 Hello, welcome to NPTEL NOC, an introductory course on Point Setupology Part 2. Today we shall continue with our study of the separation of sets, module 39. So, we have seen a number of implications and some non-implications also, but now under compact half-top space we will see that all the four axioms are equivalent. So, this is the statement, if x is a compact half-top space, then S0, S1, S2, S3 are all equivalent to each other. As I told you, we have already proved under T1 as S3 implies, S2 implies, S1 implies, S0, right, just because every point is closed, that is all we have to use. We have also proved that under Lindelofnes, S2 implies S3, the reverse implication. So, it remains to prove that under compact half-topness, S1 implies S2 and S0 implies S1, okay. So, before proceeding further, I will state a lemma which can be used again and again and of course it will be used in proving these two statements. So, the lemma is take a compact subset of a topological space x, okay, take a point P in the complement of F. Remember, F is a compact subset, okay, so I am taking the point outside that one. Suppose every point of F can be separated from P inside F, then there exists a clopon subset W of x such that it contains F and does not contain the point W. So, from point separation here for each point, for each point of F can be separated from E. So, what I have done is, I have taken the hypothesis of S2, but only for points of this compact subset. So, that is slightly generalization than assuming x itself is compact, that is all. But that will be useful for us. So, what we do? To each point Q inside F, we get a separation, that is the hypothesis. x equal to AQ per BQ, P inside AQ, Q inside BQ. Remember, this just means that AQ and BQ are both open and closed and they are disjointed. Since F is compact, F will be contained inside BQs because Q varies over F. Take the union of all BQs. So, F is inside that. Now, each BQ is clopon, in particular open. So, we get a finite cover. So, I can write that I range from 1 to n BQI union containing F. So, this I am denoting by W. So, that is my choice of this W, which is obviously open as well as closed. So, being a finite union of clopon, that is to open, it is clopon. Clearly, P is not a point of any of these BQs. So, P is not inside W also. So, over. So, compactness has helped us here. Just like in the case of Lindelof, we have got S2 implies S2. Now, we will use this lemma to prove both these S1 implies S2 as well as S0 implies S1. The first one comes very easily down. Let X be a compact half-door space satisfying S1. Let F be a closed subset of X here and P belong to X minus F. Since F is a closed ridden X, it is compact it is a closed subset of a compact set. So, we can apply the previous lemma to this closed subset F. So, what we get? We get a clopon subset W such that F is contained inside W and P is not in W. So, this is just a almost restatement. Instead of X compactness in the lemma is assumed, F itself is compact. So, because F is closed, I am getting the hypothesis. Now, let us try to prove S0 implies S1. This will take a little more time. So, S0 implies S1 means what? S0 means what? Singleton are all components. From that, I have to prove that distinct points can be separated. So, let X satisfies S0. Fix a point P inside X, look at all the points Mp, all the points Q, this Mp is all the points Q which cannot be separated from P. What we want to prove? We want to prove that every point other than P can be separated. Therefore, what we have finally proved that this Mp is only singleton P. So, there are steps to prove that. So, how do we are going that? First of all, P itself is in Mp. So, it is enough to prove that Mp is connected because the only connected subsets of S0 are singletons. If there are more than two points in any set, it is disconnected. That is what S0 means. So, what I will prove Mp is connected. So, it is singleton and then the proof is over. So, let us try to prove that Mp is connected. The first step is to prove Mp is closed. This is the strength thing. So, why do we need to prove that something is closed if it is a connected set. So, let Q be a complement of Mp. Q is in a complement of Mp. By the very definition, what does this mean? All the points here are points which cannot be separated. It is simple as that there is a separation. X equal to A bar B, P inside A, Q inside B. That is the meaning of a point is not in Mp. But then every point inside B, remember this means just B is an open subset and contains Q. Every point inside B is also separated from P by the very definition. It just means that B is contained as a Mp closure, Mp complement. So, for each point inside Mp complement, we have formed an open subset contained inside that one. That means this is an open subset by itself. It just means that it is a complement which is Mp is a closed subset. So, we have proved that Mp is closed. Now we can complete the proof that Mp is connected. Suppose this is not connected, then we will arrive at a contradiction. Not being connected means there is a separation of Mp itself. It is not separation in the whole space. It is separation of Mp because it is not connected means there are two points which cannot be separated. So, it is a non-empty separation if that is all. So, I can assume P is either inside C or inside D. So, let us assume P is inside C by changing the notation if you like. We shall show that points of D are separated from P in X itself. See, this is separation inside the subspace Mp. In general, it does not imply that these two points can be separated in X itself. So, that is what we want to prove that. That is the hardest part here itself. So, we want to prove that. Once we prove that, there is a contradiction because these points of Mpr what they are points which cannot be separated. So, now we are proving that there is a point here which can be separated. That is the contradiction. So, how do we do that? C and D are closed subsets of Mpr. So, they are closed in X also. Here where we have used that Mp is closed. The passage from the subset to the larger subset. So, you have got disjoint closed subsets inside X. Now we use X is compact after therefore it is normal. So, therefore there exist open subsets U and V. C contained in U, D contained in V, U intersection V is empty. Disjoint open subsets. That is normality. In particular, it implies that U being open, U intersection D is empty. So, U bar intersection D will be empty. Since C is inside U, it follows that boundary of U intersection C is empty. Because for any subset, every boundary consists of boundary of U bar minus U because U bar minus interior of U but U is open already. So, boundary of U intersection C is empty because C is inside U. Therefore boundary of U intersection Mp which is Mp is C union D. So, take boundary of U intersection C and boundary of U intersection D both of them are empty, union is empty. So, this means that for every point of this, let me call it as F, this boundary of U, these points are separated from P because it does not intersect this way. So, they are not inside Mp. Boundary of U intersection Mp is empty. What is empty? Everything in Mp cannot be separated. So, these points are separated from P. Boundary of U is what? It is a compact subset because it is a closed subset of a compact subset. So, now again I am using this lemma with which we started. It follows that it is a closed subset W such that this F is contained inside W and P is not in W. F is boundary of U. So, let me repeat what the kind of things we have done here. So, here is a picture. We started with the separation C here, the square, another square here D. So, this is a separation of Mp. After that, using the normality, we found out U and V which are disjoint open subsets containing C and D respectively. Then look at the boundary of U. So, that is this F that does not intersect C nor D. That is what we proved. Now, using the lemma, I can pattern this F to an open subset W. So, this shaded part that is W such that W is a neighborhood of F and it does not contain this P. So, this much we have done. So, that is what the F is contained inside W and P is not in W. So, how does this help? Now, you can complete the proof as follows. Look at G which is U minus W. So, this is U here and U minus W. You have to throw away all the shaded part. So, what you get is this G unshaded part. Clearly, that contains this entire of C. So, this G is U minus W is P U minus W which is G and this G, D intersection G which is D intersection U that is empty. G is a subset of this U and U intersection D is emptied by choice of U and V. So, G intersection is also empty. Since U is open and W is closed. So, G is also open. On the other hand, boundary of U which is F is contained inside W. So, we have G equal to U minus W. This is by definition same thing as U bar minus W because the boundary is taken care here. Boundary points are all contained inside W. I am subtracting that part. So, G is U minus W itself. You can write that U bar minus W bar and just a real expression. This used now, this is closed and this is open. Therefore, this G is also closed. So, we have found a cloak on subset. It shows that points of D are separated from T. We have got a cloak on subset containing P and D is outside. But D is inside MP. We started with a non-trivial separation. That is the contradiction. So, what we have now? Compact plus half star. All the four axioms are equivalent. So, that is one of the reasons why we studied them. However, life is not that easy. We want to study them slightly larger things. Non-compact things are very much needed. So, that is why all these differences have to come there. Most probably, we can assume Hausdorfs' actually finally in the dimension theory that we are going to develop, we are putting matrix base also no problem. So, but we do not want to put compactness. So, that is why all these. So, we will continue to study of these things a little bit now which will be useful later on in dimension theory. Take a T4 space which is written as countable union of closed sets. Each of them is satisfying S3. Then X also satisfies S3. So, let us prove this one. K and L be two disjoint closed sets inside X. K equal to G0. L equal to H0. I am starting a process here, inductive process. Inductively, we shall construct sequences of open sets GI and HI of X. GI minus 1 bar is inside GI. HI minus 1 bar is inside HI. Each CI is contained in the union of GI and HI. GI bar intersection HI bar is empty. For each I will do that. The starting thing is K equal to G0 and L equal to H0. What are K and L? They have been given to be disjoint closed sets. So, there you do not have to verify. You have to verify only the last area. So, we can then take, once we have such a sequence, we can then take G equal to union of HI. H equal to union of HI. This G will contain all the CI. This H will contain all the HI. Therefore, G union H will contain the whole of X. So, starting with K which is G0, that is also contained inside G because G is union of all the GI. Similarly, at least L is contained inside X. Finally, I want also to show that G bar intersecting H bar is empty. Once we have this one, we have this sequence, we would have proved something. So, apply S3 to G0 intersection G1 and H0 intersection C1. This we do now, both of them are subsets of C1. C1 satisfies what? S3. All CI is satisfying S3. So, I get a separation A1 union B1 equal to C1, separation of C1. A1 intersection B1 is empty and A1 will contain the first set and B1 will contain the second set H0 intersection C1. So, successively we are applying S3, S1, S2. So, now observe that A1 and B1 are closed inside X as well as closed inside X also because why? It started with A1 B1 closed inside C1. C1 is closed. So, that is also hypothesis here. Each CI is closed inside that X, union of countably many closed subset. So, they are closed subset and G0 union A1, H0 union B1 are disjoint closed subset. G0, H0 were closed anyway. Therefore, I can apply normality T4 as I have assumed for X. Now, I can pattern these things. There exist disjoint open subset G1, H1 such that this G0 union A1 is contained inside G1. H0 union B1 is contained inside H1. G1 bar intersection H1 bar is empty. Obviously, C1 which is contained inside A1 union B1 is contained inside G1 union H1. So, the construction of this sequence for i equal to 1 is over. From i equal to 0 to i equal to 1 we have done. You repeat this step with now you know depressed not by 1. The same thing you repeat for next one. You will get A2 B2 G2 H2 and so on. So, from i minus 1 to i plus 1 you can go. So, that we have G i bar intersection C i plus 1, H i bar intersection C i plus 1 inside C i plus 1. You apply the same thing to get G i plus 1 and H i plus 1. So, once you have done this you get this statement and that completes the proof. I would like to sum up a number of implications and non-implications which we have proved. Just this could be like a ready made reference you know ready recon for you. So, this is the picture. In this picture the solid arrows like this like this they indicate implies S1 implies S0 like that which is always true. So, I have put a solid arrow. A broken arrow by N here this arrow is not solid arrow broken and put N. N to indicate does not imply S0 does not imply S1. What is the example our Koratowski, Kraster Koratowski example is minus the apex point is K0. So, I have put it in a bracket to remind you what gives you this example. So, like this we have under T1. So, I have indicated with T1 S2 implies S1 under again T1 S3 implies S2 right under compact and half door just now we proved S0 implies S3. In fact all of them are equivalent anyway. Once you have got here you can come back like this. So, all of them are equivalent. So, I have shown one arrow that is enough then you can keep going. But we have also elementary example that if you take S3 may not imply S0 in general. What is the general space? The Sierpinski space remember Sierpinski space consisting of just two points. One point is closed other point is not a closed set. One point is open other point is not an open set. There are no disjoint closed sets. No disjoint non-empty closed sets to be present. Therefore S3 is automatically satisfied. But at different point you cannot separate them. It is connected space. So, disimplication is not true. And also you have proved that under Lindelofness S2 implies S3. If you have more things you can accommodate you are welcome but I think this is enough. So, by the way this set of all points is all coordinate rational inside the L2 space little L2 space. So, that came in an example of an S1 which is not S2. So, this also we have proved. You can go and see where they are there. So, this is roughly a thing that may be many other example many other things like I do not know in general whether S3 you know without T1 can you have S3 imply does not imply S2. Without Lindelof example of these all those things we have not discussed. So, lack of time also lack of interest also. That does not mean that we have completed the whole thing. So, that is not the whole idea. This is just an introductory course not meant to be comprehensive. So, next time we will start the theory of you know topological dimension theory is genuine. So, all these were more or less like a background preparation. Thank you.