Is Every Manifold Regular? A Proof Using the Hausdorff Condition

[radou]

Homework Statement

As the title suggests, I need to show that every manifold is regular.

There's probably something wrong with my proof, since I didn't use the Hausdorff condition, and the book almost explicitly states to do so.

The Attempt at a Solution

So, a m-manifold is a second-countable Hausdorff space X such that every point x in X has a neighborhood which is homeomorphic with an open subset of R^m.

Let x be in X. Let U be a neighborhood of x such that f : U --> V is a homeomorphism, and V is a subset of R^m. Since R^m is regular, for any point y in V, there exists a neighborhood V' of y such that Cl(V') is contained in V. Now, since f^-1(V')\subseteqf^-1(Cl(V'))\subseteqCl(f^-1(V'))\subseteqf^-1(V) = U, we conclude that f^-1(V') is the desired neighborhood around x. Hence X is regular.

I feel there's a bit of a misunderstanding here which is actually bugging me constantly (in a few other problems, too), so I'd like to straighten it out.

 

[micromass]

Is this an exercise in Munkres? Can you tell me which one it is, because I can't find it.

As for your proof. Regularity (in Munkres) also means that the singletons need to be closed, so you still need to show that. This will use the Hausdorff condition...

But there is something else that is bugging me. You need to show that for every neighbourhood W of x, there exists a V such that cl(V)\subseteq W. But you only show it for the neighbourhood U. I mean: you take an arbitrary neighbourhood U and you assume the existence of a homeomorphism f:U-->V. But if U is arbitrary, then you cannot assume that. You'll need to add some simple argument...

 

[micromass]

Something else, you said that Cl(f^{-1}(V^\prime)\subseteq f^{-1}(V). Why is this?

 

[radou]

It's section 36, page 227, exercise 1. Manifolds are an important class of objects, so I decided to go through this section too (it is starred, though), and some of the results are used in the study of paracompactness, too (starred too, but I think they may be worth the effort).

Yes, you're right, I completely forgot that I took a specific neighborhood U of x (the one homeomorphic to V), and not an arbitrary one. But I can correct that, take any neighborhood U' of x, and intersect it with U - the problem is solved.

But yes, it seems that Cl(f^-1(V')) need not necessarily lie in f^-1(V)... Uhh.

 

[radou]

The problem is, when I'm dealing with homeomorphisms, I get confused by the inverse image and the inverse sometimes, because of notation. But then, the inverse image of a set actually equals the value of the inverse function at this set, right?

Edit: what I said may be tremendously stupid, but I'd like to make it clear once and for all.

 

[micromass]

No, what you said is correct. If f is a homeomorphism then the inverse image is just the image of the inverse.

Now, I am sure that a direct proof of this is possible. But there is an easier proof. Think of some compactness property...

 

[radou]

No, what you said is correct. If f is a homeomorphism then the inverse image is just the image of the inverse.

Now, I am sure that a direct proof of this is possible. But there is an easier proof. Think of some compactness property...

Hm, ok...

As for your proof. Regularity (in Munkres) also means that the singletons need to be closed, so you still need to show that. This will use the Hausdorff condition...

Oh yes, of course! Since X is Hausdorff, finite-point sets are closed.

 

[micromass]

If you try a direct proof, then maybe the following property can come in handy: if f:X\rightarrow Y is a homeomorphism, then Cl_X f^{-1}(U)=f^{-1}(Cl_Y U), but you need to be careful which closure you take (that's why I added the subscripts, which indicate in which space you take the closures)

 

[micromass]

Oh yes, of course! Since X is Hausdorff, finite-point sets are closed.

[/QUOTE]

Yes, when I wrote that comment a few minutes ago, then I believed that you needed Hausdorff for that.
But, actually, any locally Euclidean space is T1. So that's not where you use Hausdorff...

 

[micromass]

Maybe it helps to take a look at the supplementary exercises on page 316...

 

[radou]

I think I got it. Is local compactness a topological property? i.e. preserved under a homeomorphism?

 

[micromass]

I think I got it. Is local compactness a topological property? i.e. preserved under a homeomorphism?

Yes, the easiest way is through using local compactness! And local compactness is indeed a topological property!

 

[radou]

Excellent! It's really easy then.

Since R^m is locally compact Hausdorff, and V is an open subspace of R^m, it follows that V is locally compact. Hence, U must be locally compact too, since it is homeomorphic to V. The condition for local compactness I used is the same as the one for regularity basically. It applies to the subspace U. From the definition of the subspace topology, our needed condition of regularity follows, and hence X is regular. And metrizable!

 

[micromass]

Hmm, I have a feeling that you're maybe going a bit to fast there...

That U is regular is very easy to show, you don't need local compactness for that: V is regular as subspace of \mathbb{R}^m, and U is regular since it is homeomorph to V.

But how do you obtain regularity of X from the regularity of U??

I think you need to show that X is local compact instead of proving that U is local compact...


It could be that I misread your proof, so I'm sorry if I did that...

 

[radou]

Hmm, I have a feeling that you're maybe going a bit to fast there...

That U is regular is very easy to show, you don't need local compactness for that: V is regular as subspace of \mathbb{R}^m, and U is regular since it is homeomorph to V.

But how do you obtain regularity of X from the regularity of U??

I think you need to show that X is local compact instead of proving that U is local compact...It could be that I misread your proof, so I'm sorry if I did that...

Oh, I forgot that regularity is hereditary when looking at subspaces... So I used a theorem to be sure, nevermind. And I even managed to forget that regularity is a topological property.

But still, I basically don't see where my proof went wrong.

First take, x in X. Then a specific neighborhood such that f : U --> V\subseteqR^m is a homeomorphism. Now, conclude that V is locally compact. Hence U is locally compact, too. This means that we can find, given a neighborhood U' of X (open in the subspace topology of U) a neighborhood V' of x such that Cl(V')\subseteqU'. Now, take any neighborhood W of x open in X. Intersect it with U, by definition of the subspace topology, this set is open in U. Apply local compactness to x and U now. You get V' such that Cl(V') is contained in U. Since V' is open in U and U is open in X, V' is open in X. And it is contained in W... Hmm but I think I see the problem now.. Cl(V') is not necessarily closed in X, right? Although it equals the intersection of some closed set of X with U...

Edit: sorry, I started off that I don't see where my proof is wrong, and I end up with not being sure about anything. This is because I think when I write right now, I find it easier for some reason..

 

[micromass]

Hmm but I think I see the problem now.. Cl(V') is not necessarily closed in X, right? Although it equals the intersection of some closed set of X with U...

Indeed, Cl(V') is the closure in U, but not the closure in X. So it's not necessairily closed. However, if Cl(V') were to be compact, then it would be closed in X!

But it might be easier to show that X is local compact, this is not that hard to show.
Now, since X is local compact and Haudorff, it follows immediately that X is regular (even completely regular!)

 

[radou]

Indeed, Cl(V') is the closure in U, but not the closure in X. So it's not necessairily closed. However, if Cl(V') were to be compact, then it would be closed in X!

Hey wait, but it just occurred to me, according to Theorem 29.2., which I was using, Cl(V') must be compact! And hence closed, as a subspace of a Hausdorff space!

 

[micromass]

Yes, that's entirely correct. Using theorem 29.2 didn't even occur to me, so good find!
Note that you used Hausdorffness two times here: in theorem 29.2 and when proving that a compact space is closed...

 

[radou]

Yes, that's entirely correct. Using theorem 29.2 didn't even occur to me, so good find!
Note that you used Hausdorffness two times here: in theorem 29.2 and when proving that a compact space is closed...

Yes, I found Theorem 29.2. to be useful on a few other occasions, too... Since it's almost the same as Lema 31.1., for regularity...

But it might be easier to show that X is local compact, this is not that hard to show.
Now, since X is local compact and Haudorff, it follows immediately that X is regular (even completely regular!)

Hmm, could we prove that X is locally compact by using the fact that X is a m-manifold to construct a homeomorphism of X and R^J. Since we could, by choosing for every pair (x, Ux) (where Ux is a neighborhood of x homeomorphic to an open subser of R^m), choose a homeomorphism fj : Ux --> V, and then obtain a continuous function F : X --> R^J with F(x) = ((fj(x))? If R^J is given the product topology, F is continuous, but it's too much (I assume) to hope that it could be a heomeomorphism. R^J is compact, by the Tychonoff theorem, and hence locally compact...But I guess I'm overcomplicating, it's just something that flashed my mind at first. I don't see an easy straightforward way right now..

 

[radou]

Ahh, I just realized that this is nonsence. Since every function fj should be defined on X, and it's not enough on the open subsets Ux...

 

[micromass]

Local compactness is shown this way:

Take x in X. We wish to find a compact neighbourhood around x.
Take a neighbourhood U around x, such that U is homeomorph with an open set V\subseteq \mathbb{R}^n. Call f the homeomorphism.
Since \mathbb{R}^n is locally compact, there exists a compact neighbourhood K of f(x) such that f(x)\in K\subseteq V. It follows that x\in f^{-1}(K)\subseteq V.
We have that f^{-1}(K) is compact since f is a homeomorphism. Also, f^{-1}(K) is a neighbourhood of x in the topological space U. But since U is open, we also have that f^{-1}(K) is a neighbourhood of x in the topological space X.
Thus f^{-1}(K) is our compact neighbourhood...

 

[radou]

OK, I get it. Actually, it's quite easy. :)

Btw, one single point I don't understand. You said since R^n is locally compact, there exists a compact neighborhood K of f(x).

I use two things when I think about local compactness - a) the definition - i.e. a space X is locally compact if for any x in X there is a compact subspace of X containing a neighborhood of X; b) Theorem 29.2.

Now, my way of reasoning would prove the same, only that I'd apply Theorem 29.2. to f(x) and V in order to obtain a neighborhood of f(x) whose closure is compact and contained in V. So not exactly directly a compact neighborhood (at least not an open one). Now, if we denote this neighborhood with K, then f^-1(Cl(K)) is compact, closed and contained in U, and clearly f^-1(K) is open and contained in f^-1(Cl(K)), and these are the neighborhoods we were looking for.

 

[micromass]

Now, my way of reasoning would prove the same, only that I'd apply Theorem 29.2. to f(x) and V in order to obtain a neighborhood of f(x) whose closure is compact and contained in V. So not exactly directly a compact neighborhood (at least not an open one). Now, if we denote this neighborhood with K, then f^-1(Cl(K)) is compact, closed and contained in U, and clearly f^-1(K) is open and contained in f^-1(Cl(K)), and these are the neighborhoods we were looking for.

Indeed, this would also be a nice proof!

 

[radou]

Indeed, this would also be a nice proof!

It's basically the same thing, I was only worrying that I misunderstood something... OK, thanks a lot for all, once again!

Edit: if I ever visit Belgium, I owe you at least a dozen six-packs of beer...

 

[micromass]

Haha! You can give them to me if I ever visit Croatia