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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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Hausdorff and paracompactness imply regularity and even normality of a space. –  Stefan Hamcke Aug 18 '13 at 22:08
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@StefanH. He's trying to prove paracompactness from regularity so that would be circular. –  Daniel Rust Aug 18 '13 at 22:09

1 Answer 1

up vote 8 down vote accepted

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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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. –  Ben Snow Aug 18 '13 at 22:29

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