# Proof that a Hilbert space is not locally compact.

The proof in my textbook is: Construct $B(x,\epsilon)$. If $x=\langle x_{1},x_{2},\dots\rangle$, defined a sequence $x_{n}=\langle x_{1},x_{2},\dots,x_{n}+\epsilon,\dots\rangle$. Clearly, $d(x_{i},x_{j})=\sqrt{2}\epsilon$ when $i\neq j$. Since the sequence $\langle x_{n}\rangle$ can have no convergent subsequence, $B(x,\epsilon)$ is not compact.

I have not come across the correlation between the convergence of a sequence and compactness yet. I have tried to contruct an explanation. Please do comment on whether the argument is sound.

Since $d(x_{i},x_{j})=\sqrt{2}\epsilon$ for $i\neq j$ we construct balls $B(x_{i},\frac{\sqrt{2}\epsilon}{2})$ around each $x_{i}\in\langle x_{n}\rangle$. Clearly these sets are disjoint. For each $y\in X\setminus \bigcup_{i}B(x_{i},\frac{\sqrt{2}\epsilon}{2})$, $$d(y,x_{i})\geq \frac{\sqrt{2}\epsilon}{2}$$ Hence $X\setminus \bigcup_{i}B(x_{i},\frac{\sqrt{2}\epsilon}{2})$ is an open set. Let us call it $U$.

We have a cover $\{U\}\bigcup \{\cup_{i}B(x_{i},\frac{\sqrt{2}\epsilon}{2})\}$ of $X$, which will not have a finite subcover. This is true for every $\epsilon\in\Bbb{R}$. Hence, we can't possibly construct a compact set around $x$.

EDIT: Also, I feel that if a space is locally compact, not every sequence in the space has to be convergent. Could someone comment on this too? Thanks again!

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A metric space is compact iff it is sequentially compact, meaning that every sequence has a convergent subsequence. The argument shows that within any closed nbhd of $x$ there is a sequence with no convergent subsequence, so $x$ has no compact nbhd and is therefore not locally compact.

Your argument is essentially the usual argument that if a metric space is not sequentially compact, then it cannot be compact.

Added: You’re quite right that a locally compact space can have non-convergent sequences: $\Bbb R$ with the usual topology is locally compact, and $\langle n:n\in\Bbb N\rangle$ is a non-convergent sequence. But if $X$ is locally compact, then every point of $X$ has a compact nbhd, and if in addition $X$ is metrizable, then every sequence that has infinitely many terms in a compact set will have a convergent subsequence.

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Awesome. Thanks Brian! – Ayush Khaitan Jul 11 '13 at 4:58
@Ayush: You’re welcome! – Brian M. Scott Jul 11 '13 at 5:01