 So, the main theorem that we proved in the previous lecture was that to check if a subspace of Rn is compact it is enough to check that the subspace is closed and bounded and as an application of this we proved that Son is compact in the previous lecture, a subspace y of Rn is compact if and only if y is closed and bounded right and as an application we saw that Son is compact right. Similarly, we can prove that Son. So, slight modifications of the proof that Son is compact will show us that On the group of unitary matrices and special unitary matrices these are all compact ok. So, today we will in this lecture we will see we will first study how compact compactness behaves with respect to continuous maps. So, let us begin with this proposition. So, let me just write that. So, behavior of compactness with respect to. So, we begin with this proposition f from x to y e a continuous map. So, the image of a compact set of compact subset is compact ok. So, let us prove this. So, let z contain in x p compact. So, then we need to show that f of z is compact. So, let us assume that. So, let me make a remark to show that to show that f of z or rather a subspace subspace is compact it is enough to show that if y prime is contained in a curve u i where u i is open in y then y prime is contained in a finite curve. So, we have seen this already on two occasions before. So, let us just prove this is easy. So, suppose y prime is compact and y prime we are given that y prime is contained in this in this curve u i's. So, this implies that y prime is equal to this is an open curve for y primes. And since y prime is compact. So, this implies that y prime is equal to there is a finite indexing set u i j which implies that y prime is contained in union. Conversely suppose y prime has this property conversely suppose y prime has the upper property as a property that if we put y prime if y prime is contained in an open cover where u i's are open subsets of y and not of y prime then it is contained in a finite sub cover. Then let us show that we show that y prime is compact as follows. First let us take let y prime be equal to union i in i w i be an open cover of y prime. Now, y prime has a subsets of all the which means that each w i is equal to y prime intersected u i for some u i open in y. So, this implies that y prime is contained in union i in i u i which implies that it is contained in because it has this property u i j which implies that y prime is equal to j equal to 1 this implies y prime is compact. So, we will repeatedly use this remark. So, if you want to show that a subspace is compact we will show that every time it is contained in an open cover it is contained in a finite sub cover of that open cover. So, we need in our case we need to show that f of z is compact. So, suppose f of z is contained in not real union let us say v i is i is open. So, then we have z is contained in this cover i in i f inverse of v i and since f is continuous this is open in and as z is compact there is a finite sets z is contained in i j which implies f of z is contained in thus f of z is compact. So, this completes a proof of the proposition and we will prove the following lemma. So, the following proposition is very useful is very useful that f from x to y be a bijective continuous map if x is compact then f is a homomorphism. So, proof. So, to show that f is a homomorphism it is enough to show. So, since f is bijective and continuous it is enough to show that the image of an open subset is open and which happens if and only if the image that is an easy check and I give you of a closed subset. So, we will show this. So, let z contain x be a closed subset then as x is compact this implies z is compact. This is because we have proved that closed subspace of a compact space is compact which implies that f of z is compact which implies that. So, once again we are under the hypothesis that all our spaces are host of. So, in a host of space a compact subspace is closed. So, f is a bijective continuous map and it takes close sets to close sets from this we can easily conclude that it takes open sets to open sets. So, or in any case from this we conclude we easily conclude f is a homomorph. So, before we started discussion on. So, this was a very general discussion on compactness and before we started discussion on compact metric spaces how compactness how we can. So, just as earlier in case of metric spaces we had given a special description of closed subspaces how we had described we had explained how to describe closed subspaces and how to describe continuous maps between metric spaces. So, in case of compactness also we can make some comments when our space is a metric space. But before we go on to that we will state this following very important theorem which we will not prove x i be compact of logical spaces for i in i. So, this capital I is now an infinite possibly infinite index set then this product x i obviously this we mean given the product apology is compact. So, this is called Tickanoff's theorem we will not prove this in this a proof may be found in Munkreis. So, we will end this lecture here.