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Why every non-discrete locally compact group contains a nontrivial convergent sequence?

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Please only use $\LaTeX$ to format mathematics. –  Arthur Fischer Dec 23 '12 at 18:39
    
in myresearch i got it –  Wreza Shafaghi Dec 23 '12 at 19:16
    
This seems like the sort of question whose answer might be found in the first couple of chapters of Differential Geometry, Lie Groups, and Symmetric Spaces by Helgason. (I don't have a copy with me or else I'd have checked.) –  Neal Dec 23 '12 at 20:50
    
It seems that non-discreteness implies that any neighborhood of identity contains infinitely many distinct elements, and compactness allows you to choose a convergent sequence out of them... –  user53153 Dec 23 '12 at 21:58
    
@PavelM No, as compact does not imply sequentially compact in general. –  Henno Brandsma Dec 23 '12 at 21:58
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1 Answer 1

This is true. The difficulty is the case in which the group is not metrizable: Pavel M (and Henno Brandsma) noted that if $G$ is a metrizable locally compact group without isolated points then every neighborhood $U$ of the neutral element is infinite, so it suffices to choose an infinite sequence in a compact $U$ and extract a convergent subsequence.

In the general case, I doubt that there is a simple proof, although I think the following argument is far from optimal.

  1. Let $G$ be a locally compact group and let $U$ be a compact symmetric neighborhood of the identity. Then $H = \bigcup_{n=1}^\infty U^n$ is a $\sigma$-compact clopen subgroup of $G$.

    If $H$ is metrizable then $U$ must be metrizable Pavel M's observation does the job.

    If $H$ is not metrizable, then it contains a compact normal subgroup $K$ such that $H/K$ is metrizable. See Hewitt-Ross, Abstract Harmonic analysis I, Theorem 8.7. Note that $K$ can't be finite (or even metrizable) because otherwise $H$ would have to be metrizable. Therefore $K$ has no isolated points.

  2. Kuz'minov proved the remarkable fact that a compact group $K$ is a dyadic space, i.e., a continuous image of the Cantor cube $\{0,1\}^\kappa$ for some $\kappa$. See Chapter III of Stevo Todorcevic, Topics in topology for a proof.

  3. Katetov and Efimov independently proved that in a dyadic space every non-isolated point is the limit of a non-trivial sequence. See Cor. 2 on page 301 of Engelking, Cartesian products and dyadic spaces, Fund. Math. 57 (1964), 287-304.

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