# Is the line with multiple origins locally compact

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.)

• 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? – Matthias Klupsch Aug 24 '17 at 7:01
• 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. – Hobbyist Aug 24 '17 at 7:03
• 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. – William Elliot Aug 24 '17 at 7:45

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.