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Recall that a (real) topological vector space is a real vector space ${V = (V, 0, +, \cdot)}$ equipped with a topology ${{\mathcal F}}$ that makes the vector space operations ${+: V \times V \rightarrow V}$ and ${\cdot: {\bf R} \times V \rightarrow V}$ continuous. One often restricts attention to Hausdorff topological vector spaces; in practice, this is not a severe restriction because it turns out that any topological vector space can be made Hausdorff by quotienting out the closure ${\overline{\{0\}}}$ of the origin ${\{0\}}$. One can also discuss complex topological vector spaces, and the theory is not significantly different; but for sake of exposition we shall restrict attention here to the real case.

An obvious example of a topological vector space is a finite-dimensional vector space such as ${{\bf R}^n}$ with the usual topology. Of course, there are plenty of infinite-dimensional topological vector spaces also, such as infinite-dimensional normed vector spaces (with the strong, weak, or weak-* topologies) or Frechet spaces.

One way to distinguish the finite and infinite dimensional topological vector spaces is via local compactness. Recall that a topological space is locally compact if every point in that space has a compact neighbourhood. From the Heine-Borel theorem, all finite-dimensional vector spaces (with the usual topology) are locally compact. In infinite dimensions, one can trivially make a vector space locally compact by giving it a trivial topology, but once one restricts to the Hausdorff case, it seems impossible to make a space locally compact. For instance, in an infinite-dimensional normed vector space ${V}$ with the strong topology, an iteration of the Riesz lemma shows that the closed unit ball ${B}$ in that space contains an infinite sequence with no convergent subsequence, which (by the Heine-Borel theorem) implies that ${V}$ cannot be locally compact. If one gives ${V}$ the weak-* topology instead, then ${B}$ is now compact by the Banach-Alaoglu theorem, but is no longer a neighbourhood of the identity in this topology. In fact, we have the following result:

Theorem 1 Every locally compact Hausdorff topological vector space is finite-dimensional.

The first proof of this theorem that I am aware of is by André Weil. There is also a related result:

Theorem 2 Every finite-dimensional Hausdorff topological vector space has the usual topology.

As a corollary, every locally compact Hausdorff topological vector space is in fact isomorphic to ${{\bf R}^n}$ with the usual topology for some ${n}$. This can be viewed as a very special case of the theorem of Gleason, which is a key component of the solution to Hilbert’s fifth problem, that a locally compact group ${G}$ with no small subgroups (in the sense that there is a neighbourhood of the identity that contains no non-trivial subgroups) is necessarily isomorphic to a Lie group. Indeed, Theorem 1 is in fact used in the proof of Gleason’s theorem (the rough idea being to first locate a “tangent space” to ${G}$ at the origin, with the tangent vectors described by “one-parameter subgroups” of ${G}$, and show that this space is a locally compact Hausdorff topological space, and hence finite dimensional by Theorem 1).

Theorem 2 may seem devoid of content, but it does contain some subtleties, as it hinges crucially on the joint continuity of the vector space operations ${+: V \times V \rightarrow V}$ and ${\cdot: {\bf R} \times V \rightarrow V}$, and not just on the separate continuity in each coordinate. Consider for instance the one-dimensional vector space ${{\bf R}}$ with the co-compact topology (a non-empty set is open iff its complement is compact in the usual topology). In this topology, the space is ${T_1}$ (though not Hausdorff), the scalar multiplication map ${\cdot: {\bf R} \times {\bf R} \rightarrow {\bf R}}$ is jointly continuous as long as the scalar is not zero, and the addition map ${+: {\bf R} \times {\bf R} \rightarrow {\bf R}}$ is continuous in each coordinate (i.e. translations are continuous), but not jointly continuous; for instance, the set ${\{ (x,y) \in {\bf R}: x+y \not \in [0,1]\}}$ does not contain a non-trivial Cartesian product of two sets that are open in the co-compact topology. So this is not a counterexample to Theorem 2. Similarly for the cocountable or cofinite topologies on ${{\bf R}}$ (the latter topology, incidentally, is the same as the Zariski topology on ${{\bf R}}$).

Another near-counterexample comes from the topology of ${{\bf R}}$ inherited by pulling back the usual topology on the unit circle ${{\bf R}/{\bf Z}}$. Admittedly, this pullback topology is not quite Hausdorff, but the addition map ${+: {\bf R} \times {\bf R} \rightarrow {\bf R}}$ is jointly continuous. On the other hand, the scalar multiplication map ${\cdot: {\bf R} \times {\bf R} \rightarrow {\bf R}}$ is not continuous at all. A slight variant of this topology comes from pulling back the usual topology on the torus ${({\bf R}/{\bf Z})^2}$ under the map ${x \mapsto (x,\alpha x)}$ for some irrational ${\alpha}$; this restores the Hausdorff property, and addition is still jointly continuous, but multiplication remains discontinuous.

As some final examples, consider ${{\bf R}}$ with the discrete topology; here, the topology is Hausdorff, addition is jointly continuous, and every dilation is continuous, but multiplication is not jointly continuous. If one instead gives ${{\bf R}}$ the half-open topology, then again the topology is Hausdorff and addition is jointly continuous, but scalar multiplication is only jointly continuous once one restricts the scalar to be non-negative.

Below the fold, I record the textbook proof of Theorem 2 and Theorem 1. There is nothing particularly original in this presentation, but I wanted to record it here for my own future reference, and perhaps these results will also be of interest to some other readers.

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