it? It's important to note that the pathways are just clear as you're walking through like I said not a lot of people are Coming down and they had to cancel their vacations and most of the people that are still here are the ones that were currently on vacation And of course your annual pass holders And like I said before I think it's a good thing that they're actually shutting down the parts to cut down the spread of The virus itself and like I said, I can use a little break and that way that when I come back I'm just gonna be like I'm so excited to be back And I'll be so happy you know what I mean the longer you go without something the more you learn to appreciate it We probably could ride the seas too since I think that's a walk on But I think we're gonna try to spend our time a little bit more wiser I only have an hour left in the park. It closes that nine o'clock and From the looks of this it looks like there's not many people looking to go soaring today either Wow I Just checked the app soaring is only at a 20 minute wait. Wow Since it's such a beautiful night tonight and the wind is blowing like very nicely I decided to grab a Bud Light and also I want to check on Journey into imagination when figment. I bet you there's not a single person waiting to ride that ride right now Nope, I don't think anyone's waiting. I Love the inside of this queue. It's actually really cool too. It's got good AC. Oh Man, literally there was nobody waiting and I didn't even finish my beer I'm gonna have to hang out up here Hi Gotta finish Said I had like another 10 seconds to finish And I'm pretty sure I've got my own vehicle like my own Vehicle I love it. This is so awesome. Look, there's so nobody coming Laboratories Just me and figment and Nigel dr. Nigel Bye Have a good day Welcome to our special drive-thru open house. I'm Dr. Nigel Channing chairman of the Imagination Institute Here's it's about listening with your imagination Now I've completely lost my train of thought With our eyes can control the eyes of the imagination But let's begin by using the eye chart to test your vision. This is not a good time It's always a good time to use your imagination. So let the good times roll Imagination I do Oh this big minute is just like spaceship Earth if you see nobody waiting you just ride I can't believe I wrote it by myself. So that means that nobody walked into the imagination Pavilion with like that. It must have been 15 minutes. Nobody came in here crazy I guess it would also be a good time if I wanted to meet uh record Ralph Wow This is pretty awesome And since I'm over here another thing I wouldn't mind doing is checking out my pal Mickey Mouse Nice Look at that Hi pal Seriously, is nobody waiting? Oh, well, thank you Hello. Hello. Hello. I like your oh Here we go I like this floor It makes it What? Magic That's crazy Nobody's Beating Mickey Mouse. He's just in there. He is literally just Waiting for people to come in Oh, and then when I come out look at that monorail shot Be a you to fold Actually, I think instead of going and checking out world showcase I think we're gonna head over to test track and see what the wait time for that is Because I feel like everything else is basically a walk on Test tracks a walk on Realistically, I highly doubt the test track is a walk on But it's definitely got to be like a low low wait. I mean, there's no way. I mean there's like hardly anybody here It says 55 minutes, but I think they're trying to pull a fast one on us Oh, yeah, 55 minutes What do you guys think? It is 832 Okay, Disney's definitely trying to pull a fast one on us. It says 55 minutes. It is 833 833 I'm walking in the standby line Seriously I Look at this I am like so shocked wait you guys actually see the result of how long it took Walt Disney world said 55 minutes, but the survey says Nine minutes It is a 41 nine minutes to get through the standby and now we're about to board crazy Of course, I did use my little trick by skipping to build a car, but you know And here we go Did you know if you raise your hands up here it goes with the aerodynamics It's so awesome. I'm so happy that I decided to do test track instead of head into world showcase. Honestly We did like spaceship earth. We did figment and test track and literally under 30 minutes almost Still says 55 minutes. And if I take a peek in there, there's no queue I bet and I also believe if I was feeling frisky, I could ride mission space without waiting as well I don't know. I don't know if I can handle the spins right now But I just want to see if there's actually anybody waiting in line. I doubt it 10 minutes it says That one's five minutes. Let's take a look see in here, huh? A little bit more intense. What do you guys think should I do it? I'm here. I might as well do it Mainly I wanted to see if there's anybody waiting And see Thank you All right. Oh, look at that Gotcha I actually thought about switching and doing the green less intense version, but I think I'm gonna do it.  I don't know what should be should be the navigator Which one Maybe I'll do both right here You'll fire the thrusters for lower organ insertion and for descent to the surface of Mars pilot on my signal I'll need you to trigger the second safe rocket. You'll also deploy the sheets commander You will be responsible for first stage separation and activating manual control for landing Engineer when it's bedtime Let me show you the sleep Okay, I kind of have an idea. I got a team here My magic band is going to be one. So we're going to set that there Good spot And the hat's going to be another one. We'll set the hat there And uh Man, I don't know. I don't have any other personal belongings on me I guess we're just going to be down in engineer I'm kind of glad I got my own one. I mean It seems really cool. It makes things easier to film like when you're on the ride when you're by yourself well I do have my hat and my magic band as my navigator I just realized my magic band glows in the dark Ha ha look at that Get out of here navigator. I have not rode that ride in probably months and that's exactly why it's just so much pressure I could do some advanced training if I wanted to Normally I don't get to play around in the little play places because there's kids in there Well, there is a sign. It says two to 12 Oh, well Oh, there is one more thing I wanted to do before I leave I got to get my pass holder magnet and I bet you I don't have to wait for that either Nope, there is a way for those magnets though And with that, I think I am done. I chose this spot to do my exit because I like the music So I had a lot of fun here tonight. Um now it's going to be probably two weeks before uh, I get to come back I'm still going to make videos though. Um, they're not going to be in the parks. I don't know what I'm going to do, but uh You know just random stuff.  That's what I like to do for fun But I mean I've done home blogs other amusement parks shopping blogs I've done a lot of it so uh everyone be safe and uh We'll see you next time. Bye


So Queen Zhuanzu is at the University of Virginia and will be talking, going back more to the marine domain on like seasonal seagrass effects on flow and sediment dynamics, specifically for a back barrier base. Hi, can anyone hear me? We can hear you all right. Yep, thanks Irina and to the CSDNS community for organizing this summer science series. My name is Qin Guangzhu, I am PhD candidate at the University of Virginia. Today I will be talking about quantifying seasonal seagrass effect on flow and sediment dynamics in a back barrier base using modeling approach. These are part of the results that we recently submitted to JGR Oceans. The causes of this study are Pat Weiberg and Matt Reinerbach. This study is founded by the NSF Long-Tone Ecological Research. Previous studies have shown that seasonal seagrass growth and senescence exert a strong influence on coastal waves. During summer when seagrass density is high, the dense seagrass canopy can result in strong attenuation of flow and waves. Therefore, the sedimentary suspension from the seabed is reduced. In contrast, during winter when seagrass density is low, the flow and weight attenuation caused by seagrass is weak. During this period, the sedimentary suspension from the seabed increased. However, mobility investigations and small-scale field measurement cannot fully resolve the spatial variations of dynamic factors in natural environment. The synergistic effect of flow, weight, vegetation, sediment interaction and metal scale need to be better understood. In this study, we apply a hydrodynamic and sediment transform model and couple seagrass effect on flow, weight, sedimentary suspension in South Bay. On Virginia's Autolantic coast, this is a barrier-based system with a shallow water depth of around one meter and the aerial image shows the distribution of a restored seagrass meadow in South Bay. We validated the model using seasonal field hydrodynamic measurement within the seagrass meadow and at the reference bear side for comparison. Then after validation, we used the model to quantify the seasonal seagrass effect. Today, I will focus on interpreting our model result. We used the depth 3D model to simulate the flow, weight and sedimentary suspension in our study area. We divided our overall model into two model domains, the large domain with a resolution with 200 meters and a small domain with a resolution of 70 meters. We ran the model in two-dimensional depth average mode with a time step of 15 seconds and the model was run for two simulation periods in January and June 2011. In order to incorporate vegetation effect in depth 3D, we implemented a depth-based vegetation model and the inflow simulation and the Suzuki vegetation weight energy dissipation model in weight simulation. This approach considered vegetation as cylindrical structures that could be characterized by vegetation height, stand diameter, shield density and vegetation jack coefficient. We used typical seasonal seagrass characteristics based on previous observations in South Bay for our model input. The first thing we are looking at is the seasonal seagrass effect. We output the model current speed on the left, the weight high in the middle and the suspended sediment concentration on the right at Bayer side and seagrass side for comparison. The results show that during winter when seagrass density is low, the seagrass method has a little effect on flow reduction, weight, significant weight high and SSC. In contrast, during summer when seagrass density is high, the dense seagrass canopy could result in 16% of flow reduction and 20% of weight reduction as well as 85% of SSC reduction in seagrass meadows. With different model scenarios, our couple model is able to separate flow and weight attenuation on sedimentary suspension. We calculated the probability density distribution of bashes just within seagrass meadows with and without seagrass effect. You can see here the y-axis is the density and the x-axis is the shear stress in log scale. When seagrass effect was not included in the model, the bashes just in the seagrass meadow is high with a mean value of 0.5 Pascal. Flow attenuation effect alone could significantly reduce the bashes just from 0.5 to 0.08 Pascal. Including flow and weight attenuation could further reduce the shear stress from 0.8 to 0.5, 0.05 Pascal. Therefore, significant reductions in bashes just during summer seagrass were mainly caused by flow attenuation rather than weight attenuation. Although low density seagrass in winter has a limited effect on flow and weight attenuation, small change in winter seagrass density could result in strong variations in sediment flux into the seagrass meadows. We plot the sediment flux in the y-axis as a function of the seagrass density in the x-axis. When the winter seagrass has a density of around 50, we can see that the sediment flux that the seagrass meadow during this period could maintain a nearly balanced sediment budget. Higher seagrass density could gradually increase the sediment input into the seagrass meadow. However, if the density is lower than 50, the seagrass area will become erosional, leading to dramatic sediment export from the system. These strong variations of the sediment flux that are associated with winter seagrass density could have a significant impact on light availability for seagrass growth and may alter the long-term sediment budget of the seagrass ecosystems. In conclusion, we use a flow-weight vegetation sediment interaction depth 3D model to investigate spatial and seasonal seagrass impact in South Bay, Virginia coast reserve. We found that large reductions in sedimentary suspension in dense meadows were mainly caused by flow rather than weight attenuation. We also found that small change in winter seagrass density could result in strong changes in sediment flux into and out of the meadow. That's it. I will end up here. Thanks.


My name is Sam Badni. I am the author of Malignan Self-Lover, Narcissism Revisited. Abusers mistreat only their nearest, dearest, and closest. Spouse, children, or more rarely, colleagues, friends, and neighbors. To the rest of the world, the abuser appears to be a composed, rational, functioning, and even benign person. Abusers are very adept at casting a veil of secrecy, often with the active connivance and aid of their victims, over their dysfunction and misbehavior. They cloak it. This is why the abuser's offending behavior comes as a shock, even to people who are acquainted with him or her. In the October 2003 issue of the Journal of General Internal Medicine, Dr. Christina Nicolaides of the Oregon Health and Science University of Portland studied 30 women between the ages of 17 and 54. All these women have been survivors of attempted homicide by their intimate partners. Half of these women, 14, confessed to have been, quote, completely surprised, unquote, by the attack. They did not realize how violent their partner can be and the extent of risk that they were continuously exposed to. Yet all these women were the victims of previous episodes of abuse, including of the physical kind. They could easily have predicted that an attempt to end a relationship would result in an attack on body and property. It should not have come as a surprise to them. Yet even the author of the research confesses to having been surprised. She says, if I had talked to some of these women before the attack, the attempted homicide, I would have counseled them about the domestic violence, but I would not have necessarily felt that their lives were in danger. Nicolaides told Reuters, now I am more careful to warn any woman who has experienced intimate partner violence about the risk to her life, especially around the time that their relationship is ending. Secrecy is a major weapon in the abuser's arsenal. Many batterers maintain a double life and keep it a well-guarded secret. Others show one face benign or touristic to an admiring world, and another face hideous, ominous and aggressive at home. All abusers insist on keeping the abuse confidential, safe from prying eyes and ears. Dirty laundry should stay at home. The victims collaborate in this cruel game through cognitive dissonance and traumatic bonding.  Many victims actually feel guilty. They have been convinced by the offender that they are to blame for the misconduct. The famous sentence, Do you see what you made me do? Other victims relabel the abuse and attribute it to the better's character idiosyncrasies. The abuser is explained away as the said outcome of a unique upbringing, childhood abuse or passing events. Abusive incidents are recast, is, priorities, the exceptions, not the rule, abnormalities, few and far between, and not as bad as they appear to be. Some victims even justify the abuse. They say that the abuser's adverts are understandable, justify temper tantrums, childish manifestations, a tolerable price to pay for an otherwise wonderful relationship. When is a woman's life at risk? Nicolaides told Reuters classic risk factors for an attempted homicide by an intimate partner include escalating episodes or severity of violence, threats with or use of weapons, alcohol or drug use, and violence toward children. Yet this list lives out what I call ambient abuse, the stealth, subtle, underground currents of maltreatment that sometimes go unnoticed even by the victims themselves, atmospheric abuse, until of course it is too late.


Thousands of migrants left the town of Tapachula in southern Mexico early on December 24 with the hope of reaching the United States, leading the caravan where migrants carrying across, a Mexican flag and a large banner that read Poverty Exodus back in Honduras, was a Santos was a security guard, but life threats from MS-13 gang members forced him to flee his country towards the United States. Migrants are heading through Mexico to the US to escape violence, economic distress and negative. of climate change, according the UN. The number crossing the perilous Darien Gap straddling Colombia and Central America has topped half a million this year, double last year's record. I'm tired of here in Tapachula, without work, going through hunger, paying home, there's no space. I come running to Las Maras, because I work as a security guard and they came to ask for the money, we didn't have any projectiles, and others didn't have the projectiles to have there, they threatened me to die, and not just for once, they took me to Trepezi, and I was afraid and decided to travel here to Mexico to see if they gave us the opportunity to go to the US.


I'm going to start our series of physical methods of microbial control by talking about heat. So we will cover the autoclave and pasteurization in more detail in several videos. This video I really want to focus on moist heat versus dry heat. Before we do that, a couple of terms you have to know when dealing with using heat and that is the thermal death point and thermal death time. So the thermal death point is what temperature do you have to cook a microbat for ten minutes to kill all of them. The thermal death time is at a given temperature how long does it take to kill all the microbes in a sample. So I personally feel like thermal death time is more a better number because it's a better comparison directly of two organisms. So just a couple of terms you have to know thermal death point versus thermal death time. All right jumping in we do so we have moist versus dry heat before I give you the examples just think about it. Both moist and dry heat can work but in generally speaking I like to say that moist heat is more effective right. I think this makes sense. Turn your oven on to 350-400 degrees let it warm up and put your arm in the oven. Don't touch anything of course and your arms are going to get warm but you take that same arm and put it in boiling water which is boiling at 212 degrees Fahrenheit and you will seriously injure yourself so please don't do that. But you'll see that it takes a much lower temperature to have a much more aggressive response with individual cells. So moist heat is able to surround and work when it comes to cells a lot better. So dry heat can be equivalent though so we'll talk about that in just a moment. So first we have boiling. Boiling is generally not considered good enough for a microbiology lab obviously you can use it at you know at home it's better to boil your water than not boil it if there's some sort of issues or you can boil baby bottles and that kind of thing. But some endospores have been proven to survive in boiling water for 20 hours and the issue is that at sea level water boils at 100 degrees Celsius and one really important example would be Clostridium botulinum takes 116 degrees to kill it.  So certainly using it's better than not using it but it's definitely not seen as effective enough to be used in a laboratory. So then we have dry heat comparing that to moist heat. So dry heat for the reasons I said before takes a lot longer in a higher temperature so dry heat can sterilize things but it takes 170 degrees for two hours. So if you have the time and things are heat stable enough to use a dry heat oven then it's very effective. But as an equivalent treatment it takes 170 degrees for two hours to do what an autoclave can do at 121 degrees for 15 minutes. So as you can see the moist heat is quicker and generally more effective because moist heat is more effective. All right the last one here is incineration so you can certainly burn and incinerate things. The key example we use in the lab would be flaming our wire loops. So it says here is destroyed by burning but the term I like to use is destroyed by oxidation so you're oxidizing, you're peeling electrons off of or burning electrons off of whatever is on that wire loop when you sterilize it. So using incinerators, flaming your wire loops with a Bunsen burner these are examples of using incineration. All right so that is moist versus dry heat and like I said before we'll come right back with a separate video on the autoclave and pasteurization. I hope this helps. Have a wonderful day. Be blessed.


Achieving highly selective separation of olefin paraffin is a significant challenge in the chemical industry. Here the authors present polydopamine-derived carbons with tailorable sub-5a micropause probed by a series of gas molecules that can realize precise-size sieving of olefin from paraffin. This article was authored by Xingzhen Du, Jiawu Huang, Matthew R. Ryder and others.


Everybody forced here ripping 2023 series one baseball jumbo six box case number nine random team style Find the results anytime late in sports cards.com. We're gonna go ahead and get started Thank you very much everybody for joining triple threads breaking next just has four teams to go and they're all on sale Orioles guardians Rockies Marlins is who they were Again all on sale Marlins and Rockies a fairly a fairly more so I'm gonna before I say anything else I'm gonna stand hasn't my hands after this guy's this is physical outside of this case is a little dirty Once I get this thing open. I'm gonna stand size my hands Doste it's not too bad the bottom is a little worse That's true Bring your own hand sanitizer apparently I was gonna say I have some my backpack nice dude my soccer jerseys. You start this guy delivered I'll have to wear those tomorrow. That's actually perfect timing The first soccer jab ever owns It's a this is not even for us one not a player Derek Jeter couldn't decide on the players I just get the one gardenist. Yeah, it's kind of strange right Taylor Maybe they does and it's not listed there currently. I Guess that could be the case usually that's not though Usually cardboard is spot-on kind of just going right off of what tops has But here we go guys again, but all base split even All base split even so we'll kind of skim through that a bit make sure to miss anything you'll get every single insert and Anything you know numbered base inserts foils silver pack cars anything. That's not a base card You'll get for your team. Oh and that it's all split even everyone's gonna have a pretty good chance at At these and rookie's rookie gold foil Cody Clemens for the Tigers David Wright welcome to the club. Not there. There's the 88 Josh young So Taylor was talking about the day the 88 just hanging out in the back somewhere Look at the number base What do you mean Nick? Short print should all be like right in the middle right here like a rainbow foil Jimmy Hurg it Over the angels there Vladimir Gutierrez reds orange foil board to 299 and A auto Marinus Luis Castillo first Marinus Luis Castillo auto I pulled Here we go aces Shane Bieber. There's the 88 again. It's Libertor Yeah, what was the one-of-one in cosmic Taylor? I for it It wasn't it wasn't black was it no Jacob the grom gold foil with in all aces Verlander that is Non-numbered. That's just backwards. All right. All right. That's that sometimes Bobby wit Yeah, the 88 that time Just hanging out about here again, do we have it? Yeah, we don't care was the wit was the way what am I doing? I guess I just assume I wasn't paying attention I just seem to be like that for the whole case Kate Cavalli rookie rainbow foil nationals Boba Shet home run challenge Yadir Molina in 2023 gold Yo Cardinals rally green greatest hits And 88 where you hiding there you go max Meyer Marlins is that a thing illegal pretty cool This oh No, why is all these backwards? What is this box? Test the breaker box gold foil nester Cortez Alex Bragman 88 Yeah, I was looking over there. I'm like, what is he talking about like? Lane send me a text Pre-sock shield auto was pulled or one of them. I think there might be a couple. I saw what you sent me There's a couple variations, right? Right green foil board pool host Cardinals to 499 gold foil.  I want to pull a game a game used NHL shields so bad It would be so disgusting. I don't think I've ever pulled one of them I mean I pulled out of like leaf and stuff, which is so cool, but out of the cup is different I got Sid the kid or something like that gold Darvish Padres 2023 rainbow foil pooholes to 299 A beret you city connect white socks. I couldn't even imagine to grom aces Yellow jade-eight. Oh, yeah, Nick. I'll be in the middle though. Yeah, we know we know where those will be Gold the 2023 Austin meadows for the Tigers rainbow foil Tyler Wells That's a tough one though because you it's a very long it's like a chant Yeah, we looked of all rainbow foil game use Jersey Boba shut to 99 blue jays Yes, trim ski 88 Ricky Henderson Yeah, so like where the golds are in this that's exactly where they'll be if you're opening this if you're opening this at home That's where your variations will be you can feel free to skip through all that other base if you'd like to but So if you pull any other number cards like a father's day or advanced stats, yeah, that's where they'll be right here Red foil board Gary Sanchez twins is 199 rainbow foil Trevor story red socks got Rizzo Shippa such a good tune. Don't say that illegal silver packs Hey, they went to the playoffs Taylor. There's a bay or rookie silver red socks Cori Lee Astros Bubba Thompson Dodgers and a Miguel Vargas Dodgers or Rangers was was Vargas. Sorry. Excuse me. No, no no Rangers was Thompson.  What's up?  Especially as a baseball player Brutal Brutal. Oh, you're wearing the the winter classic one. I had the 2016 winter classic jersey And I was like, yeah, that's just a bad memory that Jacob to Grom box topper Although I do like that jersey a lot with the brown with brown bear on the front of it Yeah, and then I was thinking about wearing the pasta jersey and I was like, you know what? I'm gonna wear a player that has actually won the cup if I'm opening the cup So I'm burgey makes logical sense makes sense. Yeah pasta stole I Think pasta has second-best odds behind McDavid for the heart. I mean it's is McDavid's I think McDavid's odds are like You could bet a hundred dollars for him to win the heart trophy and you would win ten dollars if he wins it That's how good the odds are. It's terrible. What's the hurt like the MVP for Gold of 2023 McKenzie Guardians Michael Massey rookie rainbow foil for their Royals Nice. Thanks Salty. Yeah, the next one is triple threads number 69 Which is sold out now, so there's just eight teams left on triple thread number 70 now. Nice That's infected what a one lead Vaughn habs whooped the Bruins at Gillette that year That was dude. I was that was like the one winner classic. I was like most excited I was the first one. I actually watch as a Bruins fan All right, wisdom rainbow foil. We got rookie auto Marcus Wilson mariners the 88 design Oh, oh, yeah, that lead. I forgot. Okay. I forgot. We're talking about baseball still Sorry guys. I missed it. I missed it. This is the first break of the day So I'm still warming up on everything including my mind working gold Matthew Libertor.  I can't just like slide it I think we're supposed to start those around like 650 cup first ones. Yeah 650 here is a Schedule for you Eastern 650 always break in by schedule should adjust to your local time zone I just know you're like VPN or something Rainbow full Royals and a Cunha homerun challenge to 499 Royals green foil team card Pigeon Martinez are in auto. I mean tattooed it depends on what it is, but mine is It's pretty much healed now. I just We're long sleeves. I Get it finished next Saturday Gold Tony Gonson at 2023 for the Dodgers Rainbow Robbie Ray rainbow full Robbie Ray. Yes, it's it's fine now I'll still put like lotion on it and stuff Then they get all dry Especially because it's on my elbow a Carasco gold foil now. I'm getting more added to it. What's done is done illegal So there's just more that's gonna go around it. I'm excited. I'm excited. I'll probably wear a short sleeve shirt tomorrow actually so That's probably see a little bit at least nice a Brian Bale green foil board rookie for the Red Sox numbered 407 or 499 Mets rainbow foil CJ Abrams on these city connect You know Nats red baby greatest hits Do it illegal get this get this crap Donnie's gonna love this though My brother who is you know where our fandoms align my brother and I He is moving to Philadelphia because he's a flight attendant. He's a station there He has like Eagles gear now. Oh wow Corey Lee Mother's Day pink nice 48 of 50 Astros That's a big one right there Donnie's gonna put this in a mag ever so gently Congrats Astros. I pulled one of those on the first time I opened the cup. I forget who it was though at this point It's nasty. Yeah, this is a burgey Jersey Dude, he has some he has some he has some like Philly stuff now I don't think he'll ever do like flyers, but he'll probably kind of adopt The Phillies and Eagles, which I guess is okay.  It's 2023 Cody Dodgers on the gold and Then I was making fun of him when the Eagles lost I would make fun of you guys I'm making fun of him because he just picked up the team. Yeah, it's different guys. I'm not making fun of you guys Jersey card your Stremsky Giants like well, I mean it must hurt so bad to see your Eagles finally get to another championship and lose Christopher Morel I'd make fun of him He literally lives like I think he just wants to go out and drink and hang out with all the Eagles fans, which I get pretty soon. He'll be climbing greased up lamp posts and stuff Rookie Rainbow Foil Max Castillo Royals Luis Castillo aces Yeah, he did jinx them. I'm gonna tell him that too He's like I'll never adopt the flyers. I'm like, that's fair Taryn Vaudra probably not the Sixers either because he's in love basketball and every now my job any Pascas continue, you know For the Royals there Bo Jackson Hovashar like Hovashar Luke Nick Prato And Christopher Morel for the Cubs. I told him I was like he should go to he should go to a Bruins Flyers game in Philly And just be super obnoxious. I'm like, yeah, if you ever want to get beat up, you could do that.  But the flyers almost eliminated at this point gold Katel Marte Diamondbacks of 2023 a rainbow foil Matt Chapman for the blue jays Don't beat me up God rookie gold to 2023 Ethan small Brewers rainbow foil rookie Michael Massey Royals and a mark McQuire a's auto here one of one Oh, wow Thought that I thought that color on the car was funny one of one Big Mac Go into the a's Great hit there. Congrats. A's of the Big Mac auto That's a beauty Yep, we like one-to-ones Oakland Oakland Oakland John s Big Mac and the gangster crew over here Stephen Strasburg. Oh, you guys know what that is Nolan Corbin that was from my high school.  So you don't know Yep, I Sorry, I was so on that one-to-one call Oh, I wanted to make sure because I have never pulled one like that out of a series one That's awesome though orange foil to 299 nationals team card Hassan Kim rainbow foil tips needle a JJ Bleday rainbow foil rookie Marlins Home run challenge Bobby with junior Royals. It could be a pretty cool one. I Want to get one of those right one day The home run challenge is correct. Shea Langeley has anyone ever done it got something like really big got a langle here's Ricky for the a's I've never gotten one right period Which is upsetting because I had a devour I had a couple devours ones last year that was wrong on And I had a judge one and somehow I got that wrong if I had if I had a judge one last year I got that wrong. I don't think I'll ever get it right.  I'm just terrible at it What is this? Oh nice a randy or was arena blackboarder raise 88 to 299 gold foil Alex Cobb Brett Beatty 88 Yeah, there's some really cool old-school flyers like dual autos or triple autos in the cup like some really cool ones. Oh No, it's just a card. You can like you can win some you can get some cool stuff It's not the home run challenge or something else you can do I'm thinking of something else 2023 set Lugo for the Mets Zach Davies Either way, I'm clearly bad at it. That's true legal. Yeah, you want to get the guys like you're done. That's who you want to get You're done judge Mashers, Yachty Molina rainbow foil City Connect Brandon Crawford Giants Yeah, Derby dual Derby dual could be cool too. I like to know those are a lot rarer They get the one I say well, that's cool. Yeah, I rookie want to be cool to get a rookie Harmony challenge would be pretty neat. I guess I mean nothing crazy, but Hey, you need card. You got a Blue that's pretty cool cardinals on the one-two punch to 600 copies RBI leaders here Yankees guardians and Astros judge Ramirez and Tucker Astros Jonathan B Yankees mark K and guardians John s world of die one two left three four middle five six right As they to this going left Yankees on this one left Jersey card Craig DGO Astros game use memorabilia. How do you watch when greatest hits? Sorry, Donnie It's more for That's for sure. That's what I would think but I'm just You remember doing series one where you're top story shipping, right? Yeah. Oh Goljera Rookie all-star cup team right there rookie rookie cup card 2023 There we go manners. That's pretty nice Rainbow full of Robbie Ray Machado Soto Libre Tor Kyle Stowers Bregman in a purple Marlins Jazz Chesham, Jr. That's pretty cool It's a 75. That's when Marlins and her jazz Chesham Aaron judge and a Michael Harris We go halfway through. Yeah, so the home run challenge is like you have to predict. I believe the specific date I'm not mistaken That's what I did last year. I've only done it a couple times. I'd like to get one right this year though That'd be kind of cool There it is Donnie your favorite game. Oh, yeah Christian Yelich on the box stopper I get the plate I'll do Here we go. Oh Yeah, the 425 feet one Yeah Yeah, we'll see I'm gonna do I'm gonna do some this year. I'm gonna rip a jumbo box. I like this stuff So he's serious one. I feel like serious one. You just don't like you don't like sports cards Green Gavin sheets to 499 for the white socks rainbow foil Buddy Kennedy rookie for the D-backs rookie Otto Xavion curry Guardians like that like that first name Xavion curry as it X avion the names Hey, Aron black a this Blake Manny Margo green foil number to 399 to 499 Goal foil schwindel. Nice. I know Tani home run challenge Rainbow foil Paul Seawalds to 2023 Alex Cobb Giants gold aces show. Hey, Ohtani Right back there. Definitely no Tani pack Got a clean kershaw gold foil Must grow in Darvish Wander 88. I appreciate it smack Yeah, not often. I'm breaking in a hockey jersey. I'm not gonna sleeves off left and right over here like the The penny sleeves that I had next to me knocking off the table over the place, so Special occasion we're in a hockey's your hockey jersey gold to 2023 Nathaniel low rangers rainbow foil Seth Brown aces here in Nola Greg Maddox 88 Red foil board Austin Meadows going to the Tigers of 199 Son Tundare rainbow foil city connects as a zoo key there. Yo cubbies. That's a cool one Gunnar Henderson greatest hits rookie Orioles insert That's funny Taylor. Yeah, I mean if the guy you know that you're If the guy that you're doing the homerun challenge for is like you're done Alvarez or Vladdie junior or like Aaron judge Why would you not double down? Future stars gold wander Franco nice Tampa Gold foil. I should say gold foil. I just want to get one for like Rafi Devers. I'd be kind of cool Probably go by one. That's not as cool though, right? It's not as cool The whole point is that you guess that you got it, right? I'm a foil Denny Jimenez Game use Jersey Yachty Molina Cardinals Aces Verlander Buxton Oh, there's the ambidextrous pitcher. Oh Four scoreless innings in first legis are he's an ambidextrous pitcher. That's in college That's the ultimate weapon right there. That's the ultimate dream. Yeah, that's a Not switch hitter, but a switch pitcher to 2023 Brandon Crawford Giants on the gold gold foil bellinger I hope not Kirk. I hope not a liberator To 88 That is the 288 parallel nice one Cardinals Gold foil Jonah Hyme Ali Rushman a silver pack hit there Ramirez James Altman for the Dodgers and a Brett Beatty Mets Bro Cal Mitchell Pirates rookie Yvonne Herrera Cardinals and a green Vlad Guerrero expose going to the Nationals to 99 It might not be the first but he's getting notoriety right now because he's you know doing pretty well in the In college so far so Got a cunea. It's got Ian hap rainbow foil rookie auto Nick Pratt. Oh 46 of 50 Yeah, Kansas City really good case here Let the Royals in this one That's awesome nice Nick Pratt's oh gold stars out of 50 Thomas C grass congrats That's only to make a same to you Jammer check spots after this box here. Give us into 299 aces Cardinals gold foil Matt Duffy Jimmy Rollins Moreno rookie 88 Blue Jays Fenway Park rainbow foil for the Red Sox spend we park in the back jazz Chisholm homerun challenge Marlins to 2023 gold George Kirby future stars Mariners aces Rodan Torkelson 88 to 75 Verlander aces Gold foil Urias Verlander Astros by the way Yeah, probably a met in series 2 I would say All right, got me clan a hand rainbow foil aces of Randy Johnson for us for the Diamondbacks So 2023 Jonah I'm gold for the Rangers Blue Jays Nam queue white socks Yuzu and Astros That's Houston here John of the beat and roll a die 1 2 left 3 4 middle 5 6 right to left side Astros Clayton Kershaw to 75 Patchwork of the past card. Yeah batting average isn't seen as important. So yeah, definitely not gonna see orange batting average Green foil board Lennon Sosa rookie white socks number to 499 Gold foil class a wadley-ruchman Rainbow foil nice card Orioles rainbow foil Adley-Ruchman an Emanuel class a to 76 independence day Good pack here guardians Joey Votto game used Jersey for the Reds with a pinch stripe on it Max Scherzer Mets aces Rodriguez 88 When's the last time somebody hit 400 in season the 40s was a Ted Williams was at the last time Yeah, 60 homeruns is for sure gonna happen sooner than 400 season Gold foil trend aggression pitching just too good nowadays Obviously, that's not the only thing that's different now, but yeah, that's the that's one of the main things orange Dominic Leone Giants to 299 a Rainbow or gold foil thing a rainbow foil on the Leone as well got Killian silver Cubs Buxton Derek Hall Phillies and Lendor Leblanc rookie are in Nato McGuire and J. Ghibler day. That's also like split Aaron judge Yeah, the no shift might help but even still like that's a record that haven't forever like dude the 40s What was it 42? No, he was in the war. Was it 41 last time you hit 400 45 what sport do I follow the most? That's why it's hard for me to answer It's probably baseball if on her era rookie stars auto Cardinals numbered 23 of 199 It's probably baseball, but it's really close between like The other seven sports that I have to pay attention to it's probably baseball then Football probably because you break so much of it, then it's like basketball then hockey I would think I Have to do it kind of based on like What I break the most because it makes the most sense. Wow Kyle Stowers green foil board to 499 for the Orioles thought that was a different player for a second Josh Taylor gold foil 600 OB PCs this is not gonna happen Taylor Pitching's too good man. It's too good Nice Mother's Day pink rookie Brian Servin for the Rockies 37 to 50 two Mother's Days in the case Homerun challenge Buxton twins rainbow foil Jensen when you'll cruise Perez Gold to 2023 Seth Lugo Mets a gold foil rookie Tristan Casas Red Sox. That's a nice one Triple threads is sold out baseball sold out for the night. I would not be surprised if the cup also is Hockey is sold out for the night. There are two more cases of the cup tomorrow There's two more Sunday as well, and I believe Monday, so we got a few more cases up right now guys If you miss out on the cup for release day, but we do have more Prism collegiate draft at 16 and 22 to go rookies and stars at 3 and 13. I'm act wwe down to 20 and 19 So update all those spots when we get finished with this one here Very nice. Thank you guys for filling up baseball and hockey got basketball and football and wwe as well now I don't think we're gonna do PYT Brennan. All we have is random styles up so far rainbow foil Average leaders of the National League here. Oh, let's see Mets Thomas C Dodgers Benjamin S and Cardinals Red be rolling to die one two three four five six one two left three four middle five six right It's a to go left side again. I could have just rolled for the whole case here.  I'm so excited with this stuff though Bob Gibson again a Stephen Kwan rainbow foil on the cup card rookie cup card there city connect Dodgers Walker Bueller aces Alcantra Lou Bob Yeah, it's gonna be fun Mac. I was watching a bit of it last night. I like WDI Mac a lot Dude, I want to see I want to pull a game you shield so bad to 2023 Camila DeVal Giants gold Wayne Wright gold foil Bradley Green 88. Yeah, I know vato was an OBP bees for a while, but even he wasn't up near like 500 He was in his prime vato as an OBP monster a gold to 2023 Gregory Soto Tigers Nolan Jones Ricky rainbow foil Guardians Game use Jersey Blue Jays of Vladimir Guerrero Jr. Number to 199 Blue Jays solid break here couple of Tony inserts I feel you Joshua. I feel you I feel you Joshua. I feel you Yeah, if I haven't been like Doing card related stuff or like work related stuff. I've been playing Hogwarts for sure Gold foil and she been attendee Yankees. We did yet.  Yeah, I'm Ravenclaw Vlad Guerrero rookie Cal Mitchell pirates on the silver pack Yvonne Herrera cardinals and out the enrichment Orioles The last card of the break there. Yeah, I've got like a Less than 30 but more than 25 somewhere in there is where I'm at. I Don't know the exact time damn those games. They tell you how many hours you've been on it Yeah, I'm only one of a couple hours today six hours late there Exactly what happens Oh, yeah, I'll get a couple games of cotton Like oh, I have to go to work in four hours. Well, I better Guess I better get some sleep get some food Okay, relics here Vladdy Walker Bueller Joey Vado Clayton Kershaw Yadir Molina say a Suzuki Craig BGO Brandy Crawford Mike Yastremsky CJ Abrams Boba Shets Jose Abreu autos we had Yvonne Herrera Zavion Curry Marcus Wilson Luis Castillo Nick Prato gold stars rookie out of the 50 We go Orioles Mother's Day pink Cori Lee rookie for the Astros 48 to 50 and Platinum 88 mark McGuire on card auto one of one Yeah, congrats a is one of one Big Mac good case there. Thank you very much everybody Got triple threads coming up in a bit