 Let us prove that in R that is for a, b belongs to R a less than b the closed interval a, b is a compact subset of capital R. So, we will consider this closed interval and we will start in open cover. So, let so far is a compact subset of R. So, we will prove that if script a is a collection of open set which covers this closed interval a, b then from that we can extract a finite sub cover. So, let us start with let script a be a collection of collection of open sets in capital R such that closed interval a, b is a subset of union of a, a is in script a. Now, this in particular this small a belongs to closed interval a, b hence small a will be in at least one of capital A. So, now small a belongs to capital A implies there exist say some say a1 belongs to script a such that a is in small a belongs to capital A1 and again this a1 is an open set containing a that means a is an interior point of a1 that imply will get a neighbor root that mean see, but we have to restrict. So, this open we can yeah in fact there exist here or this imply a belongs to capital A1. In fact, a1 is an open set in the whole space R in R implies there exist R such that the open interval a minus R a plus R is completely containing this a1. So, we will find a radius R this is our a plus R 1 this point is a minus R 1 R 1 is positive. So, the whole now we are interested only in the right side of a. So, now take any element in say y belongs to let say y belongs to closed a open a plus R then closed interval a comma y is a subset of open interval a minus R a plus R which is a subset of a1. So, hence this closed interval is covered by a finite I mean covered by finitely many members of script a in that case just it is covered by one single number of script a. So, then this portion is covered by a single member of script a now we will consider all those y that is now let us say say let us say C let capital C is set of all elements say x is in closed interval a comma b such that in fact x is not in x not equal to a and a comma x is covered by a finitely many members of script a meaning is that is x belongs to C if and only if or x belongs to C this imply there exist a finite sub collection there exist a finite subset script a subset of script a such that this closed a comma x is a subset of union of a a is in that sub collection script a now so what is enough to prove is our lost this b belongs to C that mean this closed interval will have a finite sub cover. So, our aim is to prove that that is let us prove that capital this b belongs to script set C that is aim is to prove b belongs to C now note that for any this set for x belongs to C if we take x belongs to C that imply x is less than or equal to b that is b is an upper bound for this imply the element b that is this number is on upper bound for the subset C of r if a subset C of r as an upper bound then l u b axiom says it will have a least upper bound. So, we will call that least upper bound as so let this imply least that least upper bound we call it supremum of C exist exist mean exist and it is finite. So, let us call let this supremum let us let us say say x c small c equal to supremum of set of all x such that x belongs to C now what will happen to this C note that already C is we want to prove that C is in small c is in capital C hence we will have to prove that that is to prove close interval a comma c is will be covered by this will be covered by a finite sub family finite sub family of script a. Now again consider this line closed interval a comma c. So, naturally will if c equal to b in fact we are through will further see will here suppose now this set now select epsilon I mean in fact for each epsilon greater than 0 that is we have this for epsilon greater than 0 already this small c is a least among the upper bound that is least. So, hence for epsilon greater than 0 c minus epsilon will not be on upper bound upper bound for C for that set capital C because small c is the least upper bound. So, this is smaller that c is c minus epsilon is not an upper bound in fact c is not equal to a because there are we have seen that is the set c is non empty because for a given a we were able to find y such that this closed interval a comma y is covered by a finite sub family of script a. So, we can assume that epsilon is such that this a is strictly less than c minus epsilon. So, then c minus epsilon is not an upper bound for C implies here it is not an yeah this is will not be sorry will not be on upper bound this will not be an upper bound for C implies there exist some x belongs to our set capital C such that this c minus small c minus epsilon is strictly less than see this not an upper bound. Hence we will get an element such that that element will be greater than this there exist an element say x in c such that x is greater than c minus epsilon in other words that is c minus epsilon is less than x. So, we will have and also c this c minus. So, we will get an x belongs to capital C. So, now, this belongs to capital C means we will have a finite sub cover for this collection closed interval a comma c minus epsilon pardon yeah finite sub cover for a comma x yeah because x is in c. So, now, this set now x belongs to C implies there exist a finite sub collection finite sub collection say script f naught of script a. So, now, again this C is in again C belongs to belongs to closed interval a comma b implies there exist c will belongs to say some a 2 or some a 2 belongs to script a correct c belongs to a 2 again a 2 is a open set that will imply c belongs to now the small c belongs to a 2 a 2 is on open subset of open subset of capital R implies there exist again say R greater than 0 such that such that this open interval c minus R to c plus R is completely contained in this a 2 correct a 2. So, now, we have the set a 2 c minus epsilon is covered by a finite sub collection of script a and what about in this again this side. So, the open c minus R so c minus R to c plus R is covered I mean this open interval c minus R to c plus R is contained in a 2. So, now, see here we started with for any epsilon we can find a say that c minus epsilon will not be covered by a we have seen that c minus epsilon for this epsilon we have where x is in c and x is and c minus epsilon is not an upper bound hence is not an upper bound. So, we will have c minus epsilon here x. So, if we see my x is strictly greater than c minus epsilon. So, in particular if we take epsilon equal to R if epsilon equal to R greater than 0 then this closed interval take here because we have the closed interval a comma x. Now, this epsilon is R this closed a comma x is covered by a finite sub family of script a. So, hence in particular a 2 closed interval c minus R is covered by a finite sub family of script a and also we have from c minus R to c plus R is subset of a 2. So, hence here we have this portion is covered by a finitely many members then c minus R to in fact c plus R is covered by 1 single a 1. So, hence the whole this in particular will imply c is in between c plus R c is in between. So, that will give. So, this will imply that epsilon equal to R greater than 0 then closed interval a comma c minus R correct is covered by a finite sub family finite sub family say script c does not matter if you want use let us say script f prime of script a then what this implies closed interval a comma c is a subset of union of a a belongs to this finite sub family script f prime then union additional member we have denoted by a 2. So, a 2. So, our sub family is script f prime union singleton a 2. So, hence this is a finite sub cover for closed interval a comma c. So, this will imply c small c belongs to capital C. Now, our aim is to prove that this c is nothing but now let us prove that now let us prove that in fact this c in fact in some that is the maximum element having this property. So, now let us prove that this small c equal to b suppose not that means suppose c is strictly less than b. So, we have a c b suppose c is strictly less than b then again we can use the same method to find a I mean again this using this that means what we can find here epsilon c for take epsilon equal to b minus c by 2 which is positive then again suppose c less than b then take epsilon equal to a b minus c by 2 and for this epsilon for this epsilon again c minus epsilon is not an I mean again this c belongs to here. So, that means c belongs to here we do not care this c belongs to closed interval c belongs to closed interval a comma b which is a subset of union of a that means c will belongs to one of the member, but that c is just let us say that suppose yeah now we are already here as we say we already proved that something right hand side not only just a comma c is covered by this if c is not equal to b then we can go to the right side. So, then suppose c is less than c belongs to c some that is what we denoted by a 2. So, c belongs to a 2 for some a 2 belongs to script a, but this imply there exist or say that open interval c minus r 2 c plus r is contained in a 2. So, that means we will get an element say up to c plus r this element this set closed a open c plus r r. So, let us say in particular r by 2 then the closed interval this will be covered by a finite of family of script a. So, this will imply that means then we can prove that this will give us c plus r by 2 belongs to capital C. So, that will be a contradiction to the fact that c is a least upper bound. So, hence this we can get a positive or if c is not equal to b hence c should be equal to. So, this will give a contradiction a contradiction to the fact that c belongs to capital C is an upper bound I mean is the least upper bound for a for script that set p. So, this give us c equal to b and we are hence c should block this set b equal to c which belongs to capital C. This imply there exist a finite sub family say script f such that closed interval a comma b is a subset of union a a belongs to script a. So, we started with a open cover for closed interval a comma b then we have got a finite sub family of script a such that that sub family covers closed interval a comma b. Hence by our definition closed interval a comma b is a compact subset of r. In fact one using real analysis what we prove that as a matrix space r is what is called a complete matrix space. That mean if you take a Cauchy sequence when a sequence in r is a Cauchy sequence mean just briefly I mean modulus of x n minus x m tends to 0. That mean that is for epsilon positive there exist a stage n naught belongs to n such that mod x n minus x m this what we call it distance between x n and x m this is less than epsilon after that stage for all n m come greater than or equal to n naught. Then we say such a sequence as a Cauchy sequence if every Cauchy sequence converges then it is called complete and also we will introduce a concept called totally bounded that is here. Zomindically if we see that it is like almost compact see in this case if you start with a a less than b start with an epsilon then consider this what is called totally bounded mean start with any epsilon greater than 0 then there exist a finite epsilon net meaning is this set there exist x 1 x 2 say x k it can be from the given set itself in this case close a b such that close interval a comma b is contained in these intervals union of x i minus epsilon x i plus epsilon i varies from 1 to n that mean for a given epsilon we can find sorry here we are denoted by n. So, a finitely many member x 1 x 2 x n from that set such that this close a b is covered by this finitely many these are all neighbor roots of x i. So, if you start with this so if you start with the epsilon then go to a plus epsilon. So, this will cover the portion close a to a plus epsilon say in between if we take like this y. So, close interval a minus now a to a plus epsilon this interval we are able to cover. So, that mean by taking x 1 equal to a. So, consider this open interval for this we have an open interval a equal to x 1 say x 1 minus epsilon to x 1 plus x epsilon that will take care of this. Now, take x 2 equal to this x a plus epsilon or a plus epsilon is same as x 1 plus epsilon. So, this again consider a small otherwise interval sender at this point and again radius this. So, that will cover up to now this is a plus 2 epsilon then that is our second point then the third point again consider a sender at this and radius epsilon then we will go up to a plus 3 epsilon that is x 3 then find n large n natural number find n large say that n a plus a plus n times epsilon comes out of b that is greater than b. So, I mean such a thing that is what we want is that is our find a natural number find n belongs to n such that n epsilon is greater than b minus a or n is greater than b minus a by epsilon which is true by Archimedean principle. So, hence this set will be called totally bounded and that is complete and totally bounded is equivalent to compact in a metric space whereas, what we have given is a direct proof using only the definition namely every open cover in has a finite sub cover. So, now, this give will give us I mean many more compact set now we have this close interval compact we have proof that is Iconov theorem for finite products. So, that is if a 1 comma b 1 compact a 2 comma b 2 compact. So, if you are in the plane here this x axis subset of say r y axis again r I mean we have a set here a 1 or let us say a 2 or a 1 b 1 here some a 2 b 2 then the this is close interval. So, the close interval a 1 b 1 Cartesian product this. So, we will get this rectangle. So, that rectangle is close rectangle is compact with respect to the product topology. So, like that we can have the any the product of rectangle or cube in r 3 and so on they are all compact sets. So, yeah let us see an equivalent formulation of compact sets there namely yeah this is rather re symbol, but it has I mean many times will be using this equivalent concept what mean this is compact if and only if a topological space that is a topological space x tau is compact a topological space x tau is compact let us say it call it theorem is compact if and only if it satisfies the following condition namely whenever and only if whenever script a is a collection of closed subsets of x whenever script a is a collection of closed subsets of capital X which has the property that it has finite intersection property which has finite intersection property which has finite intersection property that is meaning is this we call it finite intersection property f i p that is whenever a 1 a 2 a 1 finitely many members from script a then that intersection is non-empty that finite intersection meaning this imply a 1 intersection a 2 intersection a n is non-empty then we say that it has that mean whenever script a is a collection of subsets of x which has finite intersection property mean you give whenever a 1 a 2 a n is a finitely many members from script a then that intersection should be non-empty then we say that script a has finite intersection property, but if it is compact mean if a collection has finite intersection property then the intersection is non-empty that is what we say a topological space is compact if and only if whenever script a is a collection of closed subsets of x which has finite intersection property then the intersection a a from script a is non-empty this is just we use the what is called de Morgan law script a is a collection of closed subsets then the compliment set of all a compliment such that a belongs to script a is a collection of open sets. So, if the intersection is empty then see what will happen is if the intersection a a from that is we are asking suppose intersection of a a from script a we want to prove that under the given condition this intersection is non-empty suppose this is empty then take compliment both side that is we have this end of intersection a then this will imply de Morgan law is union a compliment a is from script a this empty compliment is the whole space that means this collection a compliment a is in script a that is this is our collection set of all a compliment a belongs to script is an open cover for then this an open open cover for capital X now this will have a because X is compact if we assume X is compact one way. So, now X is compact X compact implies there X is n belongs to capital N finitely many member a 1 a 2 a n from script a that is our family such that that is our sub cover is a 1 compliment union a 2 compliment union a n compliment equal to the whole space. So, again take the compliment both side this will imply so this compliment a 1 compliment union a 2 compliment union a n compliment the whole compliment equal to capital X compliment is empty but again by de Morgan law this set is nothing but a 1 intersection a 2 intersection a n this implies a 1 intersection a 2 intersection a n is empty but what is our assumption is our collection is say that it has finite intersection property meaning is whenever you give any finitely many members like this a 1 a 2 a n from script a then the intersection should be non empty. So, that gives us a contradiction we assume I mean we could get this contradiction by assuming that this will is not a I mean this intersection a a from script a is empty. So, hence that should be non empty. So, this will be a very useful criteria to check a spaces compact or not. In fact, the celebrated Iconov theorem one can prove using this I mean in compact we will start with a C what is the will not prove that Iconov theorem if X alpha tau alpha belongs to an arbitrary indexed need not be finite any arbitrary indexed. Let X alpha tau alpha be a collection of compact topological spaces then the product space then the product space I X alpha alpha belongs to J with product topology is compact. I mean whenever we say the product mean unless mentioned otherwise I mean it with product topology this is compact what we can do is one can use what is called Jones lemma that mean will we have a we have a collection we start with a collection of subsets we can use if script a is a collection of subsets of the product space then ultimately we want to prove that if we start with a collection of close sets. So, that is equivalent to again whenever it is a collection of sets which has finite intersection property then X is compact mean in fact X is compact if and only if whenever script a is a collection of subsets of X which has finite intersection property then the intersection a closer a in script a is non-empty. So, we will start with a to mean just what is an outline quickly we will start with a collection of a collection of subsets of the product space I X alpha then we will use Jones lemma that mean we will find a new collection script a naught which is larger than script a and this has the property that it is closed under finite intersection that meaning is whenever a 1 a 2 a n belongs to script a naught implies a 1 intersection a 2 intersection a n also belongs to that collection correct and second if for a subset if for a subset of the product space subset of the product space a subset the product space we call it pi X alpha and a intersect every member of every member of script a naught then a is say that a intersection b non-empty for all b belongs to script a naught then that will imply that b also belongs to yeah sorry we are using a yeah a also belongs to script a naught. So, then we will use this collection yeah to prove that such a collection will have finite intersection property and that intersection is non-empty using that we can prove the given collection intersection a a in script a is non-empty yeah thank you.