 Welcome to module 51 of point set topology part 1. So, as promised last time, let us now start the study of regular spaces and normal spaces. As before in the case of Frechier and Hausdorff, I will first state a theorem which gives you a number of equivalent conditions and then make the definition that a space which satisfies any one of the conditions will be called regular or normal and so on. That is the general plan. So, this theorem says the following conditions and topological spaces are equal. What are the three conditions? Open a closed set F and a point X away from F in the complement. There exists disjoint open sets UV in X such that X is inside U and F is contained inside D. Just like Hausdorff space, any two points are separated here point and a closed set are separated by open sets. So, you can see that this is a one step generalization of regularity. Next, for all X belonging to X and open set G such that X belongs to G, there exists an open set H such that X is in H is contained inside H bar contained inside G. So, in other words you can say that every neighborhood of a point contains a closed neighborhood. Remember a neighborhood should be such that there is an open subset contained inside that point containing the point. So, this H bar is a closed neighborhood. G was an arbitrary neighborhood. It need not be open. Actually here I should open set G that is enough because once it is a neighborhood this G can be replaced by subset which is open. That is all. The third condition is given a closed set F inside F and a point Y in outside C similar to the conditions in one. There exists open sets U and V in X such that Y is in U and F is inside V. All this is same as number one, but the last part is here it is only disjoint. Here U bar intersection V bar is empty. So, third condition is much stronger. Apparently, but the claim of the theorem is that all the three are equivalent to each other. So, let us prove one implies two. All that I do is take F equal to G complement. G was open, G complement is closed, X was in G. So, X is in now F complement. Now, apply one to get open sets U and V such that X is inside U and F is inside V and U and V are disjoint. Now, take H equal to U. You want to get something here. Since F is inside V, V open. U intersection V is empty. It follows that F intersection U bar is empty. Actually F is inside H bar is empty. This implies H bar is contained inside F complement and F complement is G. So, this is F intersection H bar is empty. 2 implies 3, take G equal to F complement. We get Y inside H contain H bar contain inside G. This is this is the property 2 where H is an open set. Now, applying 2 for X belong to H bar C, H bar is closed. H bar C is open. And X is inside that. So, what we get? We obtain an open set in H X open set say H X such that X belongs to H F contains H X bar contains H bar C. These H X are different from H. Now, take V equal to union of all these H X where X belongs to F. That to be is an open set containing F. Clearly, H X intersects H is empty for all for all X. So, V intersection H should be empty. So, V intersection H is empty. Hence, Y is inside V bar closure. Applying 2 again, we get an open set U such that Y is in U contains a U bar contained inside V bar closure. This is a property U whenever you have an open set ok. Then Y belongs to U contained inside U bar. There is such a thing. So, V bar is closed. Compliment will be open. So, you have to apply 2 at least 3 times here for the last one. For each point here you have applied 2 here. Once you have applied 2 here also ok. This means that U bar intersection V bar is empty. So, the stronger condition is obtained ok. And 3 implies 1 is obvious. So, 1 is taken as definition usually though 2 and 3 are equivalent. You can use whichever one you like because that is the easiest to verify. When you want to test whether a given topological space is regular or not, you want to test the simplest thing, easiest thing. So, for that matter I will take 1 as a definition namely given a closed set and a point outside it there are disjoint open subsets containing them X and F respectively which is similar to Hofdorfness. Only thing is instead of Y I have a closed set F ok. There is no condition on X. X could be any point not enough that is all ok. Any space which satisfies any of the conditions above and hence all the conditions above in this theorem that will be called a regular space alright. Now, I will introduce a normality. The following conditions on a topological space are equivalent. Again there are 3 of them almost parallel to 1, 2, 3 of the previous theorem. But the difference is now given disjoint closed subsets F1 and F2 there exists disjoint open sets U and U2 such that F1 is contained inside U1 and F2 is contained inside U2. So, you know immediate distance immediate difference instead of a pointed closed set I have taken 2 different closed subsets disjoint closed subsets. So, that is the difference ok. Then 2 and 3 imitate just like the 2 and 3 of previous theorem given F inside G where F is closed and G is open. There exist an open set U in between them such that F is contained inside U contained inside U bar contained inside G ok. So, this is the condition 2. Given disjoint closed sets F1 and F2 there exists open sets U1 and U2 such that FIs are contained inside UI and U1 bar intersection U2 bar is empty. So, this is similar to property 3 in the previous definition. Again the proofs are also similar. So, let us go through the proofs again that these 3 conditions are equivalent. So, how to get 1 implies 2 F as F1 and F2 as G complement you are given F and G are given G is open F is contained inside G right. Take G complement that will be a closed subset disjoint from F2. Apply 1 to get U1 and U2 ok. Then take U equal to U1 Gc which is the complement of G is F2 contained inside U2 implies U1 intersection U2 is empty. Therefore, U1 bar which is U bar is contained inside G ok. So, that is 1 implies 2. Similarly, 2 implies 1 this is identical proofs. First obtain an open set U such that F1 is inside U contained is a U bar contained is F2 complement. F2 complement is open and F1 left to disjoint means F1 is contained is a F2 complement. So, in between I can put a an open set U such that U is U bar contained inside F2 complement. Apply 2 again to this situation namely with U bar U bar contained inside F2c. In between these 2 you can put one more open set. We get an open set U1 such that U bar contained is a U1, contains a U1 bar contained inside F2c. Now, you take the complement of U1 bar, U1 bar is closed. So, U1 bar is closed the complement will be open that take test V. Then this F2 will be contained inside V ok. See here this is contained is a F2 complement. So, when you take the complement this will contain the complement update. So, and it is easily verified that U bar intersection V bar is empty ok. So, proofs are identical to the proofs for the regularity that is why I have put them together in one single place. 3 implies 1 is obvious because 3 is a stronger condition right. Go back here disjoint close of sets are contained in disjoint open sets U1 bar intersection U2 bar itself is empty implies U2 is also empty and that is all we need in one. So, 3 implies 1 is obvious alright. Now, what to do with this regularity and normality ok. So, what is normality? Anything any topological space which satisfies 1, 2, 3 of the above theorem any one of them and hence all of them ok. Normally you can take the first one as a definition, but once you are ready these two wherever you have application you can use any one of them. Proofs that 2 implies 3 in this theorem is much easier than the proof for cursing part of previous theorem because you have to do one point by point and so on and then take the union and so on. So, for normality it was easier actually yeah it is good yes ok. Also observe that normality is apparently stronger than regularity. Apparently why because point and close sets are separated in the regularity. Here any two close sets are separated, but you have to be careful both F1 and F2 must be closed here whereas, there X could be any point and the other one is closed. So, obviously we do not have the hypothesis that singleton point singleton sets are closed ok we do not have that. Therefore, normality may not give you regularity ok. So, we perceive a major difficulty in deriving regularity from normality. The problem is precisely that singleton set need not be a close set. Indeed under this extra hypothesis namely singletons are closed which is a fresh air right. If all the singletons are closed that is called a fresh air space. So, if fresh air if you have then it is easily seen that normality implies regularity. There can be other hypothesis also under which this may hold, but under fresh air we know that normality implies regularity ok. Regularity of course may not imply normality that is too much to expect, but if you say it is not then you have to give a counter example that is the only way. So, you have to give a counter examples counter examples are not all that easy alright. So, we will we will work on that one those things now ok. In the absence of fresh airness we shall later see that normality need not imply regularity which just means that we have to give a counter example of course. It is easily perceived that regularity need not imply normality under even put a fresh air condition it may not. We shall see such an example later. Same problem is there if you try to derive half darkness from regularity because it looks like I started with the comment that regularity is somewhat generalization of half darkness right. Any two points are separated by open sets is half darkness. Here point in the closed set present is separated right. So, why do not you separate two points once again? The second point may not be a closed set right. So, second point is arbitrary. So, all points must be closed that means fresh air is again. If you have fresh air plus regularity automatically it will be half darkness. Now, you see importance of fresh airness. Under fresh airness see regularity implies normality, normality implies regularity and normality and half darkness also and so on. Okay, regularity implies half darkness, normality implies regularity therefore it will imply half darkness also. So, that may be one of the reasons why fresh air even in his definition of topology itself he put that one okay. Now, let us work out these things our habit of checking whether a property is hereditary, co hereditary and so on. Fresh airness, half darkness, regularity are all seem to be hereditary. Hereditary means what? Take a subspace it should have same property okay. So, let us say regularity take a subspace and take a point and a closed set. A closed set in the subspace is what? What is a closed subset in subspace? It is some open subset and some closed subset in the larger one intersected with the original subspace. The point is already in the subspace and not inside the smaller set. So, it is not in the larger one also. Therefore, you can apply the regularity of the larger space to conclude that there are disjoint open subsets as needed. Now, you intersect them with the subspace. So, that will give you disjoint open subsets containing the point and the closed set okay. So, that was the hardest one half darkness and fresh airness you can do it easily okay. So, every subspace of a regular space is a regular space. However, you try to do the same thing for normality you will have problems. Why? Because starting with two closed subsets inside a subspace there are closed subsets in the larger one. The problem is they may not be disjoint. You start with disjoint closed subsets in the subspace. That means what? There are closed subsets in the larger one. When you intersect with the subspace they will give you the original sets. But why the original the new ones inside the larger space should be disjoint. Nobody guarantees you that they are disjoint and in fact that can happen and that way the normality may break down it may may not go down to the subspace okay. In fact, it happens that normality is not hereditary. So, again we have to construct some example for that okay counter example okay. On the other hand where are these things coming from? You know that right namely metric spaces. A metric space is fresh air, off-door, regular and normal also. Very easy to prove. In fact every subspace or metric space metric therefore every subspace of a metric space is also normal whereas subspace of a normal space may not be normal. So, that makes us think about what is going wrong, what is wrong. So, to take a new look at the metric space itself okay. So, let us take a look at metric spaces namely subspaces are also normal why okay that is the aspect. Two subsets A and B of a topological space are said to be mutually separated if A bar intersection B is empty and B bar intersection A is empty. So, this is slightly you know a generalized concept of two closed subsets being disjoint. If A and B are closed subsets then saying they are separated is the same thing as they are disjoint that is all because A bar is A and B bar is again B right. So, there is nothing more than that okay. Instead of saying closed subsets start with any two subsets putting disjointness is weaker than being saying that they are separated. See A intersection B may be empty but A bar intersection B may not be empty. So, this is a stronger condition on arbitrary subsets than saying that they are just disjoint okay. So, this is this definition is made in the light of you know the third condition in normality you will see why no you see a typical example of mutually separated subsets okay are like this 0 cross 1 upon B 1 upon 2 crossed if you take A bar and B bar they are they they are not disjoint they will have common point 1. But you take A and B are not closed also A bar intersection B bar is not yet A and B are mutually separated why A bar is 0 1 closed and 1 is not here similarly B bar is 1 2 and 1 is not here on A in A. So, they are mutually separated okay. So, this is a typical example inside Rn you can construct many many such examples. Now, what is that good for let us see now on a topological space X the following conditions are equivalent given two mutually separated subsets A and B of X there exists disjoint open subsets such that A is contained inside U and B is contained inside B. Now, you see this is definitely a stronger condition than normality why because in normality I started with A and B are closed subsets right then automatically they are mutually separated this condition is satisfied. But if you start with arbitrary subsets A and B they are mutually separated they may not be disjoint they may not be closed you cannot apply normality to get this one A contained inside U B contained inside U. So, this statement is definitely stronger than normality it implies normality I have shown because if start with A and B are closed they are automatically disjoint so you will get this one. So, condition one is stronger than normality but this theorem says that condition one is equivalent to every subspace of X is normal X is normal is fine that is that is the condition one condition one implies normality but more condition one is obviously stronger what it says every subspace of X is normal okay. So, let us prove this one every subspace of X is normal then I have to show this one also okay. So, both ways we have worked to do here. So, let Y be a subspace of X A and B be disjoint closed subsets of Y then I have to produce disjoint open subsets of Y containing A and B respectively that is my M okay. Now, what happens A and B are disjoint closed subsets okay closed inside Y not inside inside X therefore A bar intersection B which is A bar intersection Y intersection B because B is a subset of Y. So, you can put A bar intersection Y intersection B and an associativity of the intersection but A bar intersection Y what is this can you tell me what is this all the closure points of A which are already inside Y therefore they will be inside the closure points of A inside Y therefore they are in the closure of A inside Y but A is closed inside Y so it is A. So, this A intersection B and A intersection B is empty is the starting hypothesis A and they are disjoint closed sets okay similarly B bar intersection A will be also empty exactly similar therefore when you pass on to the larger space they will be separated okay therefore I can apply one to get disjoint open sets U and V in X a set A is inside U and V inside U. Now, you take intersection with Y we are through all right now let us prove the converse. Let A and B be mutually separated subsets of X put Y equal to A bar intersection B bar and the complement of that remember A bar intersection B bar may not be empty if it is empty you are in a good shape there is nothing to bother about so it may not be empty A bar intersection B is empty and B bar intersection A is empty so look at Y equal to A bar into B bar and its complement so that is a subspace then A bar intersection Y and B bar intersection Y they are closed subsets in Y after all A bar is closed inside X itself so intersection Y will be closed inside Y okay and they are disjoint now because A bar intersection Y intersection B bar intersection Y is A bar intersection B bar intersection Y but Y is just the complement of this one right so they are disjoint because our choice of Y there is nothing more than that by the normality of Y this is the hypothesis is into every subspace is normal there are open sets U and V inside Y such that this A is inside you B is inside V okay but what is A and A bar intersection B bar this is a closed set so C is complement so this is an open set of Y if A is if U and V are open inside Y they will be open inside X also you so you don't have to fat on them the same U and V will be open inside X also right therefore U and V are open inside X also A contain inside U B contain inside V so we are arrived at number one all right so the proof of not at all difficult but we have to think of you know the cleverness here instead of arguing with some points here points there and taking get confused okay so you have to think about this what subspace should I take so apply to the right subspace you get the answer very easily following this theorem we can now make a formal definition we have observed that this condition is stronger than normality so we just call it completely normal a space that satisfies one of the conditions and hence both the conditions of the above theorem is called a completely normal space okay so we shall now return to the study of normal spaces maybe this is time now so let us stop here and normal spaces will be taken up next time rigorously thank you