 Today, we will talk about the point set topological aspects of simplicial complexes. So, this is the topic for module 31. By definition, all polyadrons being homeomorphic to simplicial complex, modulus the geometric generalization of the simplicial complex, they will share all the topological properties of that. And we have seen that in the very definition that the topology on mod k is finer than the metric topology induced on it by its inclusion in i power v of those functions which take only finitely many nonzero values. Anything which is finer than metric topology will have many, many other properties. One such thing is that it must be Hausdorff. Any space which is finer than a metric space is automatically Hausdorff because the metric space is Hausdorff. So, the first theorem is obvious here. Mod k is Hausdorff. So, the next proposition here is the key for several important properties of mod k. But this key is totally obvious from the definition. Namely, every mod k is actually disjoint union of all its open simplexes. Starts with zero simplexes. By definition, zero simplexes are open. Remember, modulus of simplex consists of functions which are nonzero at most at the vertices of the simplex with various properties, summation should be zero and so on. And the open simplex consists of those which have precisely not equal to zero on that set, on the set f. Everywhere else it is zero. On that set, on all the points here, the coordinates are nonzero, that is open simplex. The zero simplex, the modulus of a singleton, nothing but the function which takes one there and zero everywhere, everywhere else, on every other vector. So, automatically the singleton modulus of singleton is both, is an open simplex as a closed simplex. So, don't confuse it with open subsets and closed subsets. A closed simplex is a closed subset also, but open simplex need not be open everywhere. If they are open inside closure of f, closed simplex f, that's so. But what I want here is that they are, the mod k is precisely equal to disjoint union of of this open simplex. Take any point here, it belongs to precisely one of them, namely its support. Look at the support, that's a simplex. Then open simplex of f4 contains the point, contains that function and nothing else will contain it. Therefore, this is a disjoint union of these things. Is that clear? Yes, sir. Clear. We will use this in a very, very subtle, very nice way. This lemma, we will use that one. Take any subset of k such that every simplex f of k has exactly one point in the interior of f from A. A intersection f has at most one point, sorry. So, some of them may not intersect. If they intersect, that point of intersection should be one and it should be the interior point of f, one of the interior points. Assume this. You see the condition on the subset A, not a condition on a mod k or f. For each f, this should happen. A intersection f can be empty or A intersection f is one single point. And that point must be the interior again. If that happens, A is a closed subset of k and it is a discrete. I hope you know what is a discrete subset of a topological space. So, proof is very straightforward. So, we use the fact that mod k is disjoint union of its open simplex f. Suppose this is true for A, the assumptions on A I have given. Now, you take any subset of B, any subset B of A. That will also have the same property. Right? It will have fewer points and so that will have same property. A intersection B may be even empty also, but it cannot be more than one because even A intersection f has one. So, B intersection f will also at most one. Okay? It has same property. In particular, what happens is if you take the closure of f, closure of f will have all the smaller open simplexes also. Namely, take any subset of f, proper subset of f, all of them will come. That is the union. Closure of f itself is a disjoint union of open simplexes. Namely, all these open simplexes are subsets of f now, including f. Therefore, there will be only finitely many of them. If this has only finitely many open simplex, each of them contributing at most one point, B intersection f with the finite subset. Any finite subset of mod k is a closed set because k is Hausdorff. So, B intersection f is closed subset of f for every mod f. Okay? But then we know that if this is the property of any subset, that B itself must be closed. Intersection with every mod f is closed and mod f means B itself is closed in mod k. Okay? So, what we have proved is that every subset of every subset B of A, every subset B of A is a closed set. In particular, A is closed and it must be discreet, but it is every subset is closed, every subset is open also. So, it is a discreet. It is a discreet topological subspace as a subset of mod k and it is closed. Okay? The topology induced from mod k on A is discreet. Okay? It is a discreet space. Every point is open in particular. Alright? Now, this lemma was an easy consequence of the easiest observation here. So, now we will derive some consequence of this lemma 5.3, which is quite completely not so obvious at all. Each mod f is compact. That is obvious because we have seen that it is homomorphic to a disc. Right? But conversely, each compact subset of mod k is covered by finitely many open or closed complexes. A closed simplex is a closed set. An open simplex is not an open set. It is not an open cover for mod k. So, don't make that mistake. If we have a compact set covered by some open or has an open cover, then it has a finite sub cover. That is not the point here. The point here is that these open sets are not at all open in mod k. Okay? Only maximal complexes will have open simplex is open inside mod k. Yet, only finitely many of them will cover it. So, this is the point. Okay? Once open simplex is finitely many of them cover it, the closed simplex will follow because you can take the closure also. So, there will be finitely many of them. So, they will cover it. Okay? So, I am going to prove this finitely many open simplex themselves will cover. Okay? So, how do I prove it? I will mod k is homomorphic to a closed disc that we have seen. So, it is compact. Now, I am going to prove something, start with any compact set L contained inside mod k. Then, we can construct a subset A of L such that L intersection open, take any open simplex is non-empty. Then, it has only a single term set. In other words, if it intersects, then I pick up one point from this one. Okay? L is given any compact set. Okay? L, we are going to construct a subset A of L. Look at L intersection F. If it is non-empty, pick up one point from this F mod F intersection F. So, that put it inside A. Did he empty? Do not worry. Okay? So, what is a subset A? A is subset of L and for each open simplex, okay? L intersection F is non-empty and I have to give that A intersection F is also one point, one point or empty. Therefore, A has exactly the same property as in this lemma, previous lemma. Therefore, A is a discrete set. Therefore, being a subset of a compact set, a discrete set must be finite. What is the outcome? Look at, for each point in A, that is one and only one F, okay, which will intersect this one. Okay? But if you take all the F, they will cover the whole of L. Therefore, there are only finitely many F1, F2, F5s like this corresponding to elements in A which will cover L. So, the first part is over. Take closures here. So, they will also cover L. So, that will give you second part here. Okay? It is not true that closed simplexes which meet another closed simplex, they will be finitely in number. That is not true. Okay? It can take a point and then there can be infinitely many edges which are incident at that point. That is allowed. Similarly, one single edge may be incident on several triangles, infinitely many triangles. Okay? Yeah? So, a simplex, you have taken any statement. Mod F is covered by mod F. So, why do you need so close simplex? Right? But this part is very important. Every compact set is closed by this one. That is the whole thing. But mod F is covered by finitely many open simplex. So, I should have stated it may be separately for these two things. For mod F is covered by mod F itself, one single simplex. That is not the beauty. Beauty for mod F is open, but the proof shows that every compact set is covered by finitely many closed sets. Close simplexes like this. Okay? The next thing is another important property, interpolation. It says that mod K is actually normal. You can also say that, okay? It is finer than the metric space. Right? So, it must be normal. But that is say, there is a catch there. Something is normal and you have a finer topology. It may not be normal. All right. Metric space is for normal. Fine. But what we have put is a finer topology. So, you have to careful here. You have to prove that mod K is normal. It's not like house darkness. Okay? So, let us not hand wave here. Let us get a proof of that mod K is normal. What is normality? Given any two disjoint closed sets, they must be, you must be able to separate them by open sets, disjoint open sets. But there are other things. Given any closed subset, okay? And a continuous function into I, you can extend it to the whole of K mod K. That will also prove that mod K is normal. So, this is the property given by T's extension theorem. If a space has T's extension property, then it must be normal and conversely. So, that is what you must have learned in point set topology. So, I am going to use that. So, starting with a closed subset K of K, okay? And a continuous function from A to I, I want to say that there is a continuous function P hat from mod K to I which extends P. If I prove this by T's extension theorem, closed set mod K is normal. Okay? There may be different ways of proving this one. But I find this one to be the simplest. So, how to construct P hat? So, again, the recipe, you have already declared how to construct, contain the functions on mod K, construct them on each simplex, each closed simplex in such a way that on a smaller simplex, whatever you have constructed is extended down the larger simplex. Okay? If a function is constructed like this, then as a function, it makes sense first of all. Continuity follows because restricted to each mod if it is continuous. So, this is the recipe we are going to follow here. Okay? This will be the first time perhaps we are doing it like this. So, what I am going to do is I will construct a family of functions P indexed with F on mod F to I for each F. F is simplex such that if you take A intersection F, this must be the given function if E. Okay? That is the first condition. If it does not intersect, I do not care, it could be anything. Second condition is if I have already constructed it on F prime which is subset of F, okay? Then the phi F that I am going to construct restricted to F prime must be phi F prime. So, this is what as compatibility. If you have constructed the function on all the vertices, then when you are extending to, when you are defining it on an edge, you should take care that on the end point is already a function that you have constructed. And once all the edges are done on a triangle when you are defining, it should be on the boundary of this triangle, it should be already the function that you have constructed. Okay? So, if you follow that, then you are done. Okay? So, inductively you can do this namely take F F. Suppose you have already defined it on the boundary which is a lower dimension simplex. You need no lower dimension simplex. If you can extend it over F, you are done. All right? So, this is easily done by induction. That is what I am telling you by F. Since each mod F is normal, okay? Because it is a Euclidean space. Each mod F is normal A intersection mod F union all the boundary, boundary of mod F that is a closed subset. On that closed subset, you have a function continuous function. So, apply that mod F is normal, you get an extension on a mod F by the same t-section theorem. Put them together, you have the whole function on mod k. Okay? Here are a few remarks and on exercises. I will go through a couple of them. One of the things that we have we have claimed that mod k as a subset of the product space here, i power v is a closed subset. If v is finite, okay? We are asked to verify this. Okay? We have defined actually, if you just read carefully, when v is finite, we have defined a topology in a particular way. Use that to see that it is closed. For any simplex complex k, this is, this is a typo I corrected it, but today I am taking the wrong slide here. Mod k cross i, okay? It is mod k cross i, it should be, triangulate mod k cross i with exactly twice as many vertices as in k such that k cross 0 and k cross 1 are sub-complex. Remember k cross i, this notation was for this prism. So, I do not want to use that notation here. k cross k is a simplex complex whereas mod k cross i is a typological space. All right? So, I want to give you another triangulation of mod k cross i in which no extra vertices are introduced. Only from k cross 0, all the vertices of k will be there. A copy of that one k cross 1, again, same thing should be there. Okay? For example, if k is just one edge, then k cross i is the square. So, then you are allowed to take only four vertices, no more vertices and triangulate it. And we have shown that there are two ways of triangulating it. Either join this diagonal or the other diagonal. There are two diagonals there. You can take any one of the diagonal as another edge and you are done. The same thing you should try to do for all simplexs. So, that is the exercise here. Okay? But there is no canonicalness here now. And hint is, choose an order on the vertices. Some, some linear order. Order means that. Okay? That is the hint. Like when you one edge, you have e1, e2 or e0, e1, whatever. So, you have to choose a order and then use that order to write down. That is, that will help. I have given you a triangulation of r. Now, you are asked to give a triangulation of rn. Okay? Do that. Give a triangulation of pn. Okay? So, hint is use a strangulation, but this is not going to help you much. But it is just a hint. That is all. This is not going to use a strangulation of the spheres and induction. Induction will be correct. But this is not quite, quite the stuff here. Okay? A strangulation will not give you a triangulation of pn. Here is another concept, which I am not very much interested in pursuing. So, I have put it in the exercise. Namely, that k be a simplicial complex, then the three conditions here are all equivalent. This is an extra condition mod k. Okay? It is not a part of every simplicial complex. The first condition says that each vertex of k belongs to finitely the number of edges. Okay? Only finitely many edges will occur there at each vertex. Second condition says that each vertex of k belongs to a finite number of simplicis. Okay? The third one says that the topological aspect. Mod k is locally compact. Okay? So, you see this is pure combinatorial condition. First two of them. The third one is topological. So, you are supposed to prove these three things. Such a thing if it satisfies any one of them, we call it k to be locally finite. Okay? A locally finite means all that you have to do is look at every vertex. There was to be only finitely many edges emanating from there. Once you immediately locally finite is done, automatically mod k is locally compact. Or, conversely, if you have started with a locally compact space, then if you want to triangulate it, okay, you have to choose a k which is locally finite. There is no other choice. Okay? Try to prove that. The next one is something about when can a simplicial complex embedded inside R n. Okay? So, it gives you some condition. f from mod k to R n is a topological embedding. Topological embedding means what? Which is continuous function 1 1 and on to the image it is a homeomorphism. If you take the image as a subspace of R n, then there will be inverse function which is continuous. So, that is the meaning of topological embedding. Show that k has to be locally finite. Second is k must be countable. The number of vertices in k must be countable. That is the meaning of countability. Automatically, number of simplexes will be also countable. Okay? The certain set is countable is the meaning of this. The third thing is dimension of k must be less than or equal to n. So, this is harder to prove right now for you. Okay? This is harder. So, I have some idea why I have given at this stage. Later on this will become easier. Okay? After we will do a little more theory. Dimension of k cannot be bigger than n if it is embedded inside R n. All right? So, I will continue here. Here it was arguably topological embedding. In this exercise, you want f to be linear on each simplex. R n has a vector space structure. A subset k, namely mod f, that has a linear structure, affine linear structure. So, a function f from mod k to R n restricted to each mod f. If it is affine linear, then we say f itself is linear. This is just the abuse of notation, abuse of language. Okay? So, here I have stated as a theorem because I have lack of time. I am not going to trade this one. So, by following the number of exercise, you can complete the proof of this theorem. Namely, every locally finite countable simplex complex of dimension k, so all these all these conditions were there in the earlier exercise if there is an embedding. Now, I am reversing. In the converse, I have put dimension of k to be very small. Namely, it is less than half, you see. Then we can have a linear embedding f from mod k to R to k plus 1. Okay? So, what are, what is the idea? Idea behind it is there in these steps. All right? So, these are steps a, b, c, d. So, let me stop here. You can read them by yourself. Thank you.