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Is the Frechet space of all real sequences locally compact? Is a Hilbert cube, viewed as a topological metric space locally compact?

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The Hilbert cube is compact hence locally compact. – azarel Sep 22 '12 at 15:13
every locally convex space which is locally compact has finite dimension. – clark Sep 22 '12 at 15:21
So: note that the Hilbert cube (inside $\mathbb R^\infty$) is not a neighborhood, so (even though the Hilbert cube is compact) this does not suggest $\mathbb R^\infty$ is locally compact. – GEdgar Sep 22 '12 at 16:47
Another example: the disjoint union $\coprod_{n\in\mathbb N} S^n$ of $n$-dimensional spheres is locally compact and has infinite dimension. – Alexander Thumm Sep 22 '12 at 17:42
What do you mean by infinite-dimensional here? – Qiaochu Yuan Sep 22 '12 at 18:06

Infinite dimensional locally compact spaces? Are you familiar with the Adeles?

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Infinite-dimensional ("Hausdorff" is part of the definition) topological vector spaces are never locally compact. The sense of "infinite-dimensional" is ambiguous, beyond this, I think.

Nevertheless, such spaces do admit "substantial" (but not open) subsets which are compact. The classic example, a "Hilbert cube" in $\ell^2$, consisting of $(a_1,a_2,\ldots)$ such that $|a_n|\le {1\over n}$ is compact, and, in fact, has the product topology (as in Tychonoff's theorem). But it is not a nbd of $0$.

For that matter, infinite (even uncountable) topological products of compact spaces are compact, by Tychonoff, if one's sense of "infinite-dimensional" goes that far. In fact, the product topology is disturbingly coarse, despite its sensible mapping properties, so I'd not count such a product as being "seriously infinite-dimensional".

The example of adeles is an instance wherein a somewhat finer topology than a product topology is put on a subset of a cartesian product, producing a locally compact sort-of-infinite-dimensional thing: let $X_i$ be a family of locally compact topological spaces, topological groups for simplicity, and suppose we have an open subset $U_i$ of $X_i$ with compact closure. For some purposes, it is reasonable to consider products indexed by finite subset $S$ of the indices $i$, where $Y_S=\prod_{i\in S} X_i \times \prod_{i\not\in S}K_i$, with the product topology. Then the ascending union of the $Y_S$, really a colimit over $S$, has a finer topology than the subset topology from the product. Nevertheless, it is (fairly obviously) locally compact.

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