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.

— 1. Proof of Theorem 2 —

Let {V} be a finite-dimensional Hausdorff topological vector space, with topology {{\mathcal F}}. We need to show that every set which is open in the usual topology, is open in {{\mathcal F}}, and conversely.

Let {v_1,\ldots,v_n} be a basis for the finite-dimensional space {V}. From the continuity of the vector space operations, we easily verify that the linear map {T: {\bf R}^n \rightarrow V} given by

\displaystyle  T( x_1,\ldots,x_n) := x_1 v_1 + \ldots + x_n v_n

is continuous. From this, we see that any set which is open in {{\mathcal F}}, is also open in the usual topology.

Now we show conversely that every set which is open in the usual topology, is open in {{\mathcal F}}. It suffices to show that there is a bounded open neighbourhood of the origin in {{\mathcal F}}, since one can then translate and dilate this open neighbourhood to obtain a (sub-)base for the usual topology. (Here, “bounded” refers to the usual sense of the term, for instance with respect to an arbitrarily selected norm on {V} (note that on a finite-dimensional space, all norms are equivalent).)

We use {T} to identify {V} (as a vector space) with {{\bf R}^n}. As {T} is continuous, every set which is compact in the usual topology, is compact in {{\mathcal F}}. In particular, the unit sphere {S^{n-1} := \{x \in {\bf R}^n: \|x\|=1\}} (in, say, the Euclidean norm {\| \|} on {{\bf R}^n}) is compact in {{\mathcal F}}. Using this and the Hausdorff assumption on {{\mathcal F}}, we can find an open neighbourhood {U} of the origin in {F} which is disjoint from {S^{n-1}}.

At present, {U} need not be bounded (note that we are not assuming {V} to be locally connected a priori). However, we can fix this as follows. Using the joint continuity of the scalar multiplication map, one can find another open neighbourhood {U'} of the origin and an open interval {(-\epsilon,\epsilon)} around {0} such that the product set {(-\epsilon,\epsilon) \cdot U' := \{ t x: t \in (-\epsilon,\epsilon); x \in U' \}} is contained in {U}. Since {U} avoids the unit sphere {S^{n-1}}, {U'} must avoid the region {\{ x \in {\bf R}^n: \|x\| > 1/\epsilon \}} and is thus bounded, as required.

Corollary 3 In a Hausdorff topological vector space {V}, every finite-dimensional subspace {W} is closed.

Proof: It suffices to show that every vector {x \in V \backslash W} is in the exterior of {W}. But this follows from Theorem 2 after restricting to the finite-dimensional space spanned by {W} and {x}. \Box

— 2. Proof of Theorem 1 —

Let {V} be a locally compact Hausdorff space, thus there exists a compact neighbourhood {K} of the origin. Then the dilate {\frac{1}{2} K} is also a neighbourhood of the origin, and so by compactness {K} can be covered by finitely many translates of {\frac{1}{2} K}, thus

\displaystyle  K \subset S + \frac{1}{2} K

for some finite set {S}. If we let {W} be the finite-dimensional vector space generated by {S}, we conclude that

\displaystyle  K \subset W + \frac{1}{2} K.

Iterating this we have

\displaystyle  K \subset W + 2^{-n} K

for any {n \geq 1}. On the other hand, if {U} is a neighbourhood of the origin, then for every {x \in V} we see that {2^{-n} x \in U} for sufficiently large {n}. By compactness of {K} (and continuity of the scalar multiplication map at zero), we conclude that {2^{-n} K \subset U} for some sufficiently large {n}, and thus

\displaystyle  K \subset W + U

for any neighbourhood {U} of the origin; thus {K} is in the closure of {W}. By Corollary 3, we conclude that

\displaystyle  K \subset W.

But {K} is a neighbourhood of the origin, thus for every {x \in V} we have {2^{-n} x \in K} for all sufficiently large {n}, and thus {x \in 2^n W = W}. Thus {V=W}, and the claim follows.