# $\sigma$ - compact and locally compact metric space

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

-

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 $i$.

-
@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 6 '11 at 2:19
oh yes, sorry, i thought that X=l2, so.. the lack of the locally compactness doesnt seem to be obvious.. – dawid Jul 6 '11 at 2:35
@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 6 '11 at 2:43