 Welcome to module 19 of point set topology part 1. Last time we had announced that we will do three important theorems in metric space theory. So, two of them we have already done last time. The third one is Bayer's category theorem. Before we even state this theorem, it is convenient to make a few definitions. Of course, one can make the statement without these definitions and so on. That is not a logical necessity. The definitions only make help in reducing the number of words we use ultimately. So, let X be a topological space and A be a subset. So, these notions I am introducing inside any topological space, not necessarily metric space. Remember that A is called an F sigma set in X if it is the union of countably many closed subsets of X. This definition is cooked up just to take care of these kind of sets because they are not closed. You know generally an infinite union of closed sets is not closed. But however in this theory what happens is countable union of closed sets becomes very important. But you cannot call them closed sets so we have put a name for it. Usually F is used for closed sets at least in German topology in those days. F was used for closed set. So, the sigma is for like countability you know like summation countable. So, F sigma stands for countable union of closed subsets. Any set which can be written as countable union of closed sets will be called F sigma. Similar to this one and dual to that is A is called a G delta set if it is the intersection of countably many open sets. So, that is like De Morgan law. So, the explanation for G delta is same thing. G was used for open subsets and delta for intersections. So, G delta is for countably many open subsets and then take the intersection. Let us continue with these definitions. If A is a countable union of no-arid and subsets. Now let us say stronger over here not just closed subsets no-arid and subsets then A is called a first category set or belonging to first category or just say first category. A set is set to first category if it is a countable union of no-arid and subsets of X. So, everything is happening in the ambient space X. If change X the nature may change. Such a set is also called a meager set. If A is not a first category then it is called second category just to make distinction between these two things that is all. Such a set is also called non-meager because it is not meager that is all. Now there is one more terminology here people are using this one. If a topological space itself is a second category you call it a bare space. After the mathematician bare. Now let us state bare category theorem which becomes very simple because of this terminology. Every complete metric space is of second category. What is the meaning of this? It is not first category. What is the meaning of that? It is not the union of countable union of no-arid and subsets of X. The only condition is that X is a complete metric space now. So, statement becomes very easy that is the whole idea. Now I have put this as BCT bare category theorem 1 because there are several versions of this one. So, I have picked up one of them simplest very simplest in terms of these definitions. So, this is every complete metric space is second category. Before going into the proof let us examine a couple of examples. Of course, every open set is G delta. There are many G delta sets which are not open. In some sense they are the next best things to open sets. See in topology we always keep studying open sets. But in a metric space G delta sets also become important. In a metric space especially every single set is a G delta set. You can write it as intersection of balls of radius 1 by N with centre X as N raise over all positivity here. In particular, every single set in Rn is also a G delta. That is precisely what we have done. In every metric space it is true. Similarly, every closed interval is also a G delta set. Because you can write it as say A to B is a closed interval. You can take A minus 1 by N and B plus 1 by N and then take the intersection. So, many interesting results on continuous real-life functions follow from this observation. Namely, let X be any topological space and F from X to R be a continuous function. Then for every R in R, the set of all points X belong to X such that f X is equal to R. This is a G delta set. Why? Because singleton R is a G delta set. You can take the inverse image of all those comfortably many open sets. They will be open. When you take the intersection, it will be intersection singleton R. You understand what is going on here? Singleton R can be written as intersection B 1 by N R. You take the inverse image of that. They are say G n's here. Intersection of that will be precisely R. So, take any continuous function into R. So, inverse image of open set is open. Inverse image of G delta is G delta. That is what I am trying to say here, more generally. So, there are some other more interesting examples. Any countable set in a metric space is f sigma by taking duality. Because each singleton is a closed set. In this countable union, each interval in R is both f sigma as well as G delta. So, without G delta, similarly we can take f sigma also. But now comes the set of irrational numbers is not f sigma. So, I have given you examples but there are not everything is f sigma or G delta. The set of irrational numbers is not f sigma inside R. For suppose it is like this, namely R minus Q is the irrational number. Suppose you write it as union of countable union of closed sets. So, that is the meaning of this is f sigma. Then, being subsets of irrational numbers, we know that each f1 will be no error dense. As observed in example 1.89, let me just show you this example which you have done fourth one error. f is a closed subset of R and it is contained in either Q or R minus Q. Then it is no error dense set. We have seen this one for f. It is a closed subset. So, f bar is f. Interior of this one is same thing as interior f bar but that is simply because f contains no intervals. Any subset of irrationals contains no intervals. So, this is what we had seen earlier. So, I am just recalling it. It is important example here. So, if you write it as union of f ns, first of all each f n is closed is assumption then each f n becomes no error dense. But then you can put more another set of countable sets namely all singletons. Singletons are anyway no error dense. They also do not contain any intervals. But now R is inside Q. So, this is a countable union. So, whole thing is a countable union of closed sets and the no error dense sets. So, that shows that this R is written as union of no error dense sets. It means R is first category in our definition. But R is a complete metric space. So, Bayer's theorem just said to you that every complete metric space is second category. So, I have given you an application, you know very simple one application of Bayer's theorem to show that irrational numbers cannot be written as a countable union of closed sets. Of course, rational numbers you can write. There is no, there is no contradiction, there is no harm. I will give you one more example. These are the easy things which are consequences of Bayer's theorem. But that is not the end of Bayer's theorem, we will see. So, let us take another example here. Take a polynomial in n variables. Of course, if you take zero polynomial, it is not interesting. It is a non-constant polynomial, let us take. Look at all the zeros of that. Zp is the zero set of p. All x1, x2, xn. When you evaluate p on that, p of x is zero. That is a closed set because p is continuous. We claim that Zp contains no non-empty open set. So, this is elementary calculus. As soon as there is an open set of Rn or Cn, it does not matter. Contained inside a set, you can study the polynomial on that open set, which is identically zero by definition. We have contained Zp, but a zero function on an open set has all its partial derivatives zero. If you compute partial derivatives cleverly, you can compute all the coefficients of this polynomial. Not only in one variable, in any variable, any number of variables. You have to do all the partial derivative, various partial derivatives. That means what? All the coefficients are zero. That means p itself is the zero polynomial, but we started with a non-constant polynomial. So, the zero set of any polynomial is no air dense. What is the consequence? The entire Rn or Cn cannot be the union of countably many zero sets of polynomials. So, that is the consequence by Bayer's theorem. The zero set of any non-zero polynomial is no air dense. It follows that An, K could be R or C, it does not matter. It has a union of countable many zero sets of non-zero polynomials. Cannot be done. So, let us prove this theorem now. We shall actually prove stronger version of the theorem. The stronger version I have put at a BCT zero, because it is, it sits over all of them. Take a metric space which is complete. Take a sequence of, you know, countable family of no air dense subsets. The complement is also dense. That is the statement. X minus union of An is dense in N. In particular it is non-empty. An empty set cannot be dense because closure of an empty set is empty. Okay? That it is non-empty, same thing as BCT one that we have seen. Because if it were empty, then X would have been union of An. That would mean that X is first category. Right? But the statement is X is second category. So, BCT zero implies BCT one very easily. So we are going to prove the stronger statement. Alright? Once again we have done the groundwork already. So this is in 1.107 it says. Let us take a look at this one. Okay? Has it come there? Yeah. Xt is a metric space. A subset A of X is no air dense if and only if each non-empty open set in X contains the closure of an open disk with disjoint from A. So there will be some BRX zero. Closure of that intersection with A will be empty. Okay? And this will be contained inside any given non-empty open set. So this theorem I am going to use again and again. Okay? So I have to go back now. Yeah. I had, I had statement here but I wanted to show that we have actually done, you might have forgotten it. A is a no air dense subset of X. Then every open set in X contains the closure of an open ball disjoint from A. So this is the statement I am going to use again and again. Okay? So this is precisely the kind of thing that we are going to be. This is dot dot dot dot dot is the no air dense set. This single line thin line indicates an open set. Inside that I can find a ball of some positive radius such that the closure is disjoint from all these points dot dot dot. Okay? A no air dense set compared with an open set. Let B1 be any open ball of positive radius you start with. Instead of any open set if I prove this one for B1 which is an open ball then it will be proved for every open set. I will produce something inside B1. So starting with an open set I can start with B1 instead. Right? That is why B1 means open ball of radius. We need to show that B1 is not contained in the union of ANs. That is enough. Okay? If B1 in open ball not contained in the union of AN that is all I am going to show. By first of all B1 contains an open ball C1 such that the closure of C1 intersection A1 is empty because A1 is no air dense. So this is the first time I am applying this theorem. Okay? 107. Now I have got C1. Let B2 be an open ball inside C1 of radius r by 2. Okay? I am making sure that the radius are going down, down, down to 0 by putting r by 2 here. Okay? B2 is contained inside some C1. Okay? C1 is a ball but I want a ball of radius smaller than this r not just smaller r by 2 or by 3 whatever. Okay? It contains again now I apply, I apply the theorem contains an open ball C2 such that C2 bar intersection A2 is empty. Now you know the game inductively suppose you have chosen Cn of radius less than r by n such that Cn is contained Cn minus 1 and Cn bar intersection n is empty. Once you have that inside this Cn you will get another one and so on. So you keep continuing this. Now we apply Cantor's intersection theorem to what? To these Cn's. C equal to intersection of Cn bars. So these are all closed subsets. Their diameters are what? Twice this one say rn, r by n. Right? So twice that was a diameter. As it tends to infinity to go to 0. So this intersection is actually a single point. It is non-empty. All that I mean is non-empty. Okay? Since I want to apply Cantor's theorem I have put r by 2, r by 2, r by n and so on r by n here. So this will go to 0. Therefore C is intersection of Cn is singleton but I want only non-empty. But they are all contained in V1. Okay? So this point is inside V1. But what is this point? This point C intersection an is empty. Okay? Why? Why? Cn. It is contained in Cn bar and Cn bar intersection an is empty. So C intersection an is empty for all n. Right? So it is in none of the an's. This point which is single point actually is not belonging to any of the an's. So I have found a point which is not in n which means V1 is not contained in that an. See if you want to prove that something is dense what you have to do? Take any open set it should not intersect the set. Intersecting the set means what this set is what? X minus union of an's. That means it should not be contained inside the union of an. It is a complement. Right? That is what I have proved. Take any open set. Okay? It is not contained inside union means what? Any open set it contains is one. It is not contained union means what? The complement intersects V1. Therefore the complement is dense. Okay? Be sure that we have proved this. Non-empty is not what we have proved. We have proved that X minus an union of an is actually dense. So that is the proof of this. Okay? So let me make a few remarks here. This theorem is very useful in function theory when one has to prove the existence of various types of functions. So this theorem something is non-empty. It is the way it is used. Okay? If you write a complete matrix space you cook up a complete matrix space. Then you cook up a sequence of close subsets which are no inheritance. They will not cover means there is something left out. So that is the existence theorem. So that is the way it is proved used in many existence theorems. Okay? Indeed proofs of several fundamental results in functional analysis use this theorem. I will quote some of them which are very very fundamental namely closed graph theorem, open mapping theorem and boundedness principle and so on so forth. So all these things come in elementary functional analysis itself. The first course in functional analysis is all these theorems. Okay? They are all using Bayer's category theorem to prove. All right? So I make this mean this thing I have already done so I will repeat this one. Above theorem is a negative tone. That means second category itself definition is it is not first category. The first category is what it cannot be. So there are too many negations there but it can be put in slightly a positive tone as follows. So I have given you those two. And then often it is how this positive it is used that that versions are used here. So let us have those versions later on in part two we shall prove a version of this Bayer's category theorem for locally compact horse-dobb spaces. There is nothing to do with metrizability. There is no matrix. Okay? So here are versions here that X be a complete matrix way. That is standard assumption. There is no other. The BC 2 says that suppose X is union of A n written as a countable union A n then the closure of A n means at least one of A n's has non-empty interior. See the hypothesis on A n's is deleted. This is just a countable union. But when you conclude I am taking closures. Right? Instead of taking closed subsets and non-empty subsets. One of them has non-empty means what? One of them is A n bar. Okay? Non-empty means what? A n is no evidence. Okay? So it is just the other way around here. So here there is no negation here. If you write like this one of them is non-empty interior. It is a positive term. Another one is intersection of a countable family of open densets is non-empty. So this is the way it will be used. So there is one element they want. So that is the existence theorem. The proof of that one implies two implies three and they are equivalent. Okay? This is very easy for you to prove. But zero is a stronger statement which will imply all of them. Okay? So here is exercise namely write down versions of BC to zero also similar to BC to two and BC to three. Okay? Take a complete metric space. Does BC to one imply that union of countable many nowhere dense subsets of X is no evidence? Pay attention to the statement there. It does not say this. Union of countable many nowhere dense sets does not fill up the whole space. This is the weaker version. Compliment is actually dense is the stronger version. But nowhere to set that the union itself is no evidence. Okay? So you have to see whether this is true. Okay? It is not stated doesn't mean that it is not implied. So I am asking whether it is whether this is implied in the from the statement. Think about them. So that is that is the exercise you have to think about and that is all. So let us stop here. Thank you.