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Let's say we call any space locally homeomorphic to $\mathbf{R}^n$ a real topological manifold. I want to find an example of a topological manifold with the above definition which is not Hausdorff and not locally compact (every point has a neighbourhood whose closure is compact).

To do this I tried to take the real line and add origins to it (I didn't bother to specify how many, just repeating the same procedure of taking $\mathbf{R}\times\{a\}$ and $\mathbf{R}\times\{b\}$ and forming the product space, identifying all non-origin points $(x,a)\sim(x,b)$ for all nonzero $x$).

It is pretty easy to show that this is locally homeomorphic to $\mathbf{R},$ and easy to see that it is not a Hausdorff space due to any neighbourhood of the origins intersecting, but is it locally compact?

I want to prove this using only the most basic topological concepts possible.

(Reference is Abraham & Marsden's Foundations of Mechanics Exercise 1.1D, page 53 here - some notes online and on Wikipedia say that all topological manifolds are locally compact since they must locally resemble a Euclidean space by definition, but their definition also commonly requires a topological manifold to be defined as a locally Euclidean Hausdorff space, whereas the line with two origins is not Hausdorff.)

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    $\begingroup$ Correct me if I am wrong, but if a space is locally homeomorphic to $\Bbb R^n$, then every point has a neighbourhood homeomorphic to $\Bbb R^n$ and this is a locally compact. But if every point of a space has a locally compact neighbourhood, then it is locally compact, right? $\endgroup$ – Matthias Klupsch Aug 24 '17 at 7:01
  • $\begingroup$ That was my first intuition but then I thought that I don't know why an author would put an impossible problem in a textbook. I don't know if there is something to do with the Hausdorff condition being dropped in some definitions but required in others. $\endgroup$ – Hobbyist Aug 24 '17 at 7:03
  • $\begingroup$ Describe the nhoods of a. It is locally compact but not locally compact Hausdorff. It gives an example of a compact set that is not closed. $\endgroup$ – William Elliot Aug 24 '17 at 7:45
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Let us note first of all the definition we are dealing with according to your reference:

A topological space is called locally compact if every point admits a neighbourhood whose closure is compact.

As it stands, the space you have in mind, is locally compact. For every non-zero point $p$, you simply take a closed interval $[p - \varepsilon, p + \varepsilon]$ and similarly for the origins, the image of $[-\varepsilon, \varepsilon] \times \{a\} \cup [-\varepsilon, \varepsilon] \times \{b\} $ is a compact and closed neighbourhood for both origins in your quotient space ($\varepsilon > 0$).

As I pointed out in my comment, for a space locally homeomorphic to $\Bbb R^n$, every point has a compact neighbourhood. However, the definition in your reference asks for a closed compact neighbourhood and this is where we can attack the problem.

Take $X = \Bbb R \times \Bbb N / \sim$ where $(x,n)\sim (x',n') $ if $x = x'$ and $n \neq 0 \neq n'$. You thus get the reals with infinitely many origins. Now, the closure of any neighbourhood of one origin contains all the origins. But these form a discrete set. A space which contains an infinite discrete subset is not compact, thus the closure of any neighbourhood of any of the origins is not compact and you have obtained your example.

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  • $\begingroup$ Ah, now I see the catch. Thanks, you've been a big help. Just so I have a better understanding of your answer, is there a difference between your identification and what I had if I made that identification infinitely many times, and if so, how? It seems that the key step in your answer is being able to get the closure of the neighbourhood of any origin contain all the others, but I don't quite see that with the identification I made in my first attempt. $\endgroup$ – Hobbyist Aug 24 '17 at 9:18
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    $\begingroup$ I do not think there should be any difference, except that the process of taking iterated quotient spaces infinitely many times seems to be formally a lot more involved. And yes, you got that key step right. This comes essentially from the failing of the Hausdorff condition: Every pair of neighbourhoods of different origins intersect nontrivially. In other words any one origin lies in the closure of every neighbourhood of any other origin. $\endgroup$ – Matthias Klupsch Aug 24 '17 at 10:01

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