Like I told you last time we started looking at a phenomena which gives us an opportunity to see how what we know or what we have learnt as theoretical apparatus to understand language helps us understand our languages. I gave you some examples from Hindi and then I want you to evaluate this phenomena with reference to languages that you speak to. We looked at then we talked about negation, negation words we discussed what negation is whether negation is a lexical item or a functional category and then we looked at how is it represented in the clause structure of a sentence that is what is its exact location in the functional domain of the conceptual representation of a sentence. Now we want to extend that and see little bit more about negation and how it intersects with other lexical categories. Now that we know negation is part of functional domain and then how it works in the language with other lexical categories. There is a particular thing in natural language which is called negative polarity item. Like we talked about negation that negation is a universal phenomena in natural language. Negation may be a matter of discussion in linguistic theory in the following way. Whether negation is a word or a suffix or a prefix whether negation conceptually belongs to functional domain or lexical domain these are the questions that one can discuss while talking about linguistic theory. However, the presence of negation in natural language is a universal phenomena. Similarly the presence of negative polarity item in natural language is a universal phenomena. I am going to show you some examples and then you will be able to evaluate whether or not we find such elements in other languages or not that is other languages that you speak. I can substantiate that it is available not only in the languages that you speak but in all the languages of the world. So as you can see the word indicates negative polarity item. So it is going to be a word which is sensitive to negation that is all negative polarity item means it is a word in a language it is sensitive to a negative element. That is all is the point what exactly it means we are we are going to look at this. So we you have seen three negation markers in Hindi how many negation markers are there in Telugu you have not thought about that more than three there are definitely three of them for sure can you tell me. So give an example of kothu penne ikada petoddu which means loudly speak loudly penne ikada petoddu which means do not put the pen here alright the second one adi phone kaadu that is not a phone. So these are the two negation markers like mat and nahi mat and nahi right what is the third one which means. Usually used in a meaning like it is not there but it is also sometimes used as a replacement for it is more like na used as a replacement for kaadu sometimes. So the first one that you said is that only restricted to imperative sentences like do not speak sit down things like that right. So without looking at more data one can also say at least for South Asian languages we have a distinction between two negative markers at least where every language is going to have a specific negation for imperative sentences which can be used in negation which can be used to negate an imperative sentence which is a tense less sentence right and then the other negation marker can be used in both types of sentences right that is true in Telugu as well. Now see the structure of language around these phenomena right there may be hundreds of differences between Hindi and Telugu right so much so that the two languages are not intelligible but when it comes to classification of a negation word it works exactly the way Hindi works see that right okay so likewise I want you to think about other languages if you happen to speak them okay now we saw this the phenomena that just we discussed with reference to Telugu how it works in Hindi then we have questioned the presence of negation in the structure of a sentence and we saw with examples last time that negation happens to occur in the functional domain and specifically below tense and above aspect phrase okay so that is so which now also pay attention to the following that it is not just a matter of a specific location negation it is also a matter of breaking the adjacency requirement between negate between tense and aspect okay and what is the implication of this breaking adjacency between tense and aspect the implication for that is these are categorically two different phenomena okay two different things and there is no strict adjacency requirement between tense and aspect so when someone says things like mai pizza khatam so ta part of the verb khata and whom which is a tense marker they do not have to be together they happen to be together true but they do not have to be together they they that adjacency could be broken okay so a look at negation and the evidence that we saw last time tells us way more than what we saw but we are not getting into too many details of that all right then let us see examples of negative polarity items okay I want you to I want to come to this one in a moment I want you to look at these two sentences the sentence in Hindi reads as Rajiv Hargij nahi bolega okay Rajiv Hargij nahi bolega it means Rajiv will not speak at all under any circumstances okay now what I am what I am trying to show you here is the word Hargij is called a negative polarity item okay this word cannot occur in in this type of a sentence if there is no nahi if we do not use nahi then we cannot use Hargij see the point therefore the second sentence the next sentence is marked with a star and ungrammatical Rajiv Hargij bolega is not a good sentence in Hindi okay and the only reason why it is not good is because there is no negative marker you put a negative marker the sentence becomes good see the point the fact that the presence of Hargij is only allowed is only warranted in the presence of negation is the reason why we call this thing a negative polarity item okay this can only be present in a sentence when there is a negative marker this is what these two sentences show you you get this point this happens in all the languages and since I was going to draw your attention to South Asian languages does it happen in Telugu it does not have to be Hargij the word could be anything else does it happen in Tamil Malayalam at this time I don't want to because I want you to look at or I want you to be able to look at your language with examples from our own languages so what you are saying is this doesn't help so far right is there a familiar word sir is there a familiar word as in Hargij is not very familiar that's fine you don't have to be familiar with Hargij look at the sentence Rajiv will not speak at all how do you say that in Telugu Rajiv asal marteladu okay now the negation remove the negation word from here is the sentence good not good there is no negation word as such there is just a morphological whatever that is remove that is the sentence good so definitely some word in this sentence is a negative polarity item whatever yeah yeah it it may not be a full word true it may be a small element somewhere so like negation the negative polarity item also does not have to be a word but there is an element of negative polarity item in this sentence because of which when you remove negation the sentence in Telugu that you just said is ungrammatical and it becomes clearer when you write the sentence properly if I knew Telugu and if we had more time I would ask one of you to do this on the board and can show you which one is negation and which one is negative polarity item understand this right in that sense the word at all also has some negative polarity meaning in it because in English also we cannot say Rajiv will speak at all can we say the sentence Rajiv will speak at all what does not mean much it means properly it means it receives an appropriate interpretation in the presence of a negative word not when we say Rajiv will not speak at all that word also is a negative polarity item to some extent I will give you more examples of negative polarity items in English too but right now take a look at this let's look at these two sentences what does this sentence mean I did not eat anything right we cannot say now please pay attention to these sentences very carefully on the on the screen on the next sentence I haven't put the star mark okay unlike the previous one do you see here on the second one there is a star mark what the star mark means is this sentence is not possible at all no star means that this sentence is possible okay but does not have the meaning that we want meaning it manekuch bikhaya it doesn't mean I did not eat anything this sentence is good with some other meaning if you raise the intonation remember we have talked about questions and intonation manekuch bikhaya right we two of us or some of us went together to some party food was not good on our way back we are we are discussing something manekuch bikhaya means I did not eat anything right it definitely means something else many it mean it could mean I eat just something randomly true it's ambiguous but ambiguity is not what we want to focus on you're right it's ambiguous but what it does not mean is I did not eat anything the word kuch bhi which you see in the bold okay does not have the meaning of anything this word receives the meaning of anything only when you have a negative item in this sentence read the first sentence carefully manekuch bhi nahi kaya right that sentence doesn't mean I ate something randomly seriously so the the negation word and this kuch bhi has some sort of dependency on one another for giving a particular kind of meaning is this is this making sense to you right yeah emphatic mark it means that we give the intonation on that no no it means the bhi that you see bhi is an emphatic marker in hindi say we it's it can be used with many other words like ek bhi when we say ek bhi right tum bhi means you too okay so bhi is an emphatic marker and when we look at the details of this composition then people say and people claim that what gives negative polarity interpretation to kuch bhi is actually bhi because kuch as a word is a quantifier kuch means something sum in the presence of negation the same word sum becomes nothing kuch means something kuch nahi nothing but with bhi it becomes anything manekuch bhi nahi kaya means I did not eat anything get get the point so in both in these four examples I am trying to show you that there are some elements like hargis and kuch bhi which are categorically dependent on negation for their interpretations for their appropriate interpretations makes make sense if we remove the negation word ek kuch bhi kata hai it becomes both make sense and both are opposite that's right so so therefore I am not putting a star mark there so the second sentence that you are giving we cannot say that this sentence is wrong this sentence is not possible the sentence is definitely possible but not possible with the intended interpretation in other words we can say without negation it has a different meaning with negation it has a different kind of meaning so if we focus on the meaning with negation that meaning is negative polarity meaning with negation it seems like there is a dependency between these two words get the get the point now to elaborate this point a little bit more I didn't mean to bring in at this point at this stage I was going to discuss this little later but since you are bringing in let me talk about this do you see I have mentioned on the top type 1 and type 2 do you see that what that means is this distinction that you see right that in one case we can put a star that is in the case of type 1 right and in the case of type 2 2 we are not able to put a star what does this mean it means that there could be two types of negative polarity item okay in one type of negative polarity items the negative polarity item is strictly dependent on negation without negation no interpretation the sentence is out in the second type which is the one that that on the screen right now it's possible a sentence is possible without negation but not the right interpretation okay so this we can say this is not very strict kind of situation whereas the first type is very strict situation you remove negation and everything is out the meaning is out okay so this is the these are the two first two points that we are making here first point there are some words some elements in all the languages which are sensitive to the presence or absence of negative elements in a sentence okay and the second point is among the elements that are sensitive to negative marker in a sentence some are strictly dependent on negation and some are not get the point so these are this is the classification introduction and classification of negative polarity item okay now I want to take you to this place do you see that the list in the one in one and two you see that the first list in one there is a set of three three things there right all three of them are elements which are strictly dependent on negation okay now I want an example from you I have already given you an example of Hargis right have you heard this phrase in English in Hindi egg foodie curry no can you give me a sentence with that can you use this egg foodie curry it means a broken penny literally it means foodie you know food now means break foodie curry is the word for a colloquial word for penny right which which together means a broken penny literally it means a broken egg a foodie broken curry has anyone seen a curry it is it is like a shell it is it is a sea animal small sea insect which is have you seen a sea shell right so it is a smallest part of that and when it is broken it is of no use so the same metaphor is used for no money not even a broken penny egg foodie curry now can you with with this much of explanation can you use this in a sentence you see you hear this sentence do you understand this sentence now what does he want to say right right he says let let me repeat this sentence again it could mean if my father did not give me a broken penny it also means my father didn't give me any money it means my father didn't give me anything right whichever meaning you like you keep that meaning but the point is we cannot say we cannot say that if you want to say that then it means the literal broken penny it doesn't mean anything or it doesn't mean any money it means a real example if this is something like a broken penny broken cell then he did not give me that broken cell get the get the point is that true sit down it I am asking him because he is he he speaks in the and it should make sense to I think I think everybody right making sense you have already seen the example of Hargis now you have the third example which means which is Moog Holna right Moog Holna literally means this one this one everyone should know Moog Holna means opening the mouth it literally means Moog Holna but it means in the in the sense in the context of language it is generally assumed that when we open mouth we say something right it means speaking about something can you use this thing in a sentence again do you see this thing or we can say which means he couldn't say anything he could not say anything again we can never say he opened his mouth I opened his mouth in front of the teacher right we cannot say these things with the meaning that I said something okay if you say I opened my mouth that means literally opening your mouth it doesn't have the meaning of saying anything or saying something the point is the first thing that you see must strictly be used with negation in Hindi to give the meaning that we want from them yeah we can we can add but it still cannot it still doesn't change much it still retains its polarity meaning negation no no ask ask Hindi speakers it cannot be get the point first of all these expressions will not be in other languages there is going to be an equivalent expression if that equivalent expression can be used without negation then they don't belong to this category then they belong to the second category I'm coming to the second category in a moment so there is a there is a strict classification one type which is going to be used only with negation and the other type which can be used without negation I have already shown you examples right so see any one of the examples in the second thing which is koi bhi kisi bhi kuch bhi ek bhi right these things will not can be used without negation and must be used with negation to give you appropriate negative polarity reading in other words again when you use the second the anyone from the list of the second item on the screen then the sentence is not ungrammatical but it may not give you negative polarity reading for the first list we just cannot use them without negation get this one more point it's not just three of them I'm just giving you three examples there could be hundreds of examples in any language so which answers your question that we do not have to have just these three or equivalent of these three what's important for us to see and understand that there is a set of these two available in all the languages if one does not and it's up to you to figure out can it be used without negation if it can be used then it belongs to the second category if it cannot be used it belongs to first category that's an important point okay that's just an observation okay now is this much clear to everybody since your friend Sandeep asked a question about questions right so let me very briefly talk about this thing and if you have a question raise your hands and I'll answer this from the board I don't have it listed here these things but look at the second second set when these or for that matter first set also when these things are allowed with negation right technically that is called licensing okay which means a negative polarity item is licensed meaning it is allowed the word licensing is just a fancy word doesn't mean much it simply means allowed and what does allow mean allowed means the sentence is good grammatical that's all if this if the sentence is not grammatical we mean not allowed not licensed so so please don't get lost in terminology it's it's very simple so we are saying a negation licenses a negative polarity item okay the first set for the first set we strictly need a negative item to license them for the second set we may not strictly need them right but what have you seen so far anything from the second set right when it is used without negation it doesn't give us negative polarity reading without negation that may be okay sentence but it doesn't give us negative polarity reading but there is one more distinction between the two categories anything from the second set can also give you negative polarity reading with question words okay the moment you question koi bhi right take koi bhi i can say koi bhi nahi aaya what does it mean nobody came it has a negative meaning nobody right i don't use negation i say koi aaya what does it mean without question koi aaya someone came you see this without negation it gives us the meaning of someone with negation it gives us the meaning of no one hold on hold on hold on slow slow slow what the distinction that we are trying to make is when we get no one reading then it's a negative polarity item okay it can get someone meaning it can get the reading of someone without negation which is all right we don't debate that but it doesn't give the meaning of reading of no one without negation okay however under certain circumstances it can get the reading of no one when you question it how do we question this thing the sentence is let me let me give you the sentence koi aaya what does it mean that's all right reading intonation it's fine but this is a question everybody agrees when i say koi aaya is this question for everybody what's the meaning did anyone come so no one or anyone these are the negative polarity reading so you can get a negative polarity reading with question also get this thing understand now someone can ask you this question when you study these things and someone can ask you just on the basis of one point you cannot make two categories right what is your basis can you give one more example one more theoretical point on the basis of which your categorization is okay do you see my question this is the second point that you can use which is you can get a negative polarity reading from the second set with question but you still don't get as negative polarity reading with question from the first set so you question the first set there is no negative polarity reading right so for example take the 40 kori example is that negative polarity reading no reading of negative polarity okay the point is the first set not only requires first the first set okay let me let me let me present to you in the following way the second set can get negative polarity reading under two circumstances one with negative word and the other with question first set can only get negative polarity items with negatives nothing else so that clearly justifies that there are two clear patterns in natural languages for negative polarity items okay and again some language may not give you two sets that may be debatable if a language does not have a clear distinction between these two sets right so in that case the categorization of negative polarity item may be a parametric affair okay but the presence of negative polarity item in natural language is a universal phenomena is a principled matter and now I can say is a the presence of negative polarity item in language is a matter of principle the presence of a negative item in a language is a matter of principle where how are parametric issues get it before we stop I want to make one more one more point with this I have already discussed this part with you where the where do negative item occur in the closet structure right there are different people like Palak Janatini Mahajan Divedi many people have studied these things and people have had different positions about the presence of negative item in the closet structure what sounds very convincing is a combination of all that not any in particular so this is the this is the point that I want to make to you so how do we we can say that a negative polarity item needs a negation right needs a negation that is true but where does it need a negation right does all negation marker license negative polarity items okay that is not that does not happen and again we do not have enough time to give you give you every single example to show you that but this will make sense to you because you have gone through that the negation marker which licenses negative polarity item must be in the C commanding position if the if the negative marker C commands the negative polarity item then it is licensing the negative if it is unable to C command then it will not license do you remember last time I gave you two types of negations what were the two types of negation anybody one was when the negation negates the entire sentence constituent negation and sentential negation sentential negation is when negation negates the whole sentence and constituent negation is when the negation negates only a particular constituent from a constituent negative position or negative polarity item cannot be licensed which is which means in that case negative polarity item is not in the C commanding domain of the negative okay therefore the NPI must be in the C commanding domain of negative polarity item as long as the negative C commands the negative polarity item the sentence is okay the interpretation of that item in an appropriate way is allowed get the point so I do not have the list I just wanted to tell you this thing and this is the issue how does it license the answer is it must be in the C commanding domain to understand the C commanding domain let us look at the structure once again negation okay the C commanding condition for licensing of negative polarity item also becomes a motivating factor for negation being located in the functional domain only when negation is high enough in this structure it can C command the elements downward see my point so if the negation is located in the functional domain it can C command all the positions that is subject position and object position both so if the negation is negation occurs in the subject position still it is C commanding remember this thing how did C command because in that case we are assuming that the subject and NP originates at the specifier position of VP okay the subject NP originates at the specifier position of VP under that situation it is in the C commanding configuration okay under that situation it is in the C commanding configuration can you can you see this from this structure if if if we have a VP V bar V if you have the subject here and object here right negation from here can C command this position okay can C command this position it can also C command this position all right therefore negation being in the functional domain helps licensing configurations right but there is a problem here we know that this subject from the negation from the specifier of VP what happens to this it moves up for understanding this thing now do you see any implication of that the subject becomes higher right so if someone asks you at this point which one takes place first licensing of negative polarity item or movement of subject licensing of negative polarity item okay licensing of negative polarity item therefore becomes an evidence for ordering of movement operations that only after licensing the element or only after being licensed by negation if the need happens to be that way subjects can move see movement of subject is for every sentence but licensing is not related to every sentence licensing is related only in the sentence where you have a negative polarity item and a negative negative word when you have a negative word and a negative polarity item then the licensing takes place first and then the movement of subject takes place okay therefore C commanding domain is the licensing condition for this and the whole licensing condition itself is another evidence for allowing movement under certain circumstances get this point when negative polarity item when the element does not need to move then there is no problem for example imagine you have a negative polarity item here instead of subject we have it an object position then this one doesn't need to move right away right however if it is in the subject position like ek bhi larka ek bhi ek bhi larka nahi aaya right ek bhi larka is in the subject position so in a way on the surfaces level you have you have ek bhi here ek bhi larka here on the surface position nahi here and aaya here then if we allow subject to be here under speck of tp then we know that speck of tp is not in the C commanding domain of negation then we need to say something else but if we allow subjects originating here then it's in the C commanding domain and we have a convincing logic to say that before it moves for anything the licensing has already taken place so to conclude this thing can I say that the licensing takes place at a conceptual level not at the surface level this this point making sense the licensing that we are talking about for which we said C commanding domain is a requirement this requirement of negative polarity item being in the C commanding domain of negation is a condition at a conceptual level alright thus I wanted you to see negation negative polarity items licensing condition with examples from our languages however it works the same way in all the languages I have shown you with negation also how and how an element is a principle thing how the same thing could be a parametric issue how licensing is a principle and also at what level of abstraction does this licensing work I do not think we can we can bring a microscope to see that level of abstraction the ones that I have been trying to show you through the examples of languages making sense alright so we we stop here with negation and negative polarity items which are examples of South Asian languages from our languages think about Telugu, Tamil, Malayalam, Punjabi and other languages and for that matter English to conclude with your question things like anybody anything or negative polarity items in English which are of the second category the first category will be something like at all a word like we can say in English he didn't say word right a word has a negative polarity meaning because it doesn't mean literally one word when someone says he didn't say word what does he mean right didn't say word you can say the same thing without negation he said a word but it doesn't mean not saying anything so such elements are negative strict type of negative polarity items in English anybody anything anywhere are the second category of negative polarity items licensing works exactly the same way it works in Hindi so you saw the examples of Telugu and Hindi that two way negation negation classification works in Telugu and Hindi right licensing works the same way now English is a different type of language SVO still the licensing works the same way you know why licensing why the same condition of C commanding domain is applicable on English too because the conceptual level that we are talking about at that level all the languages are same which is i language the the conceptual level of i language and e language underlying a structure and surface structure are the differences so we see the differences between Hindi and Telugu Hindi and English Telugu and English at surface level at the abstract level the universal principle work exactly the same way for all languages this is licensing of negative polarity item is one such universal principle which does which no language violates get it so with this i have stopped


So, the Castro Peninsula is situated on the western coast of Lemnos Island in the northern region sea. It borders the present main port of the capital, Mirina, from the north. The hilly headland consists of lava domes, most rocks being of day-side to tracky day-side composition of the lower Mayocene, and its summit is at about 115 or 16 meters above present sea level. In addition to later occupation on Castro, mainly medieval and post-medieval periods, in particular a Byzantine, Ottoman, Venetian and Genoese castle comprising important buildings and a triple wall enclosure, a larger number of earlier rock-cut features were identified. In spite of distraction due to posterior activities involving subsequent modifications, several sectors of the headland still bear vestiges of early carved surroundings. The Castro volcanic milieu was carved into ensembles of rock-cut structures in extended areas on various altitudes. Based on work in progress, the focus of this paper is to present a general overview and categorization of the Castro rock-cut features. The research, with a permit by the Greek Archaeological Service and the Ministry of Culture, includes fieldwork with surface surveying and recording, photogrammetry and topography, as well as the study of mobile finds resulting from subsurface investigation related to selected rock-cut features. Surface and subsurface data point till now towards dating the features, at least from the Bronze Age up to Hellenistic period, without excluding a larger time span. Diachronic uses may be attested in certain areas, although they are not necessarily constant, nor generalized. Radiocranology, anthropological and metallurgical analysis, as well as the study of mobile finds and organic residues, are currently in progress. The roughness, irregularity and color variability of the rocks, covered with lichens, as well as erosion and man-made destruction, complicate the task for the identification of the structures which are often invisible, even when above the surface. Besides what seem to be significant monoliths, there are various types of rock-cut structures in isolation or in combinations, suggesting utilitarian objectives, but also unknown functions. These include circular cavities, well systems and conduits, the latter sometimes along flights of steps, or connected to angles of rooms or platforms, or to other carved structures related to water-liquid management. There are two rooms or platforms, which consist mostly of two or three vertical walls that form angle and contain part of the rock floor, steps, niches and benches. Stepped ramps join them to higher or lower similar structures, on your right. Intercommunicating ensembles of smaller rooms are also attested. They are often combined, besides to niches, also to other features carved in the space which is delimited by them, such as benches, steps, vertical or horizontal post-holes and large cavities. Numerous ramps, stepped ramps and carved steps of various dimensions and numbers. Longer passageways, often suggesting intercommunication itineraries, linking carved complexes and sectors. Partly natural and partly artificial boulders or outcrops with carved small steps and niches. And a number of petroglyphs on vertical surface of boulders or niches, on walls or floors of rooms, platforms or closed cavities. The extent and complexity of the features implies a multifunctional side. Indeed combinations of rock-cut features may indicate symbolic behaviors related at the same time to concrete uses of space. For example, in an area we review over the sea, a carved main road, oriented west-east, is crossed by two descending flights of carved steps or stepped ramps, the latter directed towards the coast. At crossroads, a number of female implements and fragments of figurines, some of them wearing a polo's headgear and or enthroned, were found, possibly close to an open-air altar. Although traces of delimited spaces might constitute possible dwellings, which would be consistent with water collection structures, passageways and platforms, as well as with a harbor area. Still several rock-cut features are not adapted for habitation, being too small or not understandable. Spaces of unclear interpretation include small structures of invisible, complicated or very difficult access. Among those, there are restricted groups of small steps of obscure use as they stop in the rock and do not seem to lead anywhere. For example, two flights of steps with an approximately east-west orientation, partly diverging from each other, become narrower on their higher parts and stop into the rock without any visible issue. They are dead ends, terminating with the last small step at a carved vertical panel with a cavity. In the upper end of both flights of steps, some device or object could be embedded or inserted into dowel holes if the latter were not cut marks or labation holes. The ensemble reminds us of the so-called open-air rock-cut sanctuaries in trace, dasted from at least a late-growth age onwards, as well as the stepped altars with an upper idol-shaped seat, well known from Frigia, where they have been associated in particular to Iron Age cult, although with a long prehistory in previous centuries. Bronze Age and Iron Age seem to be confirmed by mobile finds, probably attesting ritual proceedings in front of one of such flights of steps. Besides, it would not seem a coincidence that both previously mentioned dead ends be furcate from a carved south-north passageway, which itself ends with a flight of descending steps. In fact, starting from the base of the first described step to round, one can reach another flight of steps. Towards the north-north-east, this time much wider, lower and easier to climb on than the others. This staircase leads to a large carved cube-shaped outcrop and to some other carved features. The implication is that this is again a crossroads. In fact, three ways with diverging directions are crossing at this pace. The flights of ascending steps to the west, a carved south-north passageway and the wide staircase leading to the cubic outcrop towards the north-north-east. In antiquity, such crossroads were considered as important, liminal, even dangerous places where various divinities received ritual offerings. In other cases, a number of flights of steps seem to start and to stop in the middle of the natural rock with no obvious reason. In particular, a large rock is carved on various places with narrow stepped ramps, comprising a few steps, ascending or descending tracks and passages, not easily recognizable as such. This situation shows more complicated patterns than utilitarian trajectories or simple routine, as in other Castro areas. This induced behavior in space in a complicated itinerary might suggest symbolic proceedings. Besides everyday practice, trails, paths and roads may indeed serve important functions of enactment during processions and rituals. Such paths carved into the rock could suggest open air-cald as attested at least since the bronze and iron age and the archaic period in Greece with abundant evidence of parallels, as already mentioned, also from the east and the north, on high places, hills or mountains involving processions and other ritual performances. A main feature of the Castro-coastal rock-cut site is definitely its proximity with water, in particular the omnipresence of the Aegean Sea, obviously surrounding the peninsula. This becomes obvious with the communication of the main carved ramp of the first described crossroad, more or less parallel to the coast, at the edge of the cliffs, with lower levels, as more ramps and steps are directed toward the sea, at the same time permitting intercommunication between carved rectangular spaces or platforms. These itineraries involve planning out of preset directed movement, providing access from higher levels towards areas close to the present coast and vice versa. Moreover, some areas with rock-cut features could have served diachronically as observation posts towards the entrance of the harbor and the open sea beyond, but also attest industrial activities and, possibly, ritual. The importance of the sea and the involvement of people with maritime matters is emphasized by a monumental ship petroglyph on a rock-cut structure in the harbor, also confirming its symbolic significance. Indeed, according to the current stage of research, both on fixed features and mobile artifacts, besides other aspects, the intricate maritime associations at the Castro-rock-cut headland would have involved not only concrete components, such as sea shells, lots of them, and obviously imported by sea artifacts and raw material, among others, a lead ingot, but also symbolic facets, such as graffiti with ship representations on shells. Furthermore, figurines and other female artifacts of historic periods congregating in this transitional place at the border between the sea and the rocky landscape attest an intriguing female presence. In this connection, it seems significant that in 7th century BC, Avestia, by the northern coast of Lemnos, the local goddess, also named Lemnos, was a divinity of nature, in particular protecting water, since wells and several fountain models seem to have been dedicated to it. This local divinity would be related to other goddesses of nature, such as the Thracian Bendis and the Anatolian Matar or Sivelle, but also Ekate and Artemis. Ekate was an early goddess of entryways and crossroads, also connected with doorways and boundaries. Artemis, whose cult became important in Lemnos since the classical period, is a goddess associated with boundaries and liminality, including headlands, coastal areas and offshore anchorages. In conclusion of this overview, during successive phases of prehistory and antiquity, on Castro, diachronic human action has influenced the maritime and rocky landscape, while at the same time topographic guided choices of natural forms and spaces and prompted courses of action, related both to concrete and transcendental objectives. Megalithic formations reveal a composite scenery entailing an elaborate artificial system of space occupation within nature, while carved on sandals appear indeed to have had both symbolic and utilitarian components, materializing specific concerns and purposes, and therefore also revealing human forethought and deliberation. Thank you very much.


What is up everybody Jason here for jazzy's case breaks calm We just sold out jazzy's baseball is back MLB repack guys. Remember this is a dual caser Random teams number two guys, which includes 12 total boxes plus a little hit list promo So there's some of the preview cards there guys out of the cases that you can find We did two cases last night and we did hit this cofax and the Griffey at a three Now remember guys, of course it is a repack product that we created curated by us here in-house So it's only six cards per case six boxes dual cases mean 12 total cards So again, obviously a little bit of a high risk but a very high reward if you do get something guys Remember we only made 14 total cases, so there'll be seven breaks total now when we make it in-house We like to try to give it back So there you go, right at the end of the break if you do go hit list You have an opportunity to win back $150 in break credits use on whatever you'd like or maybe use it on the next break So we'll gather all the hit list names and top two names win 150 back each and now just remember just like any repack product like hip parade Same thing with us Every MLB every MLB team is represented throughout this print run But just know though of course as we open up cases some cards might get pulled early in the print run So just remember that some teams definitely Might not be as good as they once were depending on which cases which hits are pulled In previous ones, but everything should be really nice in every case guys. So here we go. Let's do the break. So here's the dice roll Let's roll it full 12 6 and a 6 1 2 3 4 5 6 7 8 9 10 11 12 Eugene down to Kevin and then 12 times here 1 2 3 4 5 6 7 8 9 10 11 12 Detroit Tigers out of the St. Louis Cardinals All right, Eugene Lee you have the Tigers Justin with the Astros Ryan with the Mets Kev with the Brewers Jimmy you have the Blue Jays Eugene clip with the Twins Logan you have the A's Francis with the Phillies Jason with the Padres Alan Murdock with the Mariners Edward Petty with the Reds Kevin Smith of the Red Sox Jimmy you have the Braves as well as the Nationals Eugene Lee with the Rays Ed with the Orioles Johanna with the Angels Edward Petty with the Pirates Eugene Lee with the Diamondbacks Logan you have the Cubs Eugene Lee with the White Sox see you with the Dodgers Eugene Lee with the Royals Jason with the Guardians see you with the Giants Jimmy with the Marlins Eugene with the Rangers Nicholas with the Rockies Ryan with the Yankees and Kevin with the St. Louis Cardinals I know Jimmy, but you know what you're There's a lot of there's a lot of customers that do the same thing though, man.   You know, you're definitely not Chad Just just gets like a hundred thousand dollar card. That's it. He's out of the game. All right guys I'm assuming no trades in all right. Let's do this All right, also to you guys we ended up doing case 13 and 14 yesterday So basically 1 through 12 are still there. It's crazy how we got 13 and 14 So I'm gonna do a same thing quick little generator So generate one case is gonna be case 4 and Case 8 this time 4 and 8 so I'm gonna delete that also from our case list. So case 4 and Case 8 So That So let's go with case 8 first six boxes There are here we go. That's good luck. Look at that 60% of the time it works every time John Smoltz patch autograph one of one and that looks like It is a cell sheet and Jimmy will last bomb oh Joe I Think that also says start and finish. I want to say you are definitely a believer now All right, or at least you have been sorry. All right next one And Randy Johnson for the Seattle Mariners from definitive one out of five Mariners going to Allen Murdoch and actually sorry to let me actually Keep track of this so Jimmy you're out Mariners You're out. That's just one hit and still in the first case. Never know Just have a lot more Frank. Tom is the big hurt seven out of 25 Why talks going to Eugene Lee white talks Joey bottle one out of five for the Cincinnati Reds when an Edward Petty Now of course, you know, not on the Reds anymore, but back in the big leagues potentially The little minorly contract with the Toronto Blue Jays where he grew up in Toronto. So playing for its hometown team So hopefully he makes it back up. All right two more to go guys Juan Soto, how about an atomic refractor rookie PSA 10 for the Washington Nationals? Jimmy on the board again That was your pack filler one too G-man Choi. Oh It's a lot of good former players here Look at that Roburza Clemente legends relic six out of nine. That's not a cell sheet Pirates going to Edward Petty. All right guys, there you go Very very nice case. I feel like man Good names. Let's go with the next one now guys Good thing it's a dual case here Plus one signatures Tim McHarver and Bob Gibson Six are five out of seven for the St. Louis Cardinals Cardinals is Kevin Smith. You're on the board Chase de lauder nine five ten refractor ballman's first to four ninety nine And that's gonna go to the Cleveland Guardians and chase Fall off guys. Another great a car. We have a Marco Vargas SGC 1010 bowman's first chrome with bowman's first chrome There you go, man out of Chihuahua, Mexico born in 2005. Good Lord. This guy is a baby He hasn't turned 18 yet. He's still like 17.  How about a little Vladimir Guerrero junior rookie autograph and it's a PSA 10 From inception baseball one twenty three out of 125 Blue jays Vladdie Jimmy again. Oh my god, Jimmy think all your teams have hit bro Is that a cell sheet it too by any chance? I don't know No, it's not Doesn't look like it. Oh, yeah, I pulled a Clemente. I pulled out of five. I think I pulled out of dynasty It was pretty awesome. I sure and Tom Glavin one of one I guess the pen he signed it with two isn't it? Maybe I pulled out a trip, but there's I'll look it up right now. I have all those photos of my archives Another one for the Braves another one for last spot mojo and Jimmy man Jimmy you are totally killing this break, bro Jesus That's the second braves it actually All right last one here Raw car and there it is Donnie baseball transcended autograph one of one Ryan with the Yankees There you go, man. Alrighty. There you go guys. That was the break again. I'm unfortunately a lot of people won't hit this guys But you have an opportunity to win back some money top two names I'm gonna recount these really quick to make sure so of course Yankees not part of the list Braves definitely not part of the list Blue Jays not part of the list Marlins not part of the list Guardians not part of the list St. Louis not part of the list Seattle not part of the list White Sox not part of the list reds not part of the list Pirates not part of the list and then Washington not part of this. All right. We're good. All right, so I'm gonna gather up this list here and it'll be 19 names instead of 18 because technically the brave said twice Roll the dice top two names 150 back each after six times five and a one One two three four five and six Foam six six five and a one six Nicholas and Eugene you guys will get back a hundred fifty dollars and break ready each and You can use that on the next break or whatever you guys want. Appreciate it guys number three is in the store guys


What do you mean by that? When? You didn't tell me about this? I thought we were mates. Don't touch me! Don't touch me mate! Do you want to fight me? Tonight he'll be locked up. Next week he'll go to court. And next month his family will still be upset. Do you want to get bashed? Because I'll bash you! Ten seconds ago he could have let it go. Oi! It's your choice.


We have two exercises for yesterday. One, remember was we take L1, 0 1, and we take the function that I have denoted by L of x. We know that the norm of L is equal 1, so this is a linear bounded operator. Frej龍. We take an element X in E, and we assume the norm of X in 1, of course, is equal to 1, Frow. Je sredu, to je različ um vzpo雯. We want to show that it is not possible that L of X This is not a spac, to dentist. Dлиbe se kaj. takko však ima, da je tudi drugi. Čak je tudi drugi, da je tudi drugi, da je tudi drugi? Tako da je tudi drugi, da je tudi drugi. To začneko, bo hmm... so, ki mali da je tudi drugi, tudi drugi, da je tudi drugi, tako, da je tudi drugi, ki je tudi drugi, ki se počena z deltavna z 0 do delta del X je nekaj pozitivne vrte. Zatim, da je to... nekaj nekaj, da je alfa... Alfa... Ok. Zato početno, da je bilo zelo počen. na drugi... nekaj, zelo se je... Noto, da je zelo, da je zelo, da je ni... je nekaj negrej z 0, da si je zelo... t. verg. t. t.Ézolo. The integral from zero to delta, tx of t, plus the integral from the delta to one, tx of t. So we split the integral in between zero and one, between zero delta and delta one, and then here t is less than delta, so this is less than delta, delta times the integral from zero to delta. So this way we want, this is the trick to forget about the multiplication by this factor here. So in this way there is no pre-factor here, and also here t is less than one, so also in this way we are now without the multiplication by t. So this is very similar to del one norm, so this is equal to delta, integral from zero to delta, x of t, t, plus integral from zero to one, minus integral from zero to delta. So in this way we now have the absolute value of x, but more importantly there is no more, there is no more multiplication by t here. So one is equal to that, so this is equal to the integral from zero to one, x of t, t, plus, plus delta minus one, delta minus one times, thank you, times alpha, so one is equal to this, and therefore finally the l one norm of x cannot be equal to one, cannot be equal to one, because if it is one then, I mean, one plus one minus delta alpha. So this says that if I have an element x, such that l of x is equal to one, then it is impossible to have that this is equal to one. So the proof runs as follows, assume that l of x is equal to one, so we have this, and so necessarily the norm of x is larger than one, because delta is less than one, you see, and alpha is positive. So this cannot be true. So x cannot be one, so we have shown that l of x equal to one implies, so this shows that this is not a max, as we said yesterday. So the idea essentially is to find a way to remove the multiplication by t. You see, there is no more t here, so that is very similar to the l one norm, actually it is the l one norm plus something, so this is, okay. Then the other exercise was, I think it was this one, yes, so I have the span of e one, en, et cetera. So remember, any element here is a finite sum, finite combination, final linear combination of this. We will consider this with the, and do it with the l one norm. So we look, we see this inside l one with the induced norm. So the, we want, now we take the g, g was span of e one r e one, and we have the functional g of lambda e one equal lambda. I think that this is where my notation, I hope. So this is a linear function on a subspace, so this is subspace of e. This is a linear, linear, linear, and also it is bounded because we can compute the norm of g, and so the norm of g is clearly one, okay. Is this clear? The norm of g is equal one. Is the supremum of this among lambda less than one? This is supremum, just one minute. So this is just one, this is just, let me call this maybe g zero, g zero, or g one, g one, okay. So this is just a linear map on a subspace, a linear functional on a subspace of e. Now we want to show that it can be extended in several ways. No, I mean, the norm is, these are always zero, zero, zero, one, zero, zero, zero. Sorry, these are the usual notation that we have used in the course. I mean, this is one, zero, zero, zero, zero, one, zero, zero, et cetera, et cetera, et cetera. So the notation is the following, e i equal zero, one, zero, et cetera. This is in the ith position. So now I was saying that this is the supremum of g lambda e one such that lambda is less than one, and therefore this is equal to one. So now we, so fix j in between one, two, in n, say, without zero, j bigger equal than one, and define, so for instance j equal to two if you define the following, the following, my notation is this, I think, yes, define the following object. So this is equal one if i is in n and i not zero, for instance take f2, just two. So f2 with j equal to two, so f2 of a1 is equal one because i is in one, because i is equal one, f2 of e2 is equal one, because j is equal to, now, as an example, and f2 of e i is equal to zero if i is bigger than two. This is just an example, is it clear? But for any j we have fj, for instance f2. Now we have to make it linear, so we extend this fj, so let me denote it yet by fj, so fj is the same symbol for the linear extension of fj on e, I am using the same symbol, still denote by fj, the linear extension of fj, this means that if I take a finite combination of e1, e2, et cetera, et cetera, then I use the obvious way to extend it. For instance, say f2 of lambda e1 plus mu2 is just, by definition of the linear extension, just an example, lambda f2 e1 plus mu f2 e2, and by that example, this is simply lambda plus mu as an example, so this is an example, for instance, remark, any fj extends g, for any j we have that fj is equal to g on g1, on the domain, any fj is equal to g, any fj is linear by definition, I have found for the moment infinitely many linear functionals extending g, for the moment, is it okay for the moment? Now what we have to show that, what about the norm now, so I claim that not only this, I will show that each fj is continuous and norm preserving, so I claim that actually this is equal to the norm of g, which is equal to 1. So this means that if this is true, of course, any fj is finite norm, so it is bounded, not only bounded, but yeah, sorry, for any j, g1, g1, this is, sorry, it was an notation, just the domain g, g, okay, sorry, g or g1, I don't remember, because at some moment I have written g1, yesterday was g, sorry, yesterday was g, so also today is g, not g1, sorry, g. So if this is true, this means that there is an infinitely, there are infinitely many linear continuous maps keeping the norm, preserving the norm of g and extending g. So this, if it is true, means, of course, that in the Hanbanak version of the extension theorem, we cannot expect uniqueness. Okay, what do we have to do? We have to compute the supremum of, so we have to compute the supremum, so this is, by definition, is the supremum of fj, let me call it x, divided by the norm of x, such that x is in the span, the span was e, maybe, I don't remember, e, okay, we have to compute this. So what is x? So take a finite, so take a finite set of indices, set of indices, and so we have to compute fj of the sum i, divided by the norm of the sum i, okay, so take finite set of indices, take numbers lambda i, not all equal to zero, so that this element, this is x, so that this x is non-zero. So we have to compute this, so this is, I mean, this is f of x, divided by x, for where x is, so this is x. The important point here is that what is a linear span, it is finite linear combinations, so this is finite sum, not the infinite sum, it's a finite sum, finite linear combination, so now we have to compute this denominator and the numerator, so what is the denominator? It's the sum of absolute values of lambda i, because this is the L1 norm, so we are using the small L1 norm on E, so remember that we are looking at E as contained in L1, so the small L1 norm is just the sum of the absolute value of the components, okay, do you remember? So if you have say, lambda 1 E1 plus lambda 2 E2 say plus lambda 3 E3, for the I, for the EI, this is, no this is lambda 1, lambda 2, lambda 3, 0, 0, 0, okay, in components, in coordinates, do you agree? Because this is one at the first position, in the second position, this is one only in the third position, so lambda 1 in the first, lambda 2 in the second, lambda 3 in the third, okay, it's okay, and L1 norm is just the sum of the absolute value of the components, which is written here, okay? Now we have to see what happens here, so we have to look at J, now if this can be various things, depending on the relations between J and capital E, okay, if capital E contains, contains J, if capital E contains J and 1, I would say, then this is just what? Absolute value of what? Absolute value of lambda 1 plus lambda J, do you agree? Because by definition, this is the linear span, and here however, I have this lambda 1, absolute value plus lambda J, absolute value, so this is surely less than or equal to 1, and then also the other cases, it is the same, only one of these two belongs to this and the other does not belong, then still this is less than 1, less or equal than 1, and if none of these belongs to this, still this is always less than or equal than 1, so in any case, we always have that this quotient is less than or equal than 1, and therefore this is less than or equal than 1, but it is also equal 1, because it extends the functional with norm equal 1, therefore its norm is surely larger than the norm of g, and the norm of g is 1, so it is at the same time less than or equal than 1, and surely larger or equal than 1, and therefore it is 1. I think that we have to make maybe the proof of Anbanak, so no uniqueness of the extension, this is clear however, you really don't need infinite dimensions to do this. There is another corollary, sorry, before doing the proof there is another corollary of Anbanak, so the third corollary, corollary, so assume that e is a norm vector space and take x in e, then the norm of x is defined as a supremum of f, x, such that, such that, which is also a max. So, remember, if we want to write the norm of a linear operator, then the norm of a linear operator is defined as a supremum. So, if I want to write the norm of a linear operator, then I take the supremum of all possible scalar product between the, in inverse setting. In inverse setting, remember, in inverse setting, if I have a linear operator, then there is the Ritz mapping, so that we have in Hilbert, we have this representation. And therefore the norm of the linear operator is the maximum of all possible scalar product between the representing vector h0 and h, as h runs in the boundary in the unit bowl. So, this means that essentially I am taking, so if I have found the representing vector h0, then I multiply it against any h, infinite dimension in the boundary of the unit bowl, and then I take the h and the norm of the scalar product, essentially h would be, say, parallel to h0. Now, here the sort of converse. If I want to find the norm of a vector now, this norm of a vector can be computed as a supremum, and now the supremum, however, is against all covectors, say, against all equal to 1. So, it's exactly sort of dual version of the norm. So, if you want to compute the norm of a linear operator, then you have the supremum, and here you divide by x, and here you have x is varying. Now, you want to find, on the other hand, the norm of a vector, then your supremum now is over f in the unit bowl. F is varying. So, just dual, exactly dual, and this is a maximum also in this case. So, there is some gain here. So, in general, we have seen that if you want to compute the norm of a linear operator, this is a soup and not necessarily a max. But now if we do this step, in this step this soup becomes magically a max. So, it's not completely symmetric, the situation, right? It is infinite dimension, but in infinite dimension it is not. So, we gain something here. This is attained. The previous exercise shows that in general for a linear operator, this is not attained. But for indiscreet, okay? Well, fx is always less than by definition of normal a definition of normal linear operator. This is always less than or equal than x, then f times x. So, and therefore if f is non-zero, then I have that this is always less than x. So, this shows passing to the supremum that at least we have that this supremum is always less than or equal than this. Is it okay? No. So, let me repeat it. I repeat it. For any f zero, we know that we have this. Is this okay? Is this okay? This is simply by the definition of norm of f. If you look at the norm, this is a supremum of this and therefore this is immediate. So, once you have this, if this is non-zero, you can divide. Okay? Hence passing, now this is true for any f. The right hand side does not depend on f. So, taking the supremum this immediately implies that the supremum and therefore as I said we have we have not shown this equality but just for the moment this. Now, so now give an x one of the corollaries maybe cornell corollary two. Yes, let me write down corollary two here was that that was this. Don't remember the notation, please help me. So, given x0 into e, there exists not necessarily unique but there exists such that I don't remember the notation but it was such that something like this. So, now we take advantage of this, this was corollary two of ambana. So, now we have given x we apply corollary two so there is some f0 such that such that x is equal to f0 into a star such that and then f0x is equal to the product x ok. Now the point is that I want to use I want to use this f0 in order to show this inequality I want to show this inequality now so let me so I want to take the maximizer so in the maximizer here is exactly essentially is f0 not quite because it does not satisfy the constraint so let me write it for instance this we know another way to write it is fx such that norm of f is less than 1 ok. Now I want to insert here f0 can I I'm not sure because I don't know I do not control the norm this f0 is the I want to use f0 to show that this supreme move is larger than f0 scalar product with x but f0 does not satisfy this so now I force it to satisfy so I take another parallel linear parallel to f0 with satisfying the correct constraint so I just scale f0 so I take say f1 just a multiple of f0 I go parallel to f0 so this is just f0 I want it to have norm equal to 1 and therefore I divide it by its norm simply so remember x is given and so I simply divide it by this so this is parallel to f0 because it is a number times f0 and the norm of f1 is by this is 1 is clear? now f1 finally is in the constraint so now f1 is inside the unit bowl ok so let me compute the f1x now f1x is equal to f0 f0x divided by the norm of x and f0x is is so this is simply equal to x right? because f0x is x2 is x2 divided by x so now f1 satisfies the constant against x makes this and therefore this supremum finally this supremum is certainly larger than what I obtain f1x divided by x which is equal to x ok? fine now I think that sooner or later we have to go to the proof of Van Banak so which is the situation of Van Banak I don't right here now the statement is a statement of set theory but we have the notation however so I think that we have some g subspace of a vector space we have a linear g and there is also something globally defined which is p ok? so this is linear this is almost a semi norm but slightly less than a semi norm so capital G is subspace capital E is a vector space small g is linear this is sub linear and positively one homogenous just positively one homogenous and they are related by this so g is dominated by p and we want to find an extension of g in the wall space a linear extension but still the difficult part is that you could produce several linear extensions but the point is that produce one which is below p everywhere this is the point now let us start by trying to so obvious remark if g coincides with e there is nothing to do because we have already globally the linear g so we can assume that this is not true so we can assume this and therefore there is a point outside so let me denote it by I don't know x naught take x naught in e so x naught so the first the first point is now to extend in some way your linear functional to the span so you have g and then you have x naught you take the linear span of g union x naught so the first point of the proof extend the linear span of g union x naught so for the moment we do not pretend to extend keeping linearly keeping the constraint ok so what do we do we define f so any point in the linear span of g union x naught can be uniquely written if x is in the linear span of g union x naught then there is a unique way to write it as lambda x naught plus some element that I have denoted by I don't know doesn't matter any element here can be written as x plus lambda x naught uniquely for some x in g and lambda real ok any point an element of can be written uniquely as this for some x for 1x in g and lambda in r ok so so we have to define f on such kind of points and of course we want it to be linear ok so this we define as f of x plus some number alpha f of x naught lambda sorry f of x naught lambda alpha where now f is g so at the end our definition will be this and we have to find alpha where alpha equal to f of x naught of course must be chosen properly chosen so I have capital G x naught outside this is the origin now I want to extend small g on the linear span of g union x naught but the point is that is not clear that I can do this because of course but the point is to compute alpha so that this is still true is it clear for the moment the strategy so this is not we are far from the end of the proof but the strategy is first to extend at one point outside keeping the constraint so now claim so we have to choose alpha we have to choose alpha so properly so that so that f is less than p ok this is the point so now the claim is the following assume that we are able to assume that we prove the following two kind of inequalities so g of x plus alpha less than or equal than f of x plus x naught p of x plus x naught and g of x minus alpha larger or equal p ah sorry sorry, I change in the letters so this I call x but this I call in another way so g y so assume that we prove we are able to show to find alpha such that for any x and g and y in g so we find alpha we are able to find alpha such that for any x and y in g that inequality what we want to prove we want to show that this is less than p ok, our aim we want to show we want to show that g of x plus alpha is less than or equal than p of x plus x naught we want to show this I claim that if we show this then this follows is it clear so I mean we want to find alpha so that this holds equivalently find alpha so that this holds ok claim if I prove that for any x in g I have this then this holds so claim assume then this for any x in g and therefore then this and therefore we have our extension ok so now I concentrate on this and try to show that this implies this argument ok so there is no anymore lambda here this is essentially for lambda equal 1 and minus 1 is enough essentially the idea is that I have to show this but essentially if I prove it for lambda equal 1 or minus 1 then this is enough this is the claim lambda since our p is just positively one homogeneous and not one homogeneous so remember our p satisfies this lambda positive et cetera and there is not the absolute value here and lambda real it's just this for lambda positive and no absolute value so I have to distinguish lambda positive so take for instance lambda positive and I want to show that did the first one of this implies this for lambda positive ok then I will show that for lambda negative the second one implies this for lambda negative so first take lambda positive and so we have that so our assumption p of x plus x not so what do I do I think I multiply I consider x over lambda here so I take one over lambda so yes x over lambda so I have g of x over lambda plus alpha this is less than or equal than p x over lambda plus x not do you agree because I have this so if I have this g is a subspace so I can consider in place of x just x over lambda which is still an element of capital G so I insert here in place of x x over lambda and this is still true here one over lambda this is linear now I look at this it's very similar not exactly the same so let me multiply I want to show this almost the same not exactly so I just multiply by lambda which is positive the inequality remains so I just have g of x plus lambda alpha less than or equal than lambda p of x over lambda plus x not ok now simply remember that our p is positively one homogeneous and lambda is positive so now by the property of positively one homogeneous of p I just put inside and it immediately check that this is equal to p of x plus lambda x not so what I have shown is that the first of this the first inequality implies what I want to show only for lambda positive now I would prefer maybe to leave you as an exercise that the second one home work the second one implies this for lambda negative so home is almost the same trick of course now lambda is negative so when you multiply by one over lambda you have to reverse inequalities so there is some ok so home work assume for any y in g this is larger than y minus x not for any y in g then g of x plus lambda alpha is less than or equal than p of x plus lambda x not for any lambda negative ok this is home work so using the homogeneity we are reduced to the case lambda equal one and minus one ok we are reduced to find alpha that this is true now the point is is there some alpha satisfied so that these two inequalities families of inequalities are true at the same time so let us try to rewrite this so alpha must be at the same time from this alpha must be less than or equal than this minus this p of x plus x not minus g of x but at the same time alpha must be alpha must be is it correct what I have written yes sorry because this is wrong ok sorry so please correct this family of inequalities with less than or equal to ok so at the same time alpha must be less than this minus this but also larger than this minus this so at the same time g of y minus p of y minus x not so if we want to prove the claim we have to find alpha such that this is true for any x and y in g we need this for any x and y ok this is equivalent to this so there should be some separate number separating these two so let us rewrite it in another way so observe now that g of x plus g of y is equal to g of x plus y which is less than or equal than p of x plus y which is less than or equal than p of x plus x not plus p of y minus x not do we agree so let us repeat we need to show that there is some alpha so that these two inequalities are true for any pair x and y in g ok remark g is linear so take any two points x and y in g g is linear and its domain is a linear space so this is an element of the domain and I can write this this is clear I also know by assumption that g is less than p so we have this now add them subtract x not use the sub additive of p so you have this and therefore so you have separately this but this is this inequality namely for any x and y you surely know this and therefore you can put something in between by the properties of the real numbers is this clear or not its ok proof so there is alpha there is at least one alpha which makes the job so what we have done up to now is that we have found the way to extend just one point outside just in the linear span just using one point outside g as you see it from the proof the difficult part is to keep the constraint the linear extension is easy but keeping the constraint means to cleverly choose alpha otherwise you don't have the constraint so this is just the first part of the proof which is somehow constructive and now there is the not constructive part to conclude the proof so now you consider so now this is ok let me consider now the family of course I have problems with the notation p now consider following set p p is the set of all h they are domain it's domain linear subspace so now I consider an abstract object the set of all linear maps functionals defined on subspace subspace subspace containing g containing extending such that h is equal to g on g and keeping the constraint and h less than or equal than p so what is this this is a set of linear functionals abstract what we have proven up to now what we have proven that this is non-empty all the effort that we have done up to now is just to show this p is non-empty why? because f that we have constructed is an element of p because f is an element of p now the idea is to take the largest possible h what does it mean the largest possible h? h is a linear functional so now the candidate of our of our theorem is to take the maximal extension so the largest possible elements of p so this is an abstract problem of set theory unfortunately this is given by Zorn Lem but Zorn Lem says that there exist such a maximal element upper bound maybe provided that p satisfies special property so take an element c as a subset c contained in p which has the following property which is totally ordered which is the order of p I have to end of p with so when I say that h1 denotation I think h1 is less than or equal or contained let me just check denotation yes less than let me just let me just use this obvious notation this is not bad because h1 and h2 are scalar valued so maybe not bad to say that this is less than or equal than this if well if the domain of h1 is contained in the domain of h2 and h1 extend h2 h1 what does it mean h1 extends h2 significantly they coincide it means that they coincide in the smallest h1 equal h2 on the h1 it is an extension so in this way I am comparing to elements of p so inside p is a big abstract set and then saying what does it mean that one is less than the other of course not all elements of p can be compared to each other we are not as in the real numbers that any real number can be compared with any other real number not the case you have an element here in principle one is not less than or equal than the other so I cannot compare all elements one each other but I can compare some maybe one some other so I am considering a big abstract set p and I am undoing it with an order relation partial order relation in the sense that I cannot compare in principle all elements one with each other so I cannot say that I am higher than you but maybe I can say that I am higher than you I mean it is a sort of not higher maybe it is not the best way to say but well if you have say points with arrows you can order a finite set of points for instance with arrows connecting with arrows and you can say that one is larger of the other if there is an arrow in between the two with the proper orientation so for instance I can compare say this is less than this for instance but I cannot compare this and this so there is an order relation which say okay this is larger than this okay there is an arrow going from here to here with this orientation but then there is no arrow between this and this so they are not related by the order so we want so the claim is that Zorn lemma should give us the largest possible extension but to apply Zorn lemma there is an assumption the assumption is that if you take C totally ordered ordered then there is a maximal element of C so you take a sub chain totally order of P then there is a maximal element okay and then it is very easy to show that there is a maximal element because if you have HI and C consists of HI for some strange set of indices capital I so this is a family of linear maps in P E set of indices I don't know how large it is and HI is in P and then how is naturally defined of C I take simply the union so I simply take the following H so the domain of H first of all first I isolate the domain of H the domain of H is the union of all possible domains union of the domain of HI okay now I have not only to define the domain of H by define H of X how do I define it well I take a point X inside here this means that X is inside sum of these at least one and therefore I define it as X belongs to some HI some D HI so X in the union so P is non-empty P of H is this I take a point X in the H this implies that there exist an index such that X is in the HI and then I try the following definition I try the following definition and I check that this definition is well is well this is well defined because of the properties that this is totally ordered so one is inside the other and therefore there is no ambiguity in define this is well defined for a totally ordered subset C in P this is the this is a maximal element of C and therefore Zor's lemmas gives us the what we want so by Zor Zor implies that there is an upper bound that we denoted by F DF in R now what we have to prove well it remains to show it remains to show this the point is to show this that F the domain is everything well and this is again true why it is true if this is not true then this is contained in this and there is one point outside but if there is one point outside we know how to extend F outside by the previous step the concrete previous step and that is a proper extension of this because now the domain contains strictly the F the maximality of F so this proof is interesting because it consists of two parts there is one computational part in which you try to extend outside on the linear span of G and one point outside this is concrete you find alpha not terrible but not easy then there is this abstract part because it is terrible in the sense that it is not constructive there is this extension you don't know very much about it how to construct it there is but this extension of course is based on the previous step so this proof is divided into two as you can see no continuity arguments just set theory in the last part what would happen if we assume now just remark rough remark little bit rough in the sense that I don't want to go into too many details into rough remark assume that E is something more has a norm is not only a vector space now let me put it a norm and assume even that it is separable is it possible in this condition to make a rather a proof less abstract than this so why is this proof so difficult because we don't have structure we don't have norm we don't have separability we just have a vector space so when you have this you have set theory the idea would be why it is so difficult but assume for the moment that for is a smaller 2 separability is not always true in infinity is not true but very often it is true so why don't we assume and so under this condition what would be there is a natural way to do this proof one way would be take a point outside x0 outside and we know how to so let me denoted g1 the span of g union x0 and we know how to extend it call this small g1 maybe so separable means that we have x0, x1 etc etc countable number of points thenz in e countable thenz in e ok take one of these points outside and extend it on the span as we have done before but this extension we can also keep the norm because under this stronger condition we have seen that also the norm ok so the norm of g1 is equal to the norm of g ok this is one of the remarks that we made yesterday if we have a little bit more structure we will have this so g is below the norm g is below the norm now this is really a norm and so we can make the norm preserving extension as in corollary 1 on g1 ok then the idea is to do on g2 so why don't we extend now on g2 which is the span of g1 union x2 2 yeah assume also this g is continuous thank you g is continuous you have a bound using a norm the constant is given by a norm so small g is continuous and we extend it into g1 into a continuous g1 on this now I do the same and extend it in span of g1 union x2 so I end up with a continuous extension g2 and then I proceed by induction doing this on each gn plus 1 equal the span of gn union xn plus 1 so this way and I have the extension gn plus 1 now function on this span and then now the idea is now we have to globally define f so and the idea is now why don't we try to extend it by continuity extend it by continuity so take x into e take a sequence converging to x so let me indicate it by xn sequence converges to x sequence of element of the dense subspace subset so x is in the dense xn in the denotation maybe was d I don't know extended to d dense subspace subset so and define f of x equal the limit of of what do we have say gn xn something like this so this is sorry this is very rough I don't want to enter in the details just a way to say that in order to make somehow a more proof made by hand assuming normal space separability maybe we can extend on each point of the dense subs subset and then extend by continuity outside and this is also this also works I think works only under these assumptions so this is yes ah separable yes I think that this works in general it is not never also in that case you don't have uniqueness but I think that in a separable here where space you can at least I believe that this kind of more concrete proof can be done by extending outside of the union of so you have essentially you have defined your your f just only here on the union of the gn ah because you proceed inductively with a countable induction and so in the union of the gn you have defined your f out this union is dense and so at any point text you take sequence of element here extend and you simply extend it by continuity these linear functions it works I think yes but doesn't give you any uniqueness you don't I think that you don't have uniqueness even in finite dimension by the way there is no reason if you have a linear function defined on a line in R3 I think it's not yeah you have to preserve known the point is always to preserve known of course ok well maybe we continue the discussion ok


We've been trying to solve Morgan's mystery cross for a while now and we're almost there. And this is all that we have figured out so far. We know that the genes which are involved in this cross are present on the same chromosome because if they weren't then we wouldn't be getting this proportion of parental look alike and hybrids that we have gotten in this cross. Second we found out that randomly arranging these genes aren't just enough because if we were just doing that we wouldn't get any hybrids at all. And we know that that's not the case. So along with the random arrangement of genes, these genes are also undergoing an event called crossing over. And this crossing over is what is giving us these hybrids. However, a perfect crossing over or a successful crossing over will give us a 50-50 proportion and not this bizarre number that we have over here. So we know that the crossing over which is happening over here might not just be successful. But why is that? Why is crossing over not a successful event sometimes? When is it not successful? And how does it even fail in the first place? These are some of the questions that we will be answering in this particular video. So let's go. Now before we find out what is an unsuccessful crossing over, we need to know what exactly makes a crossing over successful. Well, you see this crossing over event has one major job to do. It needs to go ahead and separate the genes so that they can swap or exchange with one another. If the crossing over event fails to separate the genes, then it is not considered to be a successful crossing over event. So if you look over here, we have some blue genes and some pink genes over here on two different chromosomes. You can see that they have crossed over but then again the genes have not separated at all. So these genes are not separated. And that is what makes them a very flopped version of the crossing over event, making them completely unsuccessful or we can say that this crossing over event did not carry out the way it should because the genes never got separated and then they never got exchanged or they never got swapped. When something like this happens, this is what we're going to call an unsuccessful crossing over event. But why is it so? Why are these genes not getting separated in the first place? Well, let me ask you something. You have a chromosome over here and there are a few genes. Let's quickly label those. All right. So we have three genes which are placed on this chromosome in which A and B are super close to one another and B and C are pretty far away from one another.  Do you think the genes would be separated if the crossing over was happening right between them? For example, if the crossing over happened right between A and B, then do you think they would get separated? Yes, they would definitely get separated. And if it was happening between B and C, then B and C would get separated right. So when we say that two genes are not getting separated, it means that crossing over is not happening between them because they're so close to one another. That is why they don't get separated. But now the question arises, why is crossing over not happening between these two genes just because they're close? Well, sort of. Let's do one extra thing over here. Let's plot all the points where crossing over can possibly happen. And then we're going to see. So I'll just quickly go ahead and mark all those points. There we go. Now, we have all those possible points at which crossing over can take place. Now, I've numbered them out as well so that it's easier for us to follow. Now, can you tell me how many points can you see between A and B where crossing over can take place? Just this one over here, right? This lonesome one point over here. Meanwhile, how many points can you see between B and C? There's like 13 points over here. Quite a lot, right? I mean, compared to one. So you see crossing over between B and C is much, much simpler and easier as compared to A and B. Because between B and C, there are 13 points, 13 options, or 13 chances for crossing over to take place. Meanwhile, there's only one between A and B. So this kind of teaches us two very important things. First, if the genes are far away from one another, then the chances of crossing over or recombination or the chances of getting more hybrids increases. Because if more crossing over happens, then more recombination is going to take place and we're going to get more hybrids. So farther the genes from one another, like B and C over here, higher are the chances of crossing over to take place. Meanwhile, if the genes are super, super close to one another, like A and B over here, then the chances of crossing over decreases or they go down. So it's less likely for crossing over or recombination to take place, which means that we're not going to get too many hybrids. We're just going to get a very limited number of them. This, by the way, over here, that is the chances of crossing over to take place or the chances of recombination to take place. This is called the recombination frequency, by the way. All right, so now that you have an idea of what crossing over really looks like when two genes are close to one another or if they're really far away from one another, let's go ahead and do one thing.  Let's say that we want to cross over at this point, number five over here. Let's see what happens if we go ahead and do that. Okay then, so crossing over has taken place at our favored number five spot over here, and this is the outcome of the crossing over of the chromosomes that has happened. Now, why don't you pause the screen for a few moments here and tell me the first thing that comes to your mind when you look at this picture? Well, you see the pink genes, that is capital A and B, and the blue genes, small a and small b, they kind of moved, but they moved together, kind of like as a package, as if they're stuck together. Now, we know they didn't get separated because the crossing over didn't happen between them, like between A and B. Instead, it happened between B and C somewhere at five. Meanwhile, you can see that the C gene right below, they kind of stayed the same, nothing really happened. Now, we know that crossing over has happened because you can see the change in the color of the chromosomes, but the genes, they remain together, they never got separated. So you can say that they kind of remained stuck together, like best friends. And these best friend genes, which stay together as a unit or they move together as a unit or a package, they are called linked genes. And the most fascinating thing about these linked genes is that they don't separate, it's not that they cannot. It's not impossible for these genes to be separated out. You can do that if the crossing over point happens to be right between them. But since it is so difficult for that crossing over point to pick out that very specific point, we saw how difficult it was. So for that to happen is a very, very rare event. So most of the time, these genes, they stay together as best friends and that is exactly what we called the linked genes. Now, why am I telling you about these linked genes? What do they have to do with Morgan's Mystery Cross? Well, everything, they have to do everything with that cross. Let's circle back and take a look at that cross for a bit. In Morgan's Mystery Cross, the one thing that stumped us all was the bizarre proportion of offsprings that we were getting. There were a lot more parental lookalikes as compared to the hybrids. Now, does that remind you of something? Let's go back to our linked genes once. Now, one thing that we just learned about linked genes is that they're very difficult to separate because they are really, really close to one another. Now, what happens when genes are super close to one another? The chances of crossing over drops. And if there are lesser chances of crossing over to happen, that means recombination or the chances of recombination also drops. Which means lesser number of hybrids. And isn't that something that we have been dealing with in Morgan's Cross since forever? This is the connection that we've been looking for all this time. Now, keeping these linked genes in mind, let's take a look at what the production of gametes is gonna look like. This time when we look at the production of gametes, there is one major, major change in the diagram itself. You can see that the genes which were involved in Morgan's Cross, the body color and the length of the wing genes, both of these genes are placed super close to one another. So you can see B and L placed really close. And again, B and L are placed really close on the chromosomes right here. And because of this, B and L, they become linked genes. So we already know that if a crossing over needs to happen right between them, it's gonna be a super rare event. And over here in this diagram, that's the event that I've chosen. I have chosen a crossing over point that is right between B and L. And that crossing over has happened, as you can see right over here. Then the chromosomes, they get separated out and then the different types of gametes are formed. Now you can see that the gametes that we are getting from this combination are exactly similar to the type of gametes that we were getting in Morgan's Cross. Except, since this is an extremely rare event, the number of times this event to happen is less. And we will only get hybrids if this rare event happens. So if this rare event happens, we will get hybrids or else if this doesn't happen, we won't. So because this rare event is happening, we are getting our hybrids, but the number of times this happening is really low. So we are going to get a very low number of hybrids because of how rare this event is. Because of how difficult it is to separate the linked genes B and L. Most of the time, this separation is not gonna happen. Most of the time, B and L will stick together and crossing over or the point of crossing over will be somewhere anywhere else, anywhere on this chromosome except for between B and L. So most of the time, we are gonna get the capital BL and the small BL gametes over and over again, but every now and then, every once in a while, we are going to get the hybrids too. Except this time, the number of hybrids is gonna be less. Meanwhile, the number of parental lookalikes are gonna be so much more. And that is why we have this bizarre proportion of numbers that we have where 83% of them are the parental lookalikes. Meanwhile, only 17% are the hybrids. So you see, it was these linked genes all along that caused all the confusion. But if you're thinking that this is where the story ends, then think again. Because studying these genes gave us the power to find out the location of each and every gene in our entire genome. Now, how did that happen? Well, we're gonna find that out in another video.


And former Vice President Atikou Abu Bakr says there is no reason the masses in Nigeria are battling extreme poverty, it says especially when the country is blessed with an abundance of resources. Atikou in his Independence Day statement on Friday said, it was regrettable that despite the nation's vast oil resources, extreme poverty and joblessness have remained persistent. He called on the government at all levels to push policies that will lead to improved standards of living for the average Nigerian. Hello, hope you enjoyed the news. Please do subscribe to our YouTube channel and don't forget to hit the notification button so you get notified about fresh news updates.