 Welcome to module 16 of point set topology course. Today, we will discuss once again about interiors, derived sets and so on, just like whatever we have done for closures, etcetera last time. Start with the topological space X, A and B be any subsets, then the following three statements are true about the derived set. A contained inside B implies L A is contained inside L B, the derived set of A is contained inside derived set of A. The derived set of the union is union of the derived sets. The derived set of the interaction is context contained in the intersection of the derived sets. L of A intersection B is contained in The proofs are all straight forward, more or less similar to the corresponding statements for the closures. So I will leave the proofs to you to write down as an easy exercise. There is one more remark, just like in the case of closures, if you take infinite union, then only containment will be there, equality in this number 2 will be replaced by containment. The derived set of an infinite union of AIs is contained in the union of the derived sets of AIs, okay, so that can also be done. So now let us go through another important auxiliary result about, of course this again in terms of closures and so on, but this time it is about the so-called nowhere dense sets. Later on this we will use in proving some major theorem about metric spaces, right now this is in a general topological space. This theorem gives you five equivalent definitions of when is the meaning of that set is nowhere dense. The first statement I have included as 0 here, so 0, 1, 2, 3, 4, so there are five statements here. Start it is of set A, X minus A bar is dense in X, of course A bar being closed always, X minus A bar is open that part is easy. So the first statement here is X minus A bar is dense in X. The second statement is interior of A bar is empty, this was the statement for A being no where dense, the closure of A should have interior empty. Third statement is A bar does not contain any non-empty open set in X. The fourth one which is third here in A, each non-empty open set X as a non-empty open subset disjoint from A bar, A bar A is given, so this all statement about A bar or A, each non-empty open set in X contains a non-empty open set disjoint from A. So all these things are equivalent but we shall prove it in a systematic way, in an economic way by proving 0 implies 1 implies 2 implies 3 implies 4 and then 4 implies 0, okay, so that is the plan. So 0 implies 1, the 0 statement is what? X minus A bar is open and dense in X, okay that is the first 0 statement, look at interior of A bar, okay, interior of any set is an open set, so interior of A bar is an open set and it is containing A bar, therefore it is disjoint from X minus A bar itself but then X minus A bar will not intersect that open subset, okay, that means it will not be dense unless that open subset itself is empty. So X minus A bar is given to be dense, follows that interior of A bar which is disjoint from X minus A bar must be empty, so interior of A bar itself is empty, okay, now assume the statement 1 let us prove 2, okay, so every open subset of A bar is a subset of interior of A bar, right, so interior of A bar is definitely non-empty, interior of A bar, so first statement we have proved that statement 1 is interior of A bar is empty, A bar does not contain any non-empty open subset in X, each water out show, if it is a non-empty open subset it will be inside interior of A bar but it is empty, so that is proof 2, okay, 2 implies 3, similar if it is a non-empty open set in X, look at G1, it is equal to G intersection X minus A bar, this is an open set, it is an open set, this will be another open set, it is an open subset of G, since G is not contained inside A bar, nothing is contained inside A bar, open subset are empty, right, so G1 must be non-empty because if it is contained inside A bar then this would have been empty, non-empty there is something here, okay, so G1 is non-empty, now 3 follows because G1 intersection A is contained inside G1 intersection A bar, okay, G1 intersection A, whatever it is, is contained inside A bar, so that is, that implies the fourth statement here, 3 implies 4, 4 statement is each non-empty open subset of A contains a non-empty open subset disjoint from A, something is disjoint from A bar and disjoint from A also, okay, finally assuming this statement I won't show that X minus A bar is open and dense in X bar, open part is obvious, I will show that X minus A bar is dense in X, that means take any open subset it must intersects X minus A bar, okay, something intersects X minus A bar if it is not contained in A bar, okay, take any open subset it will, each non-empty open subset X contains a non-empty, there is another non-empty open subset that disjoint from A so that portion will not be contained inside A, so it will not be contained inside A bar either, okay, so these terminologies are just, you know, tautological one, but what is the point of doing this one? Suppose you want to deal with a non-arid ensign, then at a particular place you may be using this property, this property, this property, any one of them you can use and sometimes using just this much is easier whereas using this one will be easier at some places and so on, so this is this fifth one, fourth one which looks somewhat tedious one, okay, this is what is going to be applied when soon we are going to use it in proving Bayer's category theorem, so we come to the theorem in the metric spaces now, actually a metric space a subset A of X is nowhere then in X if and only if each non-empty open set in X contains the closure of an open disk disjoint from A, now because we are working in a metric space now we can talk about disks and disclosures and so on, okay, just now what we saw is if A is nowhere then set every non-empty open set contains another non-empty open set which is disjoint from A, every non-empty open set in a metric space contains a closed ball of some positive radius, any ball of positive radius, okay, it is the closure of the open ball, right, so that is what we get now, so start with a non-empty open set, okay, which is disjoint from A, inside that open set take a ball, okay, such that even the closure is also contained inside that, that you can do because once you start with an open set, open ball inside that you can take smaller and smaller closure balls, okay, so this comes very easily but this is what we are going to use later on, so I repeat this one by statement 4 of the previous proposition, G is a non-empty open set in X, it contains a non-empty open set G1 disjoint from A, choose an X in G1 and R positive such that Br of X is contained inside G1, then Br by 2 of X is contained in Br of X, not only that, its closure is also contained in Br of X and this Br of X does not intersect, so you just remember that interior of A bar being empty is the definition of non-empty open set because of the above theorem, any one of the five conditions can be taken to play the role of the same thing, which is what I have already remarked, okay. Now coming to the metric spaces, we have a few remarks to do about sequences, okay, start with a metric space, okay, take a subset X, okay, take any point, this X will be inside the closure of A, if filled only if you can find a sequence X in a point inside A such that the sequence converges to X, okay. The second statement is the point is in the derived set of A is a limit point, if filled only if there exists a sequence X in of distinct points in A such that X and converges to X, okay. The difference between these two is that in the first part you can take some constant sequence of X and X and X and X and converging to X, so X and X and X and X, so that is not allowed here, if X, X, X, X is a sequence that is always converges to A, that does not mean that X is inside LA, it may be just inside A, that is all, okay. So LA is cluster points have special property, so that is this one, there must be distinct sequence of distinct points such that the sequence converges to X, okay. So let us see the proof of this one, start with a point in the closure, one way up, okay. For each A in inside A, we know that if you take B1 by an X, the ball, open ball of A is 1 by N, intersection A must be on empty because X is in the closure. Therefore, you can pick up a point X and inside this sequence, inside this BN intersection A, so this will be a sequence in A, but the distance between X and X will be smaller and smaller, 1 by N, right. As n tends to infinity, this will converge to 0, so X and converges to X. The converse that if there is a sequence converging to X and the sequence is in A, then it is inside A bar, right, this we have seen several times. Now the second part, suppose X is in the, it is a cluster point, it is a limit point. We know that Br of X minus X intersection A is non-empty, for all R positive. Start with R, no R equal to R, not equal to 1. Choose X1 in Br not of X, not equal to minus X, not equal to X. Take a point X1, not equal to X, but inside A, okay. Look at the distance R1, namely distance between X1 and X divided by 2, R1 is the distance divided by 2. Inductively having chosen Xn in Br N minus 1 X minus X intersection A, as soon as choose Xn, look at this number Rn equal to dxn to X by 2, okay. Use this Rn to choose the next xn plus 1 and so on. Okay, once you choose X, now X1 is like this, X2 will be chosen inside Br1 X minus X. So, the distance between Xn plus 1 and X, okay, goes on, you know, each time the distance is less than, see R1 is something whatever, R is something R0, R1 is distance will be less than R0 by 2. Next will be R1, R2 will be R1 by 2 and so on. So, Rn will be less than R0 divided by 2 over n. So, that will come to zero, okay. Why this decision? Because look at this one, X1 is somewhere but X2, X2 will be inside, okay, inside this distance. So, distance between X and X1 is R1. Distance between X and X2 will be R1 by 2 or smaller. Therefore, they can't be equal. So, next one which we choose, its distance between X and Xn is smaller than all the earlier distances, okay. Did you say, you know, this is a distinct sequence of distinct points, that is why, okay. So, you can write down the converse of this phenomenon, it is very easy anyway. So, let us go to some functions between, you know, metric spaces, algebraic spaces, first of all and put them in the proper perspective now. Start with f from X tau to Y tau prime, set the function, okay. Then f is continuous, it is one statement. The second statement is, for every subset B of Y, which is closed in Y, inverse, f inverse of B is closed in X. Remember, what was the definition of continuity? For every open subset U of Y, f inverse of U is open. That was the statement for continuity of this one, that is A. So, B is open sets are replaced by closed sets. The third one is even much better. For every subset A of X, f of the closure of A is contained closure of f A. So, this is a forward statement. A and B were backward from starting from subsets of Y, you get something about conclusion subsets of X. Here it is everywhere wrong, okay. D is also superb in the reverse way. For every subset B of Y, f inverse of interior of B is contained interior of f inverse of B, okay. These are all equivalent, just means that you can use any one of them, by any three of them or something, to defy continuity of a function from any tuploid space to another tuploid space, okay. That is the statement. So, let us look at the proof of this one, which is not all that difficult. First, let us prove A and B are equivalent. This is just by demarcation law. If U is an open subset, X minus U is a closed set. X minus U inverse of A is nothing but, okay, sorry, this is inside Y number I am talking about. Y minus B, f inverse of Y minus B is same thing as X minus f inverse of A. So, closed subset is a closed subset. So, you just use one way A implies B implies A, because you know complement of A closed set is open subset and f inverse space very nicely under complement. Now, let us prove B implies C, okay. Take random for B and now let us prove C. C is what? Start with any subset A, okay. So, what you have to prove? f of the closure is contained in the closure of f A, okay. A is always contained in f inverse of f A, okay. F A is contained in closure of f A, therefore f inverse of that is contained in f inverse of the closure of f A, okay. Now, closure of any set is closed. The condition B says f inverse of that is closed. Therefore, A is contained in this closure, contained in this closed set. Therefore, A bar is contained inside this set, okay, because closure of A is the smallest closed subset containing the set given A. So, this is a larger closed subset. Closure of A is the smallest closed subset of A. So, this one is contained here. But this is same thing as if you put f here, f of closure of A is contained in the closure of f. C implies B is what I want to tell you, okay. So, start with any closed subset of Y. By C, we have f of closure of f inverse of B, you do not know what it is, okay. Closure of f inverse of B, f of that is by definition contained inside by condition C, closure of f of f inverse of B, which is nothing but contained in B. So, this contains closure of B. This one is contained closure of B, okay. Now, B is closed. Therefore, this is B, closure of B is B. So, f of the closure of f inverse is contained inside B, okay. This means that this closure of f B is contained in f inverse of B, okay. But then equality holds, okay, because f inverse of B is smaller than the closure. So, they are equal. But then equality holds, okay. And hence, f inverse of B itself is closed because this is a closure of that, okay. The proof of the last statement is almost similar against something like De Morgan law you have to prove. So, I will leave that as an assignment to you. Namely, f inverse of interior of B is contained in interior of F B, okay. It should not take more than two lines. You should write down, okay. So, here are some exercises A equal to 1 by n and 10 equal to 0. Take this subset 1, 1 by 2, 1 by 3, 1 by 4, etc. They include 0 also under the usual topology of compute A bar interior of A, boundary of A and L of A, okay. So, this is an easy exercise. Similarly, there are some more exercises here which you can take. Now, there is also this keratosis closure axioms, form of four axioms for interior operator, obtained topology associated trade. And through that, the operator coincides with the usual operation of taking interior, similar to keratosis closure axioms, okay. So, let us stop here today. Next time, we will study more examples. Thank you.