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First, some clarification of the definition of a manifold that I'm using:

A manifold $M$ is a Hausdorff, locally Euclidean and second countable topological space.

Now, I am trying to prove that Manifolds are paracompact, and I have established most of the details for the proof from the link above after getting so far under my own steam. The only hole I have, however, is the assertion that manifolds are regular. From what I can infer, this comes from the properties of being locally path-connected and Hausdorff, but I cannot make the leap from those two properties to the required regularity to complete the proof.

Apologies for the probably quite elementary question; I'm an applied mathematician, and am aiming to be well-read in a range of mathematical topics for my own interest, and while I have a textbook I'm working through on the topic of differentiable manifolds, there's still a certain unfamiliarity with the methods employed in certain pure maths topics.

Perhaps even a hint would be best, as I really do like to attempt to grasp these things by myself as much as possible, but I really just can't seem to get this result out... Thank you in advance for whatever help you can give me.

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  • $\begingroup$ Hausdorff and paracompactness imply regularity and even normality of a space. $\endgroup$ Aug 18 '13 at 22:08
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    $\begingroup$ @StefanH. He's trying to prove paracompactness from regularity so that would be circular. $\endgroup$
    – Dan Rust
    Aug 18 '13 at 22:09
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Note that locally Euclidean and Hausdorff implies locally compact Hausdorff. So the following more general result will answer your question.

Every locally compact Hausdorff space is completely regular. A proof can be found here.

The main idea is that the locally compact Hausdorff spaces are precisely the spaces which admit a one-point (or "Alexandroff") Hausdorff compactification. Now compact Hausdorff spaces are normal, hence completely regular. Normality need not be inherited by an arbitrary subspace, but complete regularity is.

[Note: in general I am a fan of the convention that "compact" and "locally compact" include the Hausdorff condition. So as to be maximally transparent, I am -- clearly -- not imposing that convention here.]

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  • $\begingroup$ Ah, I should've thought to use locally compact, I already proved that one earlier. I tried a lot of complicated routes to prove things but I didn't think of bringing in another property like that. Thanks so much for taking the time to answer, I'll be sure to have a look and get my head around all of this. $\endgroup$
    – Ben Snow
    Aug 18 '13 at 22:29
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Pete's answer is already sufficient for anyone who wants to work out the details of the essential lemma that locally Euclidean and Hausdorff implies locally compact Hausdorff on their own, so I wanted to add in a proof of that lemma for completeness of this page.

Proof: Let $M$ be locally Euclidean. We claim that for each point $x\in M$ there is a compact set $K\subset X$ and a neighborhood $U\subset K$ of $x$. Fix $x\in M$; since $M$ is locally Euclidean, there is a neighborhood $N$ of $x$, an open set $V\subset \mathbb R^n$, and a homeomorphism $f:N\to V$. Choose some ball $B\subset V$ centered at $f(x)$ whose closure $\overline B$ lies entirely inside of $V$, and consider the images $f^{-1}(B),f^{-1}(\overline B)$ of these two sets under the inverse homeomorphism $f^{-1}:V\to N$. Since $f^{-1}$ is an open map, $f^{-1}(B)$ is a neighborhood of $x$; since $f^{-1}$ is continuous, $f^{-1}(\overline B)$ is compact in $M$ and $f^{-1}(B)\subset f^{-1}(\overline B)$. This proves the claim.

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