Must a complete space be locally compact?

There are two versions of second category space: one is complete metric space, the other is locally compact space.

As we know, an open interval is locally compact but not complete. But how about the opposite? Must a complete space be locally compact? If not, must a complete topological vector space be second category?

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A Banach space is locally compact iff it is finite-dimensional. – martini Apr 21 '12 at 5:40
For the first question consider any infinite dimensional Banach space, for the second I think $C_c(\mathbb{R})$ (continous real functions with compact support) with the limit topology is a counterexample. – Jose27 Apr 21 '12 at 5:42
@martini May I suggest you post your answer as an answer and not as a comment? – Martin Argerami Apr 22 '12 at 22:32
@Jose27 May I suggest you post your answer as an answer and not as a comment? – Martin Argerami Apr 22 '12 at 22:32
@MartinArgerami done so. – martini Apr 23 '12 at 7:15

Interestingly enough is the converse also true. A locally compact normed space is finite dimensional. To see this we use the Riesz' lemma: If $X$ is a normed space, $U \subsetneq X$ a closed subspace, $\delta > 0$, then there is a $x \in X$ with $d(x,U) > 1- \delta$ and $\|x\| = 1$. If now $X$ is infinite dimensional we let $U_0 = \{0\}$ and choose by induction $x_n \in S_X$ (i. e. $\|x_n\| = 1$) with $d(x_n, U_{n-1}) > \frac 12$ and let $U_n := \mathrm{span}\,\{x_1, \ldots, x_n\}$. We can do so as $U_n$ is finite dimensional for every $n$ and such a closed proper subspace of $X$. Now $(x_n)$ is a sequence in $B_X$ with $\|x_n - x_m\| > \frac 12$ for $n \ne m$, hence has no convergent subsequence. If $X$ were locally compact, $0$ would have a compact nbhd $V$. We could choose $\epsilon > 0$ with $\epsilon B_X \subseteq V$. But $\epsilon B_X$ and hence $V$ contains a sequence without a convergent subsequence, so $V$ cannot be compact.