Hello, welcome to NPTEL NOC, an introductory course on point set topology part 2. Today we will have the last section on the dimension theory, module 44, local separation to global separation. Recall that our study of dimension theory began actually in the previous chapter with a discussion of separation properties which we have named S0 to S3. Keeping in mind that we are going to restrict the class of topological spaces for those which are separable and matrizable and in particular T1, we have pointed out earlier that S1, S2 and S3 are some stronger forms of house dwarfness, regularity, normality respectively. We may call this itself the first step in the passage from local to global. Moreover, the passage from SI to S3, S1 to S3, there is one S2 in between, itself can be viewed as a passage from atom to mass, atomic to aggregate or whatever you want to say, mini school to largeness and so on, a kind of local to global, that is also you can say is the passage from local to global. Recall that one of the first few results that we proved in the previous chapter is that Lindelof plus S2 implies S3. So that made S2 our central object of study. Next in this chapter we adopted S2 to represent zero dimensionality and then inductively obtained its higher dimensional versions to define higher dimension manifolds. Then came the theorem 9.13 and 9.15 which you may term as another step toward globalization. Let me just show you these steps. This was the theorem that if we have subspace of matrizable space, separable matrizable space, then x as dimension less than or equal to n if and only if given any closed subset C of x and a point outside, there is a closed subset D of x such that dimension of D is less than or equal to n minus 1 and x minus D is equal to a separated b and b are both open and closed subset disjoint with p inside a and c inside b. So we have pointed out that in the case of n equal to 0 dimension of D is minus 1 is D is empty set. This is precisely the condition S2. So similarly 9.15 you can also similar things I just wanted to show you one of the here if x is a matrizable space and x prime is a subspace of that dimension of x prime is less than or equal to n if and only if for every point p inside x now I am putting conditions on points of x all together you see I am getting condition like that. Condition is dimension of x prime is n equal to n. What is the relation between points inside x and point the subspace being dimension less than equal to n. So that step is again another step or globalization this is what we meant. So let us go back to what we are doing today. So as the inventor had termed it the success of the concept of this theory of dimension why I am going to this theory because there are other theories also hinges upon successfully strengthening the passage from local to global. This is the topic of this last section wherein we shall be able to reach our goal the final goal of proving that the topological dimension of the pyridian space R n is actually equal to n. So that I call it as a success of the theory. Here is the next step in the passage from local to global.  Given any two disjoint closed subsets C 1 and C 2 there exist a closed subset B of x separating C 1 and C 2 such that A intersection B is empty. So C 1 and C 2 are disjoint closed subsets by normality you can separate them by open set that is a different aspect. Of course that will be the starting point in the proof of this. What we are going to do is there is separation by closed subset B it does not intersect A at all. So this you can call it as really a crunching of of course we have to improve on this also of the separation properties being globalized. Globalization problem here. So let us start the proof of this.  So this step is just merely the metric space property here normal property of a metric space. So that is all we are using here.  These will be disjoint subsets of A and A is of dimension 0. So apply the S 2 property there. We have A equal to C 1 prime separation C 2 prime here U i bar intersection A is inside C i i prime. So that is the definition for that is the property for the zero dimension set. Now you enlarge the C 1 prime C 2 prime along with C i. Put F i equal to C i union C i i i equal to 1 and 2. Look at these C 1's are subsets of A C 1 C i prime C 1 C 2's are subsets of the largest space x. C i's are closed C i's are closed inside A but A is not closed inside x. So there is some problem here otherwise F i 2 would have been easily you know you can take it as closed subset. You have to take it as a closed subset. Now here this dimension 0 that is the point here. So we shall show that F 1 and F 2 are mutually separated sets which is stronger than saying just disjoint closed subsets. So I mean they are not closed subsets but F 1 bar separates it does not intersect F 2 and F 2 bar does not intersect F 1. So mutually separated subsets which is slightly weaker hypothesis than having disjoint closed subset. Disjoint closed subsets are easy to separate. So this is more. So for here we will need more than normality namely complete normality. Wait a minute. So let us have this one now. How to prove that this one namely F 1 bar intersection F 2 is empty. Indeed once you prove this one the other one is symmetrical all these things are conditions are symmetrical in the other one so that will also prove exactly same way. So once you have that use complete normality of the metric space it follows that we get open subset W in x such that F 1 is inside W and closure of W does not intersect F 2. The same thing as having disjoint closed subsets and so on. This way it is easier to state. F 1 contained inside W closure of W does not intersect F 2. We can then take B as the boundary of W automatically boundary of W is closed. So this B is a closed subset which will separate F 1 and F 2 because F 1 will contain inside W does not intersect B. F 2 does not intersect B because it does not intersect W bar at all.  If you take W union compliment of W bar that will be precisely equal to the whole x minus boundary of W union will be x minus B. So moreover if you look that is what I have shown B intersection A is empty because why this I have not shown. What I have shown F 1 and F 2 are disjoint one by one. Moreover you want to ensure that B intersection A is empty but B intersection A B is boundary of W, boundary of W intersection F 1 union boundary of W intersection F 2. Because this entire A is union of C 1 prime and C 2 prime and F 1 and F 2 contain C 1 prime and C 2 prime. So A is contained in F 1 and F 2. Therefore this intersection is intersection with F 1 union intersection with F 2. Both of them are empty. Just now we show them. So this B will serve the purpose. So we have yet to find out that one. So here is the schematic picture of what is happening. Started with C 1 and C 2 which are disjoint close subsets.  Intersected U 1 with A this is my say A is of dimension 0 that is why I have shown like dot dot dot dot dot here. Intersect that with A 1.  So they are disjoint inside A and they are disjoint close subsets. So you can separate them by C 1 prime and C 2 prime. So C 1 prime and C 2 prime may go out of U 1. They may be out of this one. So here it might come even out of U 2 also here to here and so on. So of course this dot dot dot for dot dot dot ellipses indicate that after all an open subset of subset of A is nothing but some open subset inside X intersection with A. So that is what I have shown here. Then I put F 1 equal to C 1 union this C 1 prime up to here C 2 in C 2 prime up to here. I have to show that they are mutually separated. In the picture it is obvious. You cannot use the picture to prove a theorem. You can take the help but finally everything should be purely logical. And finally what we want is this green thing W such that it is closure does not interfact this F 2 C 1 prime union C 2 prime.  So let us prove first of all that these two namely C 1 union C 1 prime C 2 in C 2 prime they are disjoint and mutually separated is what I have shown of injection. So let us do that. So it remains to prove 31. 31 is that this F 1 bar intersection F 2 is empty. The other one is similar.  First of all F 1 bar F 1 itself is C 1 union C 1 prime. Therefore F 1 bar bar denotes now everything denotes happening inside X. Closures are inside the whole space X. F 1 bar is C 1 bar union C C 1 prime bar because it is just a union of finite union. But C 1 is already closed so it is C 1 union C 1 prime bar. Therefore it is enough to check that C 1 intersection F 2 this is C 1 intersection F 2 and C 1 prime intersection F 2 they are empty. Then F 1 bar intersection F 2 will be empty. So this is the first step I have to show these two things now. Just to show one of them I have to show these two things. So let us see.  U 1 is an open subset containing C 1.  And hence C 1 intersection C 2 prime is contained inside C 1 prime intersection C 2 prime. Just now C 1 intersection A is already in C 1 prime. C 1 intersection C 2 prime is inside subset of A so I can take intersection with A itself. So that is contained in C 1 prime intersection C 2 prime and that is empty to begin with. So instead of C 1 it is in F 2 I have looked at C 1 intersection C 2 F 2 has two parts. What is C 2 and C 2 prime? So C 1 intersection F 2 will be C 1 intersection C 2 which is empty C 1 intersection C 2 prime. So this intersection is now C 1 intersection C 2 prime. Now what is happening to this one? So C 1 intersection C 2 prime is again A intersection C 1 intersection C 2 prime everything is happening in C 2 prime.  So I am more or less repeating this one here. Next to show that C 1 prime intersection F 2 is empty. So how do you see? There are two parts F 1 bar has two parts right one we showed. The second part is this one now this one I have to show. For each x belonging to F 2 we shall produce a neighborhood of x which does not meet C 1 prime then it will follows that that point is not in the closure of C 1 prime because the whole neighborhood does not intersect that. If inside C 2 there are two parts to F 2 one is C 2 and one is C 2 prime. If C 2 you can just take V equal to whole of this U 2 for all the points inside C 2 take just V because V which is equal to U 2 here is contains the whole of C 2 okay and then V U 2 is not intersection C 1 prime is empty right.  So method is the same but argument are completely different. That part takes one part if X is in C 2 prime then what do I do but C 2 prime is an open subset of A it is actually close open subset right because separation okay C 2 prime is also open inside A. So we get an open set V this time it is a different V I have to choose okay so this V is inside X open such that V intersection A is C 2 prime open subset of A. Then what happens V intersection C 1 prime is V intersection A intersection C 1 prime that is C 2 prime intersection C 1 prime okay that is contained in C 1 prime. So that is also okay so separately we have shown that C 1 prime intersection F 2 as well as C 1 intersection F 2 are empty it just means that F 1 bar intersection F 2 is empty. Similarly F 1 intersection F 2 bar is also empty and that completes the proof. Now we can state a more pliable statement and easy to remember statement start with any separable matrix space take a subset of dimension less than or equal to n where n itself is finite. Of course I assume bigger than equal to 0 because if A is empty there is no statement those things we have seen already. Given any two disjoint closed subsets C 1 and C 2 inside X there exists a closed subset B of X separating C 1 and C 2 such that the subset A intersection B okay has dimension less than or equal to n minus 1 okay. So from zero dimension we have come to arbitrary dimension here now okay so how do you do that?  Now suppose n is 0 this n is between 0 and infinity right. So n is 0 there are two cases to be handled if A is empty in which case we can take B equal to U and U and U 2 complement that is a closed subset and X minus B is just disjoint union of U and U 2 over okay that is what when this is 0 otherwise dimension of A is 0 it cannot minus 1 okay in which case the first earlier lemma which we give just now that we give you the required result. So we have the inductive hypothesis here now suppose n is bigger than 0 using a previous corollary we can write A as union of two subsets D and E where dimension of D is less than or equal to n minus 1 and dimension of E is less than or equal to 0 this is one of the some theorem that we have derived last time right. Now you use the above lemma to give a closed subset B which separates even and C2 such that B intersection E is empty I do not know what is happening to D we will come to that later but E part is empty that is the starting point with n equal to 0 but then this implies you take A intersection B that will be now contained inside D but D is of dimension n minus 1 less than or equal to n minus 1 so intersection of A intersection B is also of dimension less than or equal to n minus 1 okay. So this after hard work of lemma this comes quite easily by of course I have to use this crucial thing here namely anything which is of dimension n can be written as union now two subsets one is dimension minus 1 another one is 0. Now in the above theorem take A equal to well this was not a theorem it is a preparation does not matter take A equal to X then what do I get let X be of dimension less than or equal to n then any two disjoint closed subsets can be separated by a closed subset of dimension less than or equal to n minus 1 okay. So no question of intersecting with A because A is the whole space X now now we want to improve upon that one let X be of dimension less than or equal to n and C1, C2, C3 and similarly this C1 prime, C2 prime, C3 prime etc pair wise disjoint closed subset C i intersection C i prime is empty that is the meaning of that how many are there n plus 1 this dimension less than or equal to n where n plus 1 pairs of disjoint closed subset then there exists a closed subset B i there exists closed subsets B i how many n plus 1 of them such that each B i separate C i and C i prime and intersection of B i i rank 1 to n plus 1 is empty alright. So how do you get this one this is also easy from previous theorem applied to C1 and C1 prime you get a closed set B1 which separates C1 and C1 prime such that dimension of B1 is less than or equal to n minus 1. Now use the proposition or the theorem we get a closed subset of XA subset B2 separating C2 and C2 prime such that when you intersect it with B1 is a dimension less than or equal to n minus 2 because you already n minus 2 dimension. Now you keep on doing this you know repeat this step get a B3 such that it separates C3 and C3 prime the dimension of B1 intersection B2 intersection B3 to n minus 3 how far you can go till you have minus 1 and that is empty. So you have to have dimension less than or equal to n here and there must be n plus 1 of them then only you will actually have alright. Now you are very close to the end here okay rectangle or rectangular box minus 1 to plus 1 interval raised to n contained inside Rn okay suppose you have C i plus and C i minus denote its faces defined by equation X i equal to plus minus 1. If you just n equal to 1 this is nothing but C i minus 1 and C i minus C i plus is plus 1 that is all if n equal to 2 there will be 4 faces pair of 2 pairs of opposite faces right. So that is the way you have to take these faces defined by the equation coordinate X i equal to plus minus 1 okay. Now for 1 less than or less than n there are n pairs here suppose B i is a close subset of Jn bar which separates the opposite faces C i plus and C i minus it just means that when you throw away B i from Jn bar you get 2 open subset each of them containing C i plus or C i minus 1 of them containing the i okay that is separation. Suppose you have got these B i's like this then you want to show that intersection of B i is non-empty. The crucial thing here is what there are only n of them the previous theorem says there are n plus 1 of such things then it is empty. So together they are going to imply a big theorem for us. However the proof of this is now based on something different that we did last time namely Brauer 6 point theorem comes here okay. Let us see how pay attention to the method of proof because that can be used in several other places okay. Let D denote the Euclidean distance function in Rn for each X in U i plus I have overly told you what are U i plus here.  So this is a one of the phases. So go all the way to C i minus from U i plus do not change the ith coordinate that is all. The P i and X have same ith coordinate it is the root of the perpendicular from X to this plane. Then this is now elementary observation distance between X and B i. B i is a subset which separates the two things right. It is less than or equal to distance between X and P i and this distance is actually equal distance between X and C i minus why because P i is the root of the perpendicular and what is this distance is just one plus the ith coordinate of X okay. Similarly so what we have proved by this one is distance between X and B i is one plus X i.  Let me justify this one with a small picture here n equal to 2 okay. So this is minus 1 to plus 1 minus 1 to plus 1 the product square okay this is square and this is C 1 plus X 1 coordinate this is C 1 minus this is C 2 minus and this is C 2 plus this is my B 2 separate C 1 minus and C 2 minus take a point in U 2 plus this X take its projection on to this axis on to this plane so here is just an axis what is it its X 1 coordinate will not change what is its Y coordinate Y coordinate of this point or X 2 coordinate of this point is nothing but this X 2 coordinate here plus this one the distance sorry the distance is this distance is 1 this distance is X 2 right. So distance between X and this one is same thing as distance between X and this part which is bigger than distance between X and B 2 and that is precisely what we have to do X B i less than so X B i is a D X C i minus it is M 1 plus X now I define a function here for 1 less than or less than equal to n define F i from J n bar to R as follows if X is in U i plus just take it as distance between X and B i that is a continuous function right we know that what is this this is the minimum of distance between X and little B i where little B i runs over B i if X is in U i minus you put a minus sign there X is in B i put at 0 look at this one if X is in U i bar this this U i plus it is this one if it tends to a point you know if you take limit tends to a point inside the boundary then this distance becomes 0 right similarly this one also this will become 0 so by these are continuous this F i will be continuous because of that okay now you combine these two in equation here inequality what you conclude is that minus one is always less than X i minus F i of X is always less than good one there are two different cases you may have to work you have to work out is very easy okay you have to use this and this one according because there are different definitions are different the function F i different to define that is why okay so X i minus F i is between minus one plus one therefore if I define okay F x equal to F 1 F 2 F n and G x equal to X minus F x then what happens this is a function taking a values inside R n but G x will be taking function inside J n bar it will be always between minus and plus one all the coordinates so obviously both of them are continuous okay therefore you have got a function from the closed rectangle J n bar to J n bar you can apply Brauer 6 point theorem so we get a point X such that G x equal to X what of that mean F x equal to 0 what of that mean what of that mean F i of X equal to 0 for all i what of that mean X must be inside each of this D i which means X is in the intersection okay so that is the statement here okay this theorem is proved now as I told you now you can combine this with the previous theorem you get a wonderful result now type thing namely dimension of J n bar has to be equal to n okay let me go through this one it is not so clear if not dimension is always less than equal to n because of what we know that part we have already proved okay dimension of R n less than equal to n and this is a subspace so that part we have already proved so dimension must be less than equal to n minus one as soon as n minus one by our theorem 9.38 whatever we have proved today it means that for each i 1 less than equal to n there exist n closed subset B i which separates C i plus and C i minus such that the intersection is empty but Brouwer's point theorem applied now just now the previous theorem says this is unempty so that is the contradiction to this theorem right 9.39 therefore dimension has to be equal to n R n being a larger subspace of containing J n or J n bar already so this dimension is also n alright so we have proved that not only this one can you can take now any open subset any take any open subset which is homeomorphic to some D n, G n bar and so on contains something etcetera right any open subset will contain some copy of this one therefore all open subsets will be of dimension n inside R n all non-empty open subset so here is a remark topological dimension whatever you have defined okay is invariant for the class of separable matrizable spaces we have not defined it for arbitrary spaces that is one point you have to remember thus we may also conclude that you know the Brouwer's invariance of domain which is a weaker form of this one weaker form of B i d so I am going to give that namely R n and R m if n is not equal to m cannot be homeomorphic because we have just shown that dimension of R n is n dimension of R m but dimension is a homeomorphism invariant okay this is weaker form of Brouwer's invariance domain of course the real Brouwer's invariance of domain is the following which is which is a stronger form of this one namely take any two non-empty open subsets non-empty subsets of R n suppose they are homeomorphic then if one of them is open the other one is also open and that is why the name invariance of domain the word domain was used more often than just open subsets in the older days so being open is same thing as being a domain and that is an invariance so that is why it is called invariance of domain okay unfortunately we are not able to touch this one we have come very close but there is still a big gap here okay so the proof of this will take us much deeper into dimension theory which is beyond the scope of this course original proof due to Brouwer uses another topological dimension theory namely Lebesgue covering dimension in the modern times it is fashionable to prove this using homology theory so there are many proofs of this word it is great theorem but a proof using only simplisical approximation which is actually part of it is there in Horowitz-Jewelman implicitly you may see my book so here is a easy exercise for you deduce 9.42 from 9.43 namely this I said this stronger form I said why so you assume this and prove this one okay we have proved it using the whole thing using the all our you know first two different chapters that we have studied so carefully but assuming this you prove this one that if you are okay alright so next time we will start a new topic thank you


Hello, many of you know that here in Marburg we have been inverting our classroom for several years now. We have been using digital teaching and learning scenarios for some time, and we are no longer sage on stage but guide on the side, helping our students to deepen their knowledge in our in-class scenarios. So for quite some time now, we have been inverting our classroom, where content delivery and content acquisition are carried out online in a self-guided fashion, and the in-class phase is primarily used for practicing and deepening. This model goes back to the influential publication by Maureen Legh, Glenn Platt and Michael Treggelier who suggested the term inverted classroom. In a second publication they also used the term classroom flip, and later terms such as flipped classroom were added by other authors. Why do we have such a mixture of terms? Shouldn't we have one term since the idea of inverting the two phases of teaching and learning is essentially the same? Well, I want to argue in the following that there are in fact two different settings for inverting the classroom. One is the high school setting and one is the university setting. Let us look at these in more detail. There are several parameters that distinguish the teaching scenarios that are used at high school from those used at most universities. Take for example the average age of the target groups. Whereas most high school kids are younger than 18, the standard student at university has come of age and is in most cases older than 18. Or take the class size. High school classes are generally smaller than those at universities, where you can have classes with several hundreds of students. As a consequence, university teachers often do not even know their students' names. Furthermore, class time is different. And normally much shorter at high school, where classes of 45 minutes, for example, are the rule at least here in Germany, at university it's 60 minutes or even more, giving us more time to practice and to deepen particular problems. Another difference concerns the rhythm. At university, teaching and learning is normally organised according to a weekly rhythm. At school the learning units come in daily or at least several times per week. And what about the use of a learning management system? Well, all universities have one and most colleagues normally use it. Whereas at school it's still the exception rather than the rule. And then we have the possibility of using tutors at university. A common method, but this is not known at school. And last but not least, there are differences between schools and universities as far as the technical equipment that can be used in the classroom is concerned. In other words, there are different parameters affecting teaching and learning scenarios at school or at university. Thus there are also different conditions for the realisation of an inverted classroom. Hence it can well be justified to use different terms for what I think are slightly different models. Let's look at these different models or variants. Inverting the classroom at the level of high school has commonly become known as flipped classroom. That is a scenario where simple videos are used in phase one. And the homework, well, homework in inverted commas is done in phase two in class. At the university level we refer to the model as the inverted classroom where now the material available in phase one is more complex. Classes are much larger and tutors may also be added to the in-class phase. A special variant that we use here in Marburg adds a mastery component to the model that is a system of electronic tests by means of which our students can show mastery prior to each in-class meeting. Using the inverted classroom mastery model we bridge the gap between phase one and two in order to find out how well prepared our students join us in the in-class phase. So we have two main variants, the flipped classroom for high schools and the inverted classroom for universities. Whereas it would be problematic for the flipped classroom to integrate permanent testing prior to each in-class phase, this is no problem for the inverted classroom. Its weekly rhythm at universities makes it possible to add a component where students are offered online tests to show mastery prior to each in-class meeting. And this is then the inverted classroom mastery model. Well, here is a little bit more information about these models, especially about the literature associated with them. The idea of calling the high school model flipped classroom originally goes back to Wesley Baker and a publication around the year 2000, but it was taken up by the two North American high school teachers, Aaron Sams and Jonathan Bergman, who described the development of their model in their 2012 book. And they also published an article about it in our first inverted classroom conference volume. The label inverted classroom originally goes back to Maureen Lake and her colleagues. We already talked about this. Well, we have used this model since 2006 and have taken up this label in our 2012 publication about e-learning at German universities. Well, and the addition of the mastery component, where on the basis of formative electronic assessment, students can demonstrate mastery, but sometimes fail to do so. This was first mentioned in our second inverted classroom conference volume. It was awarded Excellency in Higher Education in 2013. And in fact, I took it up in my new book, which came out a few days ago. So here is an overview of the publication starting with Lake and her colleagues and a number of publications where I could make some contributions and the high school variant by Aaron Sams and Jonathan Bergman. They all suggest that the inversion of the traditional phases of education leads to significant learning effects. So to sum up, essentially we all refer to the same thing. A classroom where the traditional phases of teaching and learning have been swapped to use a neutral term here. At high school, they simply have been flipped with almost no additional components between the phases leading to the flipped classroom. At university level, the phases have not just been flipped in a one-to-one fashion, but they have been inverted, either fully or partially leading to several variants and stages in between, from simple tests to complex mastery components. So we believe that the two terms flipped and inverted have both their justification, flipped classroom for high schools and inverted classroom for universities. Thanks for listening and see you again.


Welcome back to Kids Fun Science. My name is Ken. Today's experiment is the hot air balloon experiment. As always, adult supervision is required. What you need for this experiment is a tissue paper or a plastic bag, a glue stick, scissors, and a hair dryer. Welcome back to Kids Fun Science. My name is Grace and Ken. And today we are doing the hot air balloon experiment. How heat rises. So Grace is setting it up on the base here. The metal wire gets hooked on these clamps right here. And then you don't need this, but this just helps hold it down. And then I'm going to take a hair dryer with hot and we're going to fill it up and then we're going to launch it off from there. It's going up. Grace is there. And we're going to give it a countdown. 5, 4, 3, 2, 1, go. The ceiling, there it goes. It's about 20 feet. Now as the cool air starts to come out of the balloon, it will slowly descend. Just like a real hot air balloon. Grace is coming around. It's coming up. Success. Nice. Let's do it one more time. So here's a homemade tissue hot air balloon that I made and I show you at the end of the video how to make that. I'm also using a heat gun, which is very hot. You could also use a hair dryer. But I had to come outside because the inside air was too hot with the hot air. There's my cat making a cameo and so therefore the tissue paper weighed more than the plastic bag and it wouldn't rise inside the house. So with the cool air outside and the heat gun adding a little extra heat, I probably kind of got away with the hair dryer. We'll see if we can actually launch this and there it goes. I don't know if I'm going to get this back. It's coming out. Launched over the house. Alright, here's the instructions on how to make your own hot air balloon out of tissue paper. So here's the dimensions here. It's in inches and centimeters. Go ahead and freeze that. I bought two packs of 12 sheets of tissue paper, 20 by 20 or 58.8 centimeters.  Then you're going to take a glue stick or you can use tape and then you just kind of put it right at the end and then you repeat this five more times.  As you can see, I'm just doing about a third of the paper at a time and you kind of just pull it up to arc all the way there and it should even out right there. Then you're going to repeat that five more times and here's your end result. This is your hot air balloon and the hole in the bottom is a little too big. So I made a seven inch circle and then I kind of just take it on with some painter's tape so I can take it off later. Here is my first attempt with the tissue hot air balloon as I ended up going outside because it was too hot inside with the heater on and the hot air and it just did not want to flow because of the weight of the tissue paper. I hope you enjoyed this video. If you did, please share and like. Thanks for watching.