 So, welcome to this lecture this is lecture 11. So, so far so far we have defined topological spaces and continuous maps between them. Next we will define continuous maps in terms of limits we want to define continuity in terms of limits in a metric space. So, we will have to define metric spaces and then limits and then finally, give a criterion of when a map is continuous in metric spaces in terms of limits and that is what is the theme going to be the theme of the next few lectures. So, we begin with a discussion on closed subspaces. So, we are still not we haven't reached metric spaces, but we will reach there soon. So, definition. So, this is a general discussion the definition of a closed subset is very easy. So, let X be a topological space a subset Z contained in X is said to be closed if the complement X minus Z is an open subset of X. So, recall what we mean by this. So, X is a topological space. So, it has a topology tau and Z is said to be closed if when we take the complement that is in tau. So, let us make a remark one easily checks the following properties of closed subsets these follow from the definition of a topology. So, the first property is that the empty set phi and the full set X are both closed. The second property is that let Z 1 up to Z R be finitely many closed sets then their union Z 1 union Z 2 union Z R is closed. And third property is that let I be a set and suppose we are given a collection of closed subsets of closed subsets then the intersection of these closed subsets which is a subset of X is closed. And all these three properties all the above properties follow from the analogous properties for open subsets from the respective properties of open subsets that is one remark. And the second remark is that suppose we are given a collection of subsets suppose we are given a subset S contained in the power set of X which means we are given a collection of subsets of X which satisfy the above three properties. So, S satisfies these properties 1 2 and 3. So, then let tau be the collection of those U in X such that X minus U is contained in S belongs to S sorry. So, then tau defines a topology. So, in other words what this remark tells us is that to define a topology on X we should either it is enough to specify the collection of open subsets which should satisfy the three conditions which came which we saw in the first lecture either that or we can specify a collection of subsets which satisfy these three conditions these three conditions here and that would specify a topology ok. So, this remark is left as an exercise proof of this remark prove this. So, this is left as an exercise and another simple statement which is again left as an exercise which just follows from the definition. Let X and Y be topological spaces let F from X to Y be a map of sets. So, then F is continuous if and only if if and only if for every closed subset in Y Z contained in Y the subset F inverse Z is this is a subset of X is a closed subset. So, the main point behind this exercise is this easy check that F inverse of Y minus Z is equal to X minus F inverse. So, we have to use this well known property of the inverse image of map. So, in one of the lectures we saw after we introduced various ways of specifying topologies we saw several examples here. So, let us just revisit those examples and see which of those subsets are closed and which are open ok. So, the first example or one of the examples we had seen was the set S 1 right this is those X comma Y in R 2 this is that X square plus Y square is equal to 1. So, can we say something about the set is this set open or closed. So, the answer to this is quite easy. So, first we have to make the following observations. So, recall that we proved that the set of continuous maps is closed under a multiplication and addition and also that we also proved that the projection maps the projections from a product right. So, therefore, this map X comma Y maps to X is continuous. So, just for fun let us just make. So, we have both these maps X comma Y goes to X and X comma Y goes to Y is continuous right. So, we can make the graph of this map X comma Y goes to Y. So, let us say this is going to the Y axis. So, if we take a small open subset here right. So, the inverse image of that is going to be this strip over here if this is A and this is B. So, then this is going to be 0 comma A and this is going to be 0 comma B. This point here is 0 comma A this point here is 0 comma B right and we have to take this open. So, the inverse image of this interval is this is this tube is a strip if I may say so and clearly this is open because given any point over here we can always find a square like this right. This is a picture of the projection onto the Y coordinate and similarly we can make a picture of the projection onto the X coordinate. So, here the inverse image of A comma B will be this open strip right. So, this point is A comma 0 this point here and this point is B comma 0 and clearly this is also open because given any point here we can always find a right. So, now both these predictions are continuous. So, therefore X comma Y goes to X square is also continuous because this is simply the map let me call this map P1 and let me call this map P2. So, this is simply the map P1 square right. So, if you want so X comma Y maps to P1 X comma Y and the square of this. So, both these are continuous similarly X comma Y goes to Y square is continuous and addition is also continuous. So, therefore this implies that so the results we had proved in the previous class X comma Y square is a continuous map and now inside R the subset 1 is a closed subset right because what is the complement of 1. So, this is 0 and this is the point 1. So, we are just deleting this point. So, if I take any X over here then I can always find a small epsilon such that the neighborhood the epsilon neighborhood around X is going to be contained in R minus. So, in other words for every R minus 1 for every X in R minus 1 we can find epsilon positive such that this epsilon neighborhood around X is contained in R minus 1 right. Therefore R minus 1 is open which implies the subset which contains only 1 is closed right. So, this in turn implies. So, let us call this function give this function name f. So, since f is continuous this implies that f inverse of a closed subset is closed. So, let me just write that of a closed subset is closed which implies that f inverse of this set single to 1 is closed right, but what is f inverse 1? f inverse of this single to 1 is exactly those X comma Y by definition in R 2. So, is that f of X comma Y is equal to 1 this is same as saying that X square plus Y square is equal to 1, but this is precisely equal to the set S 1. So, therefore, we see that S 1 is a closed subset. On the other hand we could have proved this fact directly from the definitions because we can look at S 1 S 1 is this yeah. So, if we take any point in R 2 minus S 1 if we cannot we could have directly shown and it is not very hard that we can find a small ball or a square of radius epsilon which is completely contained inside the complement. So, we could have proved that directly as well right. So, let us take the second example another example which we had seen was the spheres S n right this is those X naught up to X n in R n plus 1 such that X naught square plus X 1 square plus X n square is equal to 1. So, is S n open or can we say we just phrase it like that is S n open or closed in R n plus 1 right and by the same reasoning as above yeah. So, since the function f from R n plus 1 to R given by f of X naught up to X n is equal to X naught square plus X n square is continuous once again because the projections are continuous and therefore, the squares of each of the projection is continuous and when we add then that is continuous yeah. So, this implies that. So, once again f inverse of this single to 1 is a closed subset R n plus 1. So, which implies that S n is a closed subset of R n plus 1 right and this is this is exactly equal to S n by the same reason as the previous example. So, here are two examples another example we had seen was the subset G l and R right this is equal to those A in n cross n matrices. So, is that determinant of A is not equal to 0. So, again can we say if this is open or closed. So, let us look at the determinant map from M and R. So, as an example let us look at the determinant of a 2 cross 2 matrix matrix. So, let us say this is X 1 1 X 1 2 X 2 1 X 2 2 right. So, we know that the determinant is X 1 1 X 2 2 minus X 2 1 X 1 right. So, each of the projections are continuous. So, each X i j is continuous and the product of continuous maps is continuous. So, this implies that this is continuous yeah and similarly this is continuous ok because the projections are continuous and this function is the product of two projections. So, for instance this function is this first function is X 1 maps to X 1 1 X 2 2 right. So, this function is continuous. So, therefore, so in this case this continuous and in general we know that right. So, in general since the determinant is a polynomial in the coefficients a matrix right. So, this determinant function is actually a polynomial function in the in the entries here. So, therefore, and each coefficient is the projection map right and the projection is continuous. Therefore, polynomials and projections are continuous. So, therefore, the determinant map right and now note that gl and r is simply the determinant inverse of this subset r minus u right. So, determinant is continuous and as r minus 0 is open. So, this implies gl and r is an open subset of m and r. So, another example we can take is the set SL and r right this those a in m and r. So, is that determinant of a is equal to 1. So, this is clearly. So, this is closed as SL and r is equal to determinant inverse of 1 and yet another example we saw was the set of orthogonal matrices right. This is those a in m and r says that a transpose a is equal to identity right. So, we claim that this is a closed subset. Now, in each of the above examples showing that these subsets are closed directly is could be somewhat complicated and we see that once we use once we obtain these subsets once we can realize these subsets as inverse images under continuous maps of certain subsets. So, then it becomes a lot more easier. So, here also we are going to do the same thing. So, first we have to look at this map from m and r. So, let us call this map f. So, the map f is a goes to a transpose a right. So, let us see what this does for a 2 cross 2 matrix. So, we have x 1 1 x 1 2 x 2 1 x 2 2 right. So, this is going to map to the transpose of this times. But this is equal to x 1 1 square x 2 1 x sorry x 2 1 square this is x 1 1 x 1 2 plus x 2 1 right. So, notice that on the right hand side each entry is a polynomial in the entries of a right and m and r here has the product topology which is the same as a standard topology. So, if we view m and r as r n square right. So, f is continuous if and only if each f i j is continuous right. So, f i j. So, in our example f i f let us say what is f 1 1. So, f 1 1 is precisely this coordinate right. So, f 1 1 of this matrix let us denote this matrix by x for simplicity of notation right. Similarly, f 1 2 of x is equal to this coordinate that is x 1 1 x 1 2 plus x 2 1 x 2 2 right and similarly we have f 2 1 of x this is x 1 2 x 1 1 plus x 2 2 x 2 1 and finally, we have f 2 2 x. So, all these each of the coordinate functions. So, each of the coordinate functions coordinates is a polynomial in the entries of x. In other words each of these coordinate functions we can write it as a polynomial obtained using the projection maps. Each of the projections is continuous and since we take product and add etcetera that is also when we continue. So, therefore, each of these coordinate functions is a continuous function which implies that the map the f is a continuous function right because. So, we can view f as a map from MNR to MNR well standard topology or the product topology it does not matter because as we saw the product topology on RM is same as the standard topology on RM right. So, therefore, so this shows that right and now so note that. So, this is for the 2 cross 2 example yeah. So, in the n cross n situation. So, in the n cross n situation or let me not write this. So, you can you can write down explicit coordinates for general n right. So, what are Fij for general n in terms of xij. So, this is an easy exercise. So, you can just write this down yeah in any case this is show. So, this shows that f is continuous in general yeah. So, f is from MNR to MNR and it is continuous. So, clearly this ONR is equal to f inverse of the identity matrix right. So, MNR contains the identity matrix yeah and therefore, to show that to show that ONR is closed in MNR it suffices to show this singleton identity in MNR is a closed subset. But this is a very general result. So, we will write this as lemma. So, let A 1 up to A M in R M. So, let us denote this by A right. So, then R M minus this point A is a closed subset ok. So, proof this is left as an exercise. So, we will end here.