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Is the following sentence is true?

Each complete, separable and $\sigma$ - compact metric space is locally compact.

I suppose (but I'm not sure) it is a truth, becouse it was evidently used in the paper of Łukasz Stettner "Remarks on Ergodic Conditions of Markov Processes on Polish Spaces"(108 p.) which I am studyng now.

full text of this work - http://www-bcf.usc.edu/~lototsky/InfDimErg/Stettner-InfDimMarkProc.pdf

Dawid C.
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1 Answers1

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For a counterexample let $e_i, i=1, 2, \ldots$ be the standard unit vectors in $\ell^2$, and $X$ the union of the line segments $L_i$ joining 0 to $e_i$ for all $i$.

Robert Israel
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  • @dawid: By definition $X = \bigcup L_i$ (a subset of $\ell^2$). The more interesting question is: Why is $X$ not locally compact? (look at $0$) – t.b. Jul 06 '11 at 02:19
  • oh yes, sorry, i thought that X=l2, so.. the lack of the locally compactness doesnt seem to be obvious.. – Dawid C. Jul 06 '11 at 02:35
  • @dawid: A neighborhood of zero must contain a sequence of the form $t, e_{i}$, $i=1,2,\ldots$ with $t$ small. This sequence has no convergent subsequence. – t.b. Jul 06 '11 at 02:43
  • Is it complete? – Henno Brandsma Sep 07 '19 at 05:30
  • Yes. Consider a sequence $x_n$ in $X$. If it is Cauchy, $|x_n|$ must converge. If $|x_n| \to 0$, then $x_n \to 0 \in X$. If $|x_n| \to r > 0$, then use the fact that $|r e_i - r e_j| > r$ for $i \ne j$. – Robert Israel Sep 08 '19 at 01:27