 Welcome to Morbules 14, last time we had studied quite a bit about some examples and so on of various notions like open set, closed sets all right, interior, boundary, no air dance and so on right. So today we shall start taking them up one by one. The foremost one is about closed sets which are just nothing but complements of open sets right. So they will be as important as open sets after all. So let us start putting various properties in one single place as many as possible. It is not possible to list all of them exclusively okay but it is possible to do quite a bit of good job here. Start with any topological space. Let A and B denote any two open subsets, any two subsets of x sorry. Then we have a number of property 1, 2, 3, 4, 5, 6 up to 10 I have listed here. They are all easy properties. Let us go one by one, go through one by one. The first one is the closure of empty set is empty. So that is obvious thing why because closure is defined as whenever you take a point all points which have a pouring property, whenever you take a point and a neighborhood of that it should intersect the set but the set is empty. Everything is intersection is empty there. Therefore no point will satisfy this property. Therefore closure is empty. A is contained inside a bar because a neighborhood of a point of A will always intersect it. So that is also easy. The third one if A is contained inside A sorry, A intersection B is empty and B is open then A bar intersection B is empty. Once again take a point which is in B I want to show that it is not in A bar. Then A bar is in B will be empty right. So take a point in B then B is an open set. So it is a neighborhood of that point and we have assumed that B intersection A is empty. Therefore by the very definition of closure the point is not a point of A bar. So every point of B to take is not a point of A bar. So it is empty. So you see all these things are very easy one by one but when you club them together they may become more difficult that is why we are going through them slowly okay. The third fourth one is A bar itself is closed. Showing that A bar is closed what I should do take a point in the complement namely I should show that the complement of A bar is open okay. Take a point in the complement in the complement means what it is not a point of A bar. That means the negation or that is no closure point what is negation there exists an open neighborhood of X which does not intersect A right. So that is a neighborhood if something is in the complement means right. So whole complement see all those points will have the same property. Once X belongs to an open set U U intersection A is empty all the points of U have that property for the same reason the entire of U must be in the complement therefore we have shown that the complement is open okay. Similarly now if A is closed then A is equal to A bar. A is already contained inside A bar I must show that A bar is contained inside A what is the meaning of that every closure point is inside A or conversely what I should show A bar is contained that means any point which is not in A is not in the closure that is what I have to show right. It is not in A because A is closed not in A means what it is in A complement but A complement is now open clearly A complement intersection A is empty therefore we have found a neighborhood of every point inside the complement of A what is that neighborhood 8 A complement itself which does not intersect A therefore they are not points of A bar. So fifth is also pro namely A is closed means A is A bar. The sixth one if A is contained inside B A bar is contained inside B bar this is also as easy because starting with a point in A bar and a neighborhood of that point by the very definition it intersects A but B is larger so it will intersect B also therefore the point must be inside B bar okay so that is the property sixth here. Now A bar bar the closure of the closure is the closure itself A bar bar is A bar that is the statement 7 here how do you do this one you can go on taking point wise and so on but already you have done enough work here so this is a corollary consequence of that namely look at what we have done A is contained inside A bar here therefore when you take the closure what does it give you this sixth one it says that A bar is contained inside A bar bar okay one way you hold but the fifth one says the closure of the closure is already closed so it is it is equal to A bar so therefore the two are equal okay so you can use this one A bar is closed then A is equal to A bar so A bar bar is A bar that is directly the seventh one okay says that A bar is the smallest closed subset containing A first of all A bar contains A right here so it is a closed subset containing A now you take any B which contains A bar then you take bar again that A double bar which A bar itself will be contained inside B bar but B bar is B because I take any closed subset which contains A bar so A bar is the smallest closed subset so that is the seventh one the ninth one and tenth one are quite useful especially this one says what is happening to the union under closure so this is like you can say that the two operations of taking union in closure they commute each other first you take the take the union and then take the closure the same thing as first take closures of both of them and then take the union but in the ninth tenth one it only says he's contained in here you have to be cautious equality is not assured here it may happen but it is not true in general that is the meaning of this okay let us work out nine so first of all A is contained in the union therefore by sixth show A bar is contained inside the bar of A and B similarly B bar is contained inside so one way is clear namely right hand side is contained in the left hand side okay you have to show left hand side is contained in the right hand side for that you have to work a little harder so let me this time to use this all these things I have written down carefully you can go through them okay I have already done but the ninth one let us go to a ninth one now okay namely one way we have already done namely A bar is contained in the union because A is contained inside this one so bar will be contained inside a bar similarly that so union is contained inside union of the bars is contained inside bar of the union so we want to do the other way so start the point in the closure of the union we want to show that X is either in A bar or in B bar that is the meaning of union if X is in A bar no problem we are done so assume X is not in A bar then we must show that X is in B bar okay what is the meaning of X is not in A bar this means that there is an open subset V such that X is inside V V intersection A is empty okay now use this fact to show that X is in B bar okay how take any open set which contains X put W equal to U intersection B okay then W is an open set and X belongs to W now use the fact that A union B X is inside A union B bar that means that this W will intersect A union B okay if you intersect A union B what is A union B intersection VW it is A intersection W union B intersection W one of them must be empty one of them must be non-empty because it is empty it is non-empty okay but A intersection W is already in A intersection B we started with A intersection V equal to empty here so one of them is empty the other one must be non-empty what is the other one B intersection W but now B intersection W is contained inside B intersection U which is bigger therefore this is also non-empty so that is what we wanted to show starting with any open set containing X we wanted to show that U intersection B is non-empty so that that will prove that X is in B bar so what we have done assume that X is in the closure of the union assume that it is not in one of them closure of one of them then it is in the closure of the other okay so that proves right actually sir one question please here A union B is subset of A bar union B bar yes if I take both sides closer then right side is A bar union B bar is closure union of closures what I have started A is contained inside union B this is done other side other inclusion other inclusion A union B is subset of A bar union B bar I take both sides closer but you don't know you have used your A bar union B bar is closure of A bar is closure B bar is closure union of closure is closure union is closure yeah then bar will be equal to that also can be used sir yeah yeah you are sure thank you sir yeah you can do that way also very good yeah okay so let us come to the intersection take a point in the intersection if U is a neighborhood of X then we have U intersection A intersection B is non-empty because X is inside the closure but if this intersection is non-empty both U intersection A and U intersection B must be non-empty therefore for every neighborhood of X both intersection non-empty means what X is in both A bar as well as B bar so the closure of the intersection is contained in intersection of the closure okay but you have this you have this equality inclusion map here okay by repeated application of this you can always get finite union right actually 9 to 1 let's go to 9 to 1 here it's the equality you can take finite union okay there will be equality but if you take infinite union will this work equality usually that is what is going to be problem because just now as one of you pointed out infinite union of closed section may not be closed so you can't come back okay it may happen that they are equal but in general that is not true right so if you take arbitrary union the closure of the union definitely contains the union of the closure but if you take first separately closures and then take the union it will be contained in that solve but may not be equal may not be equal okay we are plenty of examples for example here is an example take each singleton where the point varies over the q okay that is a r is just one point a r is singleton r then you take this collection the union of all this a r is just q right and what is the closure of each point points are closed inside the metric space so their closures are they are itself so union closure is just q right but first you take the union that is q then take the flow here is with the whole of r okay so here I am taking each a single ton rational number this closure is that single rational number itself when you come here it is the whole of q its closure we know is the whole of r okay we know from elementary real analysis that q is dense in r all right similarly the tenth one here okay what is the tenth one a intersection b bar is containing equality may not be true here even when there are only two points and two sets for that what do I do I take a equal to q and b equal to complement of q rational numbers irrational numbers intersection is empty here so closure is also empty but what is the closure of a it is the whole of r what is the closure of b it is the whole of r so intersection is r so this is empty and that is the whole of r so that gives you that equality need not hold all right so that is roughly you know familiar to be close subsets okay so let us now go to something some more examples here about closed sets a hyperplane in l hyperplane l in k power n you can talk of r n or c n does not matter to think of r n no problem then every coefficient set retry will be real numbers that is all so hyperplane is given by a linear equations like this okay this is actually if you put r that is also called hyperplane this is called hyper surface okay equal to you can put any some real number or complex number here that will be hyperplane equal to r is the hyper surface hyper subspace hyper means what just one dimension lower surface subspace is linear subspace okay so if you write f as f of x 1 x n is this linear map a 1 x 1 a plus a n x n etc this is nothing but a linear map from k n to k okay we have seen that such a linear map is continuous because adding two continuous functions is continuous scalar multiplication by the continuous function is continuous you use it iteratively but you know 0 is a closed set inside k so inverse image of a closed set is closed under a continuous map because inverse image of an open set is open and a continuous map so this is what we have already seen right therefore what it emits to saying that each such l is a closed subset of k n easy examples now you can use finite intersection of closed sets is closed to get that if you have finitely many equations like this the common zeros will be also closed so those are all the vector subspaces of k n okay take any vector subspace of k n so suppose it is of dimension say m then you can write it as n minus m equations common solutions of n minus m equations only finitely many of them anyway so they are intersections of four dimension one subspaces okay so they will be also closed okay these are elementary uh examples but they have to be properly understood same explanation will be we will hold for these are half spaces like you can say x1 is positive x1 is greater than or equal to 0 or y1 is greater than or equal to 0 and so on so here a1 x1 an xn is greater than or equal to r greater than or equal to r that means what it is the inverse image of the closed interval r to infinity open inverse image will be a closed set so such hyper planes are also closed so intersections of these hyper planes a hyper half spaces is also closed okay so for example you can take all points which are bigger than or equal to uh x coordinate is bigger than equal to 0 and y coordinate also bigger than or equal to 0 bigger than equal to 0 both of them are closed sets intersection will be closed set what is it it is the closed first quadrant right so like this you can construct lot of them moreover you can go to other things like look at the closed disk drxd in any metric space d is little d is metric metric drxd set of all points y such that distance between x and y less than or equal to r right similarly the sphere what is sr it is those points where in the distance between x and y is equal to r okay so those things should be closed also why this time you have to use this fact namely fixing any y or fixing any x whichever whichever one you want to fix fix why x going to dx y this is a continuous function on the topology of the space namely tau d to the real numbers okay it is a continuous function from x to zero infinity what is the topology on this one it is the metric topology here okay so these things we have seen that they are continuous function therefore closed balls the spheres they are all closed subsets okay any questions so let us consolidate these things let us have any question if you have any questions you should ask now okay hello sir yeah so in in the hyper plane equation instead of zero it can be replaced by some real number yeah so definition this is actually this is hyper plane fine but this is actually a passing through origin when they pass through origin they are called hyper hyper subspaces okay okay so i wanted to concentrate on one thing instead of floating r first so if you can float r singleton r is also closed the same argument holds so that i have used in the second part if you are going to put r it is fine okay so let us stop here until next time thank you