 Welcome to mobile 26. Today our topic is discussion of union of spaces. We start with a set X which is written as disjoint union of sets. This is one special case. Suppose each Xi is given a topology tau i. Consider tau equal to all subsets of X such that A into section Xi is in tau i for every i. A being subset of the disjoint union, it is union of subsets from each Xi. So those are the A intersection Xi's. All of them must be open in the corresponding Xi. That is the meaning of A intersection Xi is tau i. Take all such A, put them together in one single family. That tau is a topology on X. So this is called disjoint topology. I start with the subsets here are all disjoint, mutually disjoint. That is the meaning of this symbol. Cup union I am always using for disjoint union. Verification is obvious. I mean easy. Empty set belongs to this. Union belongs to this because each Xi will be given tau i. So X, if you take A equal to X, A intersection Xi will be Xi. So they are there. So if A1 and A2 are there, then A1 intersection Xi, A2 intersection Xi are there. So A1 intersection A2 intersection Xi will be there for each i. So intersection is there and so on. You can verify that. We are on straightforward modifications. What is important here is Xi tau i becomes a subspace of X tau. Why? Because what is an open subset here? If something is open subset here, I can just take that open set, whatever it is, don't put anything from any other Xi at all. Its intersection with Xi will be the same set. Intersection with any other Xi will be empty. Therefore, such a head is inside tau automatically. So each Ai belonging to tau i is already inside tau. So Xi tau i becomes a subspace of X tau. Because once Ai is there, Ai intersection Xi will be just Ai. Not only that, each Xi will be now open inside X and therefore it is also closed inside X because each Xi is a complement of other Xi. So this is the easiest picture of when you take disjoint union. This is also interesting but more interesting cases will be when things are not disjoint. That is where what you call as patching up topological spaces comes into picture. So I will not use the word patching up because patching up has different meanings at different places. So what I want to do is more interesting case is when union is not necessary disjoint. Can we still put a topology of meaningfully, you can always put a topology in discrete or discrete and so on. It has something to do with each subspace if there is already a topology for each of these subsets. So let us go step by step. Let us consider the following situation where X is union of Xi now. I have put around a cup here. So this is not necessarily disjoint. Now each Xi, X is simply topological space. Suppose that each Xi is given the subspace topology. X is topological space. These are subsets. So we know what they mean by the subspace topology. Further suppose that A contained inside X is open in X. If filled only if A intersection Xi is open in Xi for each Xi. So this is an extra assumption I am making. Start with a topological space X which is written as union of Xi and give a subspace topology and then assume this condition. It is not clear why this should be true at all. So this I have copied from whatever we did here. So that is why I am trying to put that condition here and see what happens. So all these depth are it is better to call it a name so that we do not have to keep on telling all these things every time. So in such a situation we will say X has the topology coherent with respect to the collection of subspaces Xi. So this is the definition of coherent. Coherent with collection of subspaces means this much. Maybe later on we will improve upon this definition and so on but right now this is definition. It is easily checked that X has coherent topology with respect to Xi in the following cases. Namely the first case when X itself was a disjoint union. We have actually copied that property of the disjoint union topology here. So there it is obvious. That is another easy example wherein each Xi is an open subset of X. So I recall take X, Xi are some subspaces, union is the whole thing X. But now I am assuming each Xi is open inside X. Then what happens? Suppose A intersection Xi is open in Xi. Automatically it will open in X also that is what we have seen. So each A intersection Xi is open in X. Their union will be A so that is also open. So the condition coherency whatever we have there I have verified it here. So these are two examples, easy examples of coherent topology. Let us introduce another important concept here which is relevant in this context, appropriate in this context and that will allow us to have more examples. Let A be any family of subsets of a topological space X. We say A is locally finite at a point X. It is like continuity definition. It is point wise I am defining first. Locally finite at a point. If there is a neighborhood U of X such that U intersects only finitely many members of A. Each point when you are given the neighborhood you can choose. The neighborhood automatically depends upon the point. The number of members which intersect also depends upon the point as well as the neighborhood. Once there is a neighborhood, all smaller neighborhoods will also satisfy that. So that is why I can put just a neighborhood then you can make it open neighborhood also because every neighborhood contains an open neighborhood. So there is no harm in putting neighborhood. If this happens at every point of X then you call A locally finite. That is why I have indicated the similarity between definition of continuity. Continuity at a point then continuity at all the points is continuous function. That is what we have defined for continuous functions for example. So similar local finiteness at each point and then local finiteness on the whole of X. So now I will give you an interesting example of coherent topology, the third example in SA. Now suppose X i is a locally finite collection of closed subspaces. So this is the extra condition. I am assuming X is equal to union of X i that is given already. So this is extra condition. But each time they are closed also. Each of them is closed also. Then local finiteness. So two extra condition 5 minutes. Then I claim that X i has the coherent topology with respect to X i. So let us check this one. I have to check it point wise. That is what I have to do. Sorry I have to coherent topology I have to prove A if A is such that A intersection X i is open in X i. Then I have to show that A is open in X. It is not point wise. Point wise is the definition of local finiteness. So start with the subset X, subset A of X such that A intersection X i is open in X i for each i. This just means that A intersection X i is equal to U i intersection X i where U i is open in X. This is the definition of subspace topology. The X i's are subspaces of X. Now given X belonging to A. So now I have to do local. I should produce an open subset of X. The set X belongs to that open set and that open set is contained inside A. That is what I have to do. So use the local finiteness and choose an open set U in X such that X is in U and U intersects only finitely many members of this family. You can just for a while you can just label them as X 1, X 2, X n. All right. So this is a picture. This X is there. I have found a U. This U intersects only finitely many of them all other X i's are far away. Out of which look at X where does X belong to? Here X belongs to X 1 also X 2 but it does not belong to X 3, X 4, X 5, other things. Okay. So that is important for me. So what I do? Start with a open set U such that this intersects only finitely many among the stem by relabeling if you want X 1, X 2, X n in that not in that order. You may assume that X is inside X i Y equal to 1, 2, 3 up to K and not in other X i's for i bigger than K. If it happens that it is inside all of the X i's it does not matter. It must be in one of them after all. Because it is inside U and U intersection union of all X i's is U out of which you have taken all those which intersect that one. So it must be inside 1 of 3. So some of them will be there. This part is non-empty. This part may be empty but it does not matter. So now we put V equal to our U whatever you have chosen intersect with finitely many of these U i's. What are U i's? U i's are open subsets of X such that X i intersection U i is a given open set. That is what A i's. X also must belong to all these U i's for i 1, 2 up to K. That follows. That follows. No, no that follows. You see what is that A intersection X i, it is an element of A first of all. If it is inside U i also. X is not in U i. X is in X i only. No, no. X must be inside U i. The ones which are in U i, if it is in X i and A, it will be in U i. This is equality. Yes. Make it U i and X i or whichever way. It is same thing. Correct. Thank you. Okay. So first of all it must be there. This X 1 X 2 X K is at least greater than or equal to 1. That is important. Otherwise this will be empty. You will not get any neighborhood. U intersection i r into 1 to K U i and throw away all these X i's. K plus 1 to n. Throw away means what? You have to compliment. Compliments are open subsets because each X i is open. Each X i is close subset. So union is a close subset. The compliment is also open subset. That is why you are intersecting the tag. It is the same thing as that V is an open subset now. Since X is not here and X is here, X is here. X will be inside V and V is contained inside A now because what is A? A is a union of A intersects all these X 1 X 2. It is contained inside X 1 X 2 X n itself and it does not interact with this part but it will be inside this one. So this V will be inside A. So history for every X inside A so V A is open. A typical way coherent topology is used is the following namely patching up continuous functions which you have been using in analysis. In one variable calculus you usually use the other one namely a continuous function is given in first closed interval and in the second closed interval the two closed intervals intersect only at one point. There the function is just defined that's all but then automatically it will be continuous on the whole thing. Only the function should be defined properly means in the intersection the value should be the same. So there you are taking finite intersections of closed sets but usually if you have in arbitrary union of open sets then this is possible. This will be used only when you go to R2, R3 and Rn and so on in the I am talking about in analysis. In one variable analysis you don't meet this one quite often. X period topological space X is union of Xi's and the topology on X is coherent with the family of subspecies of Xi. Now I am not making any assumption whether Xi's are open or closed and so on. Those three examples were there disjointness, openness or closeness with local finance. Under these three we know it's coherent but suppose any family's coherent like this has coherent topology and so on then given any function X to Y, saturated function okay and Y is in a topological space. F is continuous if and only restricted to each Xi's continuous is one line proof. How do you prove our definition of continuity? Take an open subset of Y say U, F inverse of U must be open inside X. How do you prove X is open? Use coherent topology intersect with each Xi you show that it is open. What is the intersection of F inverse of U with Xi? It is nothing but F restricted to Xi take the function take inverse of that of U. So suppose you call this as Fy then Fy inverse of U is nothing but F inverse of U intersect with Xi because Fy is nothing but F itself but restricted to Xi okay. So we have coherent topology has a property that something is open if filled only with intersection with each Xi is open in X. Thus coherent topology provides a method of constructing continuous functions on the whole space from continuous functions given on each piece okay. So this is what you have to remember the actually three cases we have now I said both cases because disjoint union was introduced as a motivating example okay. So both cases discussed in the previous remark are instances where the above theorem is applied in constructing continuous functions. Even you can include disjointness of that is obvious anyway nobody mentions that one but that is the starting point of all this discussion. Now let us move to somewhat more general situation. Like XPS said which is written as union of substrates how many I do not know whether they close no open I do not know just union of Xi but each Xi has a topology to Y. Now you see what I have done coherent topology I start the topology on X. Here I am just given topology on each Xi that is all okay. Can we put a topology on X which is coherent with the collection Xi to Y. Remember I repeat this one what is the meaning of topology you must have topology on X then the given topologies to Y must be subspaces after that that coherent condition is there one more namely something is open in tau if and only intersection with each Xi is open in X you know open in Xi. So all these can we do there is no topology given on tau given on X can we find such a tau is a question okay. So this is a really like a fundamental question but it does not seem to have an if and only if answers okay but that is lucky actually it will give you we can have theorems wherein there will be useful theorems suppose this happens then this is true suppose it is happened then that kind of theorems are more useful than if and only if theorems quite often. So we can can we put this so we would like to have at least some partial solution of this problem if not for one okay here is one such there are two conditions here actually so I have clubbed them together so do not get afraid because it is going to be too much okay. So let us go through slowly most of the things are repeated here X is union of Xi each Xi is an topology okay so that much is under underlying hypothesis now assume that the following compatibility conditions are satisfied there are two conditions here each of them has two of them okay in the in the brackets there are another set of conditions that is like that. So for each pair ij Xi intersection Xj is open this one condition is openness you replace by close that is another set of conditions so you can call it as cc1 and cc1 prime if you like okay each pair ij Xi intersection Xj is open in both Xi and Xj so this is the first condition the second condition is for each pair ij the two topologies on Xi intersection Xj induced from Xi tau i and Xj tau j must be the same these two there are two topologies they are the same okay see Xi intersection Xj is subspace of Xi it is all subspace of Xj okay both of them if there is a topology on X they will be subspace of the same X then by the reflect by the transitivity condition we have proved automatically this would have been the same I want to put a topology on on X then this condition is a must so that is why I have put this condition you understand because of the transitivity of substituting subspace topology on the intersection you can come from Xj or Xi two different ways they must be the same okay right now there is no topology on X I do not have that one but if I want to put it this is a must so better start with this condition of course this condition may not guarantee so I have some more conditions also here first one okay so the conclusion is now there exists a unique topology tau on X which is coherent with the collection Xi tau i so answer is positive so it is characterized by the property that each Xi is open inside X or in the other part when you have taken closed it will be closed subspace of X so as I have told you the condition cc2 is a must but this one this one is not a must we do not want everything to be open or closed or something we just want them to be you know subspaces and coherent topology is all that why Xis are some spaces you want them to be subspaces of a common topological space in the very first case what was that distant union we had such a thing right so the other two cases I am taking if all of them are such that intersections are open in both of them or intersections are closed in both of them then it is possible so this is the next case that we have okay the proof is very easy the only statement is long all that I do is exactly same way I defined for the disjoint union I am going to define tau tau is all a containing side such that intersection with each Xi is inside tau i all right when you verified that tau is a topology when the case that they are disjoint that disjointness was never used for the verification it is just the phenomena that if a is written as union of intersections then it is the same thing as intersections of unions and so on that is what that is what you have to verify okay so this is topology is no problem the above definition of tau is forced on us by the condition of coherence because if if something is open here a intersection Xi okay must be inside tau i for every i that is the coherency okay so I have to put that and I have put that much only that is enough that is the point therefore the uniqueness follows the tau must be like this if it coherence okay what is required here is that the verification of tau is a topology which I have already done whatever condition verification condition is forced on us we have put just that much that gives you coherence automatically but why topology that comes much easier that is a general phenomena okay so sometimes easy proves will you will stun you so better spend a little more and half a minute more on that so that we are not making any obvious mistakes okay so you have this one there is two other things which I have to prove suppose each Xi is such that Xi intersection Xi is open inside both Xi and Xj then I have to show that Xi itself is open similarly for close but that is very easy again because to fix i I want to show that Xi is open right what is the coherency condition intersection with each Xj must be open in Xj j equal to i also allowed X intersection Xi is X i itself that is open X intersection Xj is open in Xj is the condition given in cc1 that is all similarly open you replace by closeness that will give you cc1 the close for the closeness okay the finally one more thing I have to share why the topology original topology tau i is the subspace topology the definition of subspace topology is what a subset is open if and only if it can be written as intersection with some open set in X with respect to Xi okay so how do you check this one look at the definition of tau itself okay what is tau take any tau take any member of tau a intersection Xi is inside Xi is a condition so when you restrict it it is already contain tau they are elements of tau i you have to show that there are no more elements in tau i right so take an open set inside tau i that means something belong to tau i then I might show that it is a intersection you know something where this a is coming from a a sorry a is a is an open subset of X all right so that is the converse part suppose a equal to the X i is open in Xi in the first case you have two two different cases separately in the first case we have shown that Xi is open a is open in Xi Xi is open in X so a is open in Xi okay so it follows that each tau i is contained inside tau restrict to Xi and conversely okay in the second case Xi is closed in X right take take demorgan law use complement Xi minus a is closed in Xi because I started the a open inside Xi so Xi minus a is closed in Xi and Xi is closed in X so this is closed in X therefore it is complement X minus Xi minus a right it is open in X but what is this set you have to see that this set is precisely a is equal to Xi intersection with this one this is a closed subset this is open subset now intersection with this one will be precisely this one Xi intersection a X minus Xi minus a okay he is purely sectarian thing a is a subset of Xi started so it is contained in Xi and it is contained inside this X minus X minus a you are throwing away and then throwing away Xi itself Xi minus a itself so a will come back okay only a will remain only a will remain is the point so this intersection is precisely equal to a if this is the case a belongs to tau of X tau restricted to X okay so that proves the coherency of this topology completely there are three conditions we have to verify what were there that each Xi a tau i is the subspace topology right whatsoever tau is a topology restricted to each Xi gives you the original topology tau i okay and then depending upon closeness and openness you have to show that each Xi is open or each Xi is closed two different cases so here is an example one of the useful examples in this situation is when X the disjoint unit itself okay then the necessary condition cc1 and cc2 are automatically satisfied so we have tau the coherent topology with respect to tau i in particular each Xi is both open and closed okay so i have already indicated i am just repeating this here here is the place wherein each Xi is both open and closed will come because intersections are all empty which are both open and closed okay so let us stop here today