Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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?

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.