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Math Help - Topological group/locally compact subgroup

  1. #1
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    Topological group/locally compact subgroup

    Proposition: Let G be a Hausdorff topological group. Then a subgroup H of G that is locally compact (in the subspace topology) is moreover closed. In particular, every discrete subgroup of G is closed.

    I've been struggling over the proof of this all day. I have been reading a proof in Fourier analysis by Ramakrishnan & Valenza and I understand it all except for this one point. It starts off by using local compactness of H and letting K\subset H be a compact neighbourhood of the identity element e in the subspace topology of H which is fine. Then it concludes therefore there is a closed neighbourhood U of e in G such that K = H \cap U. How on earth does one achieve this? I mean, its easy to find a closed set U with that, but how do I also make sure U is a neighbourhood of e?
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    I've been able to provide a fix for the proof given in the book. I'm fairly sure the author made a thinko. The proof can be fixed since it is enough to prove the theorem for H dense in G. For instance if we can prove that H is closed in \overline{H}, then its also closed in G.

    So assuming that \overline{H} = G its easy to prove the statement I was having trouble with:

    Let K\subset H be a compact neighbourhood of e\in H (in the subspace topology H).

    That K is a neighbourhood of e in H means that there exists an open neighbourhood U \subset G of e such that U \cap H \subset K.

    Because U is open it is easy to show that for any X \subset G we have that U \cap \overline X \subset \overline{U \cap X}. In particular we have U \cap \overline H \subset \overline{U \cap H} \subset \overline K = K.

    Since H is dense in G this means that U \cap \overline{H} = U \cap G = U \subset K and so K is a closed neighbourhood of e in G such that K\cap H = K.
    Last edited by Aradesh; October 27th 2010 at 11:13 AM.
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    I've worked out that the book was right after all - though the reasoning is not obvious (to me). The claim is: Let G be a Hausdorff space and H \subset G is a locally compact subspace. If x \in H then there is a closed neighbourhood C \subset G of x such that H\cap C is a compact neighbourhood of x in the subspace topology of H.

    Proof of claim: As H is locally compact we have a compact n'hood K\subset H of x. This means that K is compact in both H and G and there is an open U \subset G containing x such that H \cap U \subset K. Now H is locally compact and Hausdorff and this means that it is a regular space. So since H \cap U is an open n'hood of x in the regular space H, there is an open n'hood V \subset G of x such that H \cap \overline V \subset H \cap U. It is easy to check that C = K \cup \overline V \subset G is such a closed n'hood of x that we desired.

    Phew.
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    Quote Originally Posted by Aradesh View Post
    I've worked out that the book was right after all - though the reasoning is not obvious (to me). The claim is: Let G be a Hausdorff space and H \subset G is a locally compact subspace. If x \in H then there is a closed neighbourhood C \subset G of x such that H\cap C is a compact neighbourhood of x in the subspace topology of H.

    Proof of claim: As H is locally compact we have a compact n'hood K\subset H of x. This means that K is compact in both H and G and there is an open U \subset G containing x such that H \cap U \subset K. Now H is locally compact and Hausdorff and this means that it is a regular space. So since H \cap U is an open n'hood of x in the regular space H, there is an open n'hood V \subset G of x such that H \cap \overline V \subset H \cap U. It is easy to check that C = K \cup \overline V \subset G is our desired closed n'hood of x.

    Phew.
    This is good, but don't you meant to say that x has a precompact neighborhood?
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    I'm not quite sure what you mean - I'm not asking that C be compact for instance. To see where I got stuck and why I wanted to work this out see on page 8 proposition 1.6 in google books link.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Aradesh View Post
    I'm not quite sure what you mean - I'm not asking that C be compact for instance. To see where I got stuck and why I wanted to work this out see on page 8 proposition 1.6 in google books link.
    Ah, no. My problem was missing the Hausdorffness of the space. For, as I'm sure you know there are multiple definitions of locally compact (every point has a compact neighborhood and every point has a precompact neighborhood being two) which aren't in general equivalent, but are equivalent in Hausdorff spaces.
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