Basic Topology 12 | Intermediate Compact and Open Sets [dark version]

Added: 2026-06-02

[00:00:00]Hello and welcome back to basic topology the video series where we talk a lot about open closed and compact sets and indeed in today's part 12 we will talk about intermediate compact sets which means that I want to put a compact set in between two different sets and this result is something we can use later on to prove Urizison's lema however as always before we start with the details I first want to thank all the nice people who support this channel on steady here on YouTube or via other means and please don't forget that you can use the link in the description to download the additional material for all the videos. So for example there you find the full PDF version of this video course here and then without further ado let's immediately fix the assumption of today which is a locally compact housetop space. This means in the space we always have compact neighborhoods and the separation property. And why these two properties are important you will see later in the proof of the statement.

[00:01:05]But first let's formulate the statement which starts with an open set U in our topological space X. And moreover in addition to that we also assume that a compact set inside U is given. And now the statement of the theorem will be that we always find two intermediate sets and the first one is an open set that still contains K. And then the second set is a supererset of that but also compact and still contained in U.

[00:01:35]So you could say the statement is that we can make the compact set K larger without leaving the open set U. and making the set larger here is described with the open neighborhood of the whole set K. The whole thing seems reasonable but as we will see the proof will require some work. But before we go to the proof let's first clearly formulate this statement here. It's a proposition that holds in a locally compact housetop space. So X is our topological space and then we fix an open set U and a compact set K. And most importantly here is that K lies inside our open set U. And then as stated in the picture, we find another open set that still contains K.

[00:02:22]And in order to keep it simple, let's call this open set A. And now actually a cannot be equal to the open set U because we want that the closure of A is a compact set. Indeed, this should be the compact set that still lies completely in U. This means now we can just write down a whole chain of subset relations. So first k should be contained in a and then the closure of a could be bigger than a. But most importantly we still have that the closure is contained in u. And that's it. These are our two intermediate sets an open one and a compact one. So in order to show this proposition we have to construct this open set a and the closure of a. And for this construction, we can definitely use our locally compact property because then we have compact neighborhoods. So this will be the first idea to make our compact set K larger. So we just fix a point X from the compact set K. And then we know that we have a compact neighborhood of X. And there we need a good name. Let's call this compact neighborhood C with index X. And now, please don't forget by definition of a neighborhood, we also know that there's an open set that contains X and is still completely inside CX. And now this open set around X I want to call VX. And here the index X is important because we know that this picture here works for every point X in the compact set K. So there you see this is where we already use the locally compact property of our topological space. In fact, it's quite important that we have the existence of both sets vx and cx and both sets are neighborhoods for x. So the point x lies in both of them. And moreover cx is always a super set of vx which means we immediately get an open cover and a compact cover for our set k. So we can just write that K is covered by the union of the sets Vx and the sets CX.

[00:04:35]And now since K is a compact set, we can also go to a finite subcover. And this is what we can use in the next step.

[00:04:43]There we just have a finite cover for K.

[00:04:46]And to write it down, let's simply enumerate the points. So let's say we have I that goes from 1 to N. So we only need end points from K to cover the whole compact set K. And then obviously this also works for the super sets CXI which implies that here on the right hand side we still have a compact set.

[00:05:08]In particular we know in the house of space it's also a closed set as well.

[00:05:14]And there you might think that we are already done because we could take this union of open sets as our set A.

[00:05:22]However, this can only work if this union here stays inside the given set U.

[00:05:29]And obviously in general, we cannot guarantee that. But we definitely have it in one case. Namely, in the case that our open set U is just the whole space X. Naturally, under this assumption, our union cannot leave the open set U. And then we immediately have our chain of subsets just by taking A as the union of the V's. And moreover the closure of A is then given as the union of the closures of the V sets. And this immediately implies that we have a closed set here that is contained in a compact set. So we simply know by part five that this one is also a compact set. And these are all the conditions we wanted to satisfy for our set A.

[00:06:15]However, as already stated, this whole construction here only works if the set U is large enough, namely the whole space X. In the general case, there's simply also something outside which we have to take care of. This means we have to consider the complement of U in X. So in the picture here you could just put in the space x around u. So what we have there around u is the complement of u and we know it's not empty in this case.

[00:06:48]So we definitely have a point p which does not lie in u and then we can use the housetop property to separate points. This means for any point q in the compact set k we can find open neighborhoods that don't intersect. So formally this means we have P in the set U C and Q in K and then we find the open neighborhoods for both points. This means we have an open set where P lies in and an open set where Q lies in. And now the housetop property tells us that we can always find open neighborhoods that are disjoint. And maybe let's call the one open set B and the other one D.

[00:07:32]And to make it clear that the points go into this choice, we put them into the index. And now this whole construction of the open sets helps us again because we get an open cover of the compact set K again. Indeed, we can just go through all the points Q of K and take the sets DPQ.

[00:07:53]So clearly we cover our compact set with open sets and then we know we have a finite sub cover. So this is not so complicated. We can just enumerate finitely many points again and let's say we go from one to M. So we have finitely many points in K and we call them QJ.

[00:08:13]And this also implies that we can look at finitely many neighborhoods of the given point P in the complement of U. In fact, what we can do is to look at the finite intersection of these open sets.

[00:08:28]By the definition of the topology, we know this is still an open set and therefore a neighborhood of the point P as well. And moreover, it still has to be disjoint to all the D sets.

[00:08:41]Therefore, maybe let's simplify the notation and let's call this intersection just B with index P. And on the right hand side here, the union of the D sets, we can just call D with index P. This makes everything simpler because we can just say that the intersection of both sets here is empty.

[00:09:02]So this is how it looks in the picture.

[00:09:04]And of course this is quite helpful because we can do it for any point P in our complement of U. So in some sense you could say that the house of property extends to a compact set K. Indeed, for any point P, we find an open neighborhood and we find the whole open neighborhood for the set K such that both sets are still disjoint. So let's quickly formulate it as a fact. For any point P, we find disjoint open sets and we call them BP and DP and their DP just enlarges our compact set K. And obviously the important part here is that these two neighborhoods are disjoint. And now clearly we can also go to the closure of the set DP and then we know by the disjoint property that P cannot lie in the closure of DP. So you can say that P has a safe distance from the closure of DP. However, this does not mean that DP has no intersection with UC. In fact, even in our sketch before, we see that the P goes over the set U. Therefore, if we look at the closure of DP, what we get out of this intersection is a closed set that might not be empty. Okay. And now I want to combine this closed set with the covering we had before for our compact set K. So there please recall there we had the compact neighborhoods called CXI and there we already know they nicely cover the whole compact set K. And also there we cannot guarantee that the whole cover stays inside the set U. However, this is all not a big problem because we can just look at the intersection as before. In other words, we also put this compact set into our intersection.

[00:11:01]Therefore, what remains in this intersection here is just the small part outside of the set U. And in order to keep it simple, let's call this new set WP. And because we have an intersection of closed sets, it's definitely a closed set as well. And since it's also contained in a compact set, we conclude it's also compact. And as already mentioned, because P has a safe distance from the neighborhood DP, we know that P is not an element of WP. However, this immediately implies if we look at the intersection of all possible sets WP, we get an empty intersection. So we have an infinite intersection here where P goes through all possible points of the complement of U. And now this intersection has to be empty because otherwise there would be a point P that lies in its corresponding set WP. This is clear because any point in the intersection has to lie in all the sets WP. Okay. So here we have an infinite intersection of compact sets and there you might already guess this can be reduced to a finite intersection simply because if we go to the complements of the sets we have open sets and an open cover of the whole space. So in particular we also have an open cover of any compact subset of X.

[00:12:31]And now if you want we can just fix any point P 0 and cover the compact set WP 0. And indeed in this video we have done this already many times. We can find a finite sub covering. So let's also say here the compact set is covered by M elements where J goes from 1 to M. But here please keep in mind on the right hand side we still have the complements of the W sets. So the meaning of that is that WP0 lies completely in the complements which implies that the intersection with the original sets WPJ has to be empty. So you see suddenly our infinite intersection is just a finite intersection with M +1 elements. So by going back to the definition of WP, it means that we only need finitely many of the D neighborhoods. So let's simply put the definition of WPJ into this intersection. So we just write DP0 intersected with DP1 and so on. And then finally in the end we also have the intersection with the complement of U.

[00:13:46]And now the conclusion from before is that this finite intersection is already the empty set. Hence we can just split this set up into two parts and then the left hand side has no intersection with the complement of U. In other words, this is exactly what we want. We have a set that lies completely in the set U.

[00:14:09]Moreover, it's an intersection of closed sets. So the result is also a closed set. And since this closed set lies inside a compact set, we also know it's a compact set as well. And there we have it. This is one part we wanted. And now we can finally define the set A.

[00:14:30]However, please recall the set A has to be an open set. So we have to change the sets from above a little bit. So first instead of the compact set CXI we can use our open neighborhoods we called VXI and on the other hand we don't need to use the closures of our neighborhoods because we can just use the open neighborhoods themselves.

[00:14:55]So we have a finite combination of open sets here. So the result is definitely also an open set. Moreover, we also know that every part in the intersection here is a super set of our compact set K.

[00:15:09]Therefore, K is contained in the set A as we want to have it. And therefore, the only thing missing is that the closure of A is also compact and lies completely in our open set U. However, there it's easy to check that we can pull the closure to the parts of the intersection and we get a super set.

[00:15:32]This is a general result and good enough for us because we don't need the equality here simply because we know that this set here with the closures is contained in this closed compact set.

[00:15:46]This means it's also compact and it has no intersection with the complement of u. This means this is the compact set that we want that lies completely in our open set U. And that's it. This is the whole construction of our two intermediate sets. The one is the open set that extends our compact set K. And the other one is the compact set that still lies in our open set U. So now we have proven this nice result. And what we can do with that I show you in the next videos. So, I really hope I meet you there again and have a nice day.

[00:16:24]Bye-bye.

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