 Hello, welcome to NPTEL NOC course on point set topology part 2. We will begin chapter 4 with certain other notions of compact. Other in the sense we have studied compactness itself. Clothe related was Lindelof property. Then we have local compactness and para-compactness, right? So now we shall study some other notions here. These notions have been selected by just by maybe you can say personal taste. Or rather one's belief that these are the more useful ones in analysis and topology. Central analysis and topology. When we study function spaces which may not be compact, okay? Yet have certain other features of compactness, okay? So that is what we would like to study. Like para-compactness was one of our obsession. We shall also study these properties for metric spaces wherein many of them come together, okay? Then we shall give a criterion for metric spaces to be compact, okay? Lastly, we give a very important application of this topology of function spaces namely what are known as Ascoli's theorems actually. We will do only one of them. So that is the general idea of general plan for chapter 4. So let us now begin with this module 15, start with any topological space. Then I am going to introduce three different notions of compactness here. The first one is countable compact. So I may just denote it by cc in the following section altogether. So countable compactness, if every countable open cover forex admits a finite sub-cover. This is similar to Lindelow, but in contrast, Lindelow property gives you any cover to countable cover. So countable compactness gives you countable cover to finite cover. So you can see that Lindelow plus countable compactness implies compactness, okay? That is one easy consequence of this definition. The second one is limit point contact which has something to do already with our experience with metric spaces namely dealing with sequences, dealing with limit points. So if every infinite subset of x has an accumulation point or what is called a limit point that is called limit point compactness. This is also called Bolgenow-Westras property. Bolgenow-Westras property, partly is also reflected in the next one, but we will call it as sequential compactness and denote it by sc. If every sequence in x admits a subsequence which is convergent. In fact, in standard real analysis, whenever you have this property, every sequence, bounded sequences of sequences convergent and so on you prove, you state it as a Bolgenow-Westras theorem. But this property is not Bolgenow-Westras property. So you have to make it a slight distinction here. Bolgenow-Westras property, if you want to refer it, it is limit point compactness. That is why I have called it limit point compact, not to confuse with the standard Bolgenow-Westras theorem. So countable compactness, somewhat similar to Lindelow, but somewhat dual to that one. Limit point compactness, sequential compact. In a metric space, you may have seen that compactness implies other three properties. These are the standard properties of compact metric spaces. So we have extracted these properties and made it a definition in the general case. In general, this may not be true. Compact spaces may not imply any of these things. We will have to be careful here. Something may be here, some other thing may not be so. So let us go through this carefully here. So let us check these things are fresh, one by one, in the general setup. Start the compact space. It is clearly countable compact. The first part is very nice, because after every open cover, we will admit a finite sub-cover. Countable covers automatically admit a finite sub-cover. So that is compact space implies it is countable compact. Now take the second one, that a be an infinite subset of x. If a has no accumulation point, then it follows that a is a closed discrete subset of x. In the closure of any set a, there are two types of points, those which are points of a as self, otherwise they will be accumulation points. If they are not accumulation points, they will open subsets of x. So these are the three types of points in a set. So if a has no accumulation point, then it follows that a must be a closed discrete subset. In particular, being closed, it is a compacting. I start with a compact space. But in a compact discrete subset, we must be around the finite many points. So this is the contradiction, we started with an infinite set, therefore x must have a limit point. So 1 and 2 are fine, compactness implies 1 as well as compactness implies 2. Maybe 3 is also true, and then you are lucky, you have to be careful, ok. So finally, let x n be any sequence in x, where x is a compact set. If the image of the sequence is a finite set, then there will be a subsequence which takes a constant value and it is convergent. On the other hand, if the image of the sequence is infinite set, we just do not know how to extract a convergent subsequence out of it, ok. And indeed, it happens that compactness does not imply sequential compactness. This may be somewhat surprise for you, so you have to be careful with this property. So let us first understand an example also here, let us verify an example here. Make sure that compactness in general need not imply sequential compactness, ok. So do not confuse it with Rosino-Astras properties all the time, so that is the whole idea, ok. So here is an example, if G2 is standard notation for the space consisting of minus 1 plus 1 in the multiplicity notation or in the additive notation, the group consisting of 0, 1, 0, 1, right, additive group. So for us, it is just a discrete space with two points that solve, we do not care about the group structure here, though I have used the notation, this group with two arguments, ok. So take Z2 cross Z2 cross Z2 for many copies, take p power n, namely this n is the natural number, the power set of it, ok. I have a some, you know, deliberate some intention here, denoting like this, the power set of n is nothing but the cardinality of c, ok, cardinality of r, right, one could have just written c here. So but I want to use this specific description of this cardinality, so this is the power set of n, all subsets of the natural numbers. So you take that many copies of Z2 and take the correlation product, the product space, what is the topology on Z2, discrete, ok. So indexing set is power set of n, ok. So by Ticknoff's theorem, we know that X is compact, ok, product of any sequence of, product of families of compact with compact here, same set is there, same space is there which is a compact, ok. We shall claim that this space is not sequentially compact, ok. So suppose fn is a sequence in X defined as follows, to show that it is not sequentially compact, I have to produce one sequence which has no subsequence which is convergent, right. So I will give you an example, namely take the sequence fn defined as follows. What is fn? fn's are functions from power set of n into Z2, right. So fn of a subset a, what is a, a is a subset of natural number, is equal to 1 if this n, the nth term, the index n belongs to a, ok, otherwise define fn of a equal to 0. For example f1 of a, f1 of say 2, 3, 4, 5 will be 0, f1 of 1, 1, 1, 1, 2, 3, 4 that will be 1 because 1 is there and so on, ok. Now let fnk be any subsequence, ok. Take a to be, after taking the subsequence you have take a to be n1, first one, n3, n5 and so on, just take the odd, you know, the odd numbers here, I mean in the indexing, n1, n3, n2k plus 1, take f of those, so take those numbers first, form in the subset a of its natural numbers. Then the sequence obtained by projecting fnk on the a coordinate, see after all the coordinates of this one, this product are indexed by p power, you know, pn, power set of n, so the 8th coordinate will be what, fn i of a, right, so what is fn, fn1 of a, n1 is there, right. What is fn2 of a, fn2 is not there, n2 is not there, so it will be 0. What about fn, fn3 of a, n3 is there, so it will be 1. So this sequence is alternating sequence 1, 0, 1, 0, 1, 0 and so on, ok, indefinitely. Hence this is not convergent in 0 to, anything which convergent in 0 to must be eventually a constant here, right. So, being the projection map which continues, if fnk were convergent to something, fn i of a would have been convergent, so since this is not convergent, fnk is not convergent, fnk is totally arbitrary subsequence of fn, so we have proved that this space is not sequentially compact, ok. Now, let us try to do some other implications on the positive side, ok. The first thing is, countable compactness implies limit point compactness. Under T1 axiom, the converse also is true, so we can just say 1 implies 3, 1 implies 2 limit point and under T1 axiom they are equivalent, ok. You can see that slowly we are bringing them nearer to the metric spaces, so T1 spaces are slightly closer to metric space, metric spaces are very strong, they are house dogs, normal and so on, ok, so completely normal actually, so let us see how one proves it. The whole thing is, now you are not supposed to use any distance and so on, you have to do everything purely topologically, so let x be countably compact and a be any finite subset, if possible suppose a has no accumulation point, this means that a is a closer discrete subset, that is what we have seen earlier also, being close subset of x, a itself is countably compact, so this is an easy thing to see that close subsets are close subsets of a countably compact space is countably compact, just like close subset of a compact spatial compact, exactly same proof will work, clearly there exist a countably infinite subset b of a, because we started with a infinite subset, then b is also countably compact and discrete, countably you know it is a countably subset already, it will be automatically countably compact, because started with no accumulation point is discrete, but b has a countable open cover little b, b belong to b because this discrete which has no finite sub cover, so that is the contradiction, so any infinite set must have a limit point, if it is countably compact, let us prove now the converse under the axiom t1s for x, so x is a t1 and limit point compact, now you take any countable open cover un for x, if possible suppose there is no finite sub cover, what is that mean, u1 it does not cover it, so there will be x1 which is in the complement of u1, then you can find an x2 in the complement of u1 union u2, then you can take x3 in the complement of u1 u2 union u3, each time you can make that the new element is taken distinct from the perennial choices, if you cannot do that that means already you have exhausted the whole thing, that is the whole cover, since no finite sub cover is there you can always choose new elements here, so you get an infinite subset xn, a equal to xn, xn belong to n, it is an infinite subset, remember that the nth term does not belong to u1 union u2 union un, that is all we have to figure, now use the limit point compact property x must x be a limit point of this infinite set, then xn will x itself will be in one of the un's because it is covered by or whole x is covered by un, so x will be in one of the un's this limit point, but then x is a t1 space, it follows that un intersection a itself is an infinite set, so this is where we are using the t1 property, a limit point if it belongs to an open set intersection with original set a here must be infinite set, but we know that none of xk, k bigger than n are inside un, that is the quantity, so each un will contain only finitely many of these, so that is the quantity, so what we have proved, we have proved that under t1 axiom the first one and second one, countable compactness and limit point compactness are the same, next similar result sequential compactness implies limit point compactness, the three implies two now, under axiom t1 and first countability something more the converse holds, one implies two always and three implies two always, so two is the weakest one, but under t1 as one and two are same under t1 and first countability two and three are equivalent, that is the theorem, so let us prove this theorem also, this is also equally easy, xb a sequentially compact space a b a infinite set, then we can define a sequence xn in a of distinct points, an infinite set have a sequence of distinct points, that a be a limit point of some subsequence, then it easily check that a is actually a accumulation point, so infinite set and distinct points is important here, so very straight forward there is nothing hidden here, now conversely let now xb a t1 and first countable and x is a limit point compact space, now start xn be a sequence in it and a equal to set of all xn n belong to n, so the sequence is different remember the set is different, the sequence x1 x2 xn you have to write all the infinite terms, a set may have only finite terms, a is all xn such that n is at, so this is a, this is the image of the sequence, that is what it is, if a is finite, then clearly there is a subsequence nk such that xnk is equal to a, it may be eventually constant then there is no problem, otherwise there will be a repetition after some stage, so that is what it is, that means that there is subsequence nk such that xnk is same as a for some a and that subsequence is convergent, so if a is finite is no problem, suppose now a is infinite, then let a be an accumulation point of infinite substance, I must have an accumulation point, now let un and belong to n be a decreasing family of neighborhoods of a which form a local base at a, so this is where t1 nest is used, it is a countable local base as soon as you label them then you can take u1, u1 intersection u2, u intersection u3 and so on, so you can make a decreasing sequence of neighborhoods, that will form a local base at x, since x is t1 it follows that a intersection un is an infinite set, this we have used earlier also because a is a accumulation point of capital A, so a intersection un is infinite set for each n therefore we can choose a subsequence like x r1 belonging to u1 intersection a, now there exists r2 bigger than r1 such that x r2 will be inside u2 intersection a, having chosen rn then we choose rn plus 1 such that this rn plus 1 is bigger than rn and x r1 n plus 1 is inside un plus 1 intersection a, so distinct and larger and larger index has been used here, so that we have got a subsequence inside a, un plus 1 intersection, it follows that x rn is a subsequence of xi which is convergent to a of course, because all these un plus 1, un plus 2 etcetera they become smaller neighborhoods of this a, so first countability as well as t1 s is used here to extract a subsequence out of a sequence ok, in the case when the sequence is infinite, if the sequence is consists of finite number of elements then there is no problem that will always have a subsequence which is convergent. The next theorem is a sequentially compact t1 s pair is countably compact, now I am going from 3 to 1 ok, sequentially compact t1 s pair is countably compact, this proof is not at all new at all after seeing the proofs of the previous theorem, I feel that you should be able to extract a proof on your own, so try your hand ok, this will be left as an exercise, so as an assignment to you. Next theorem is if a is count if x is first countable ok, compact space then it is sequentially compact, right in the beginning we saw that compactness does not imply sequential compactness, however under first countability compactness implies sequential compact ok, so that may be the reason why compact metric spaces are sequentially compact you see, because metric space are always first countable ok, so this way the essence of all these properties comes out rather than just you should study the amount metric spaces ok, so how do we prove this, start with a sequence x n in a first countable space x and suppose this has no subsequence which is convergent ok, this assumption starting with a sequence which has no subsequence which is convergent, we first claim that every point in x has a neighborhood u such that x n is inside u only for finitely many n inside n, there may not be any point that is fine, if at all ok I can choose some open for each point I can choose some open set only finitely many elements from this x n will be there ok, suppose this is not true what is the meaning of this is not true I am claiming that for every point something, so not true means there is some point that x belong to x with such there is at least one point such that every neighborhood of x, every neighborhood u of x infinitely many x n's will be there no matter what how small you choose that u to be ok, so this is the denial of our claim, now let b n be a countable local base at x, so now first countability enters into it, choose n 1 such that the first one x n 1 is inside b 1 ok, now you choose next one, so that it is bigger than the indexing is bigger n k n k plus 1 is bigger than n k, so that x n plus k x n n k plus 1 is inside b k plus 1 and so on ok, after having chosen n k and x n k inside b n. So, this is possible since there are infinitely many n such that x n is inside b k plus 1, so you have chosen only finitely many next one can be taken to be bigger than that, now it is clear that this is a subsequence ok and this subsequence will convert to x because they are inside smaller and smaller neighborhoods of this local base ok, thus we have proved that for every point x in x there is an open set u x such that x in u x and x n is in u x for only finitely many n inside n, u x neighbor all and for each point as a neighbor all and for each point as a neighbor all where in the given sequence only finitely many of them will be there, now look at the open cover u x x belong to x, this has a finite sub cover because x is compact, see compactness enters only now, so that would mean that the sequence x n has only finitely many point right, where are they they have to be somewhere, but they are contained inside finitely many u x 1, u x 2, u x k each of them contains only finitely many, therefore there only finitely many point and hence as a constant subsequence which is a contradiction. So, we have proved actually something stronger if all this happen the sequence as a constant subsequence ok, we wanted to prove only that it is convergent, so I repeat what we have proved is now the first countable compact space is sequentially compact, so you must be satisfied now that whatever you have learnt in metric space is is correct ok, even if you have not learnt it now all these together will give you whatever you have learnt in the metric space because metric space is t 1 as well as first countable that is all, here is a picture which will help you to remember what are all the implication compactness, countable compactness, limit point compactness, starting compactness add first countability you can go to sequential compactness, start is sequential compactness you can always go to limit point compactness, put t 1 s and first countability you can come back, start from here you can always come to limit point compactness, but under t 1 s you can go back also ok, so this is all what we have proved so far. We have never gone back here right, so that is our next step what will imply this one that is the kind of thing we would like to do all right, I think that is enough for today so next time we will go into deeper into the this one namely we will bring now metric space is and criteria which will finally give you will allow you to go from somewhere here to here and so on all right, thank you.