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For a (possibly infinite-dimensional) vector space $V$, I thought about the following topology $\tau$: Let $O \in \tau$ if every $x \in O$ has the property that for every $v \in V$, there is an $\epsilon > 0$ such that $x + \alpha v \in O$ for every $\alpha$ with $| \alpha | < \epsilon$. This gives a topology indeed (right?). I think that in finite-dimensional vector spaces, this gives the standard topology (right?).

What about infinite-dimensional vector spaces? Is there a name for this topology? To me, this looks like a very generic way to define a topology on a vector space, but I've never seen this definition of a topology. Is there any connection to locally convex topologies? (I'm not very familiar with this concept) Is this topology important in any application?

Tom Jonathan
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    This is called the finest locally convex topology on $V$ (the topology induced by all linear functionals) and appears e.g. in Mackey's work on locally convex spaces. It is discussed e.g. in Kelley-Namioka or Schaefer's book on topological vector spaces. Sometimes sets that are open in your sense are called "algebraically open", e.g. section 2.2 here. – t.b. Apr 05 '12 at 08:56
  • @t.b.: You and Michael are right: I missed the dependence of $\epsilon$ on $v$. – Brian M. Scott Apr 05 '12 at 08:57
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    @t.b. according to exercise 1.9.9., page 133 here the family of algebraically open sets can disagree with the usual topology even in the finite dimensional case and is therefore usually not a vector space topology. – Michael Greinecker Apr 05 '12 at 09:13
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    In my previous comment I overlooked that you didn't have a condition of convexity: the algebraically open convex sets form a base of the finest locally convex topology, which is the same as the one described by Harry in his answer. @Michael: yes, you're right. – t.b. Apr 05 '12 at 09:13
  • t.b.: Thanks, this gives me a notion to look up (if convexity is added). @MichaelGreinecker: Thanks, this shows why this is not a neat topology. So I was wrong, thinking that it reduces to the standard topology in finite dimensions.

    I would actually accept your comments as answers, but this is technically not possible, right?

     – Tom Jonathan Apr 06 '12 at 02:12
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    @Tom: I wouldn't be able to add much to what Harry said. I suggest that you accept his answer. – t.b. Apr 08 '12 at 14:34
  • I think the finest locally convex linear topology is indirectly defined in many works which deal with inductive topologies, i.e. for l.c.t.v.s's ${X_\alpha}\alpha$ with linear maps ${f\alpha}\alpha:X\alpha\rightarrow X$ there is a unique finest locally convex linear topology $\tau$ on $X$ which makes the $f_\alpha$ continuous. That result works also when the index set is considered empty, i.e. $\alpha\in\emptyset$, and gives you the desired result. – fweth Feb 01 '15 at 16:40

1 Answers1

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This is not an answer. Rather, let me explain another canonical way of endowing vector spaces with a topology.

Let $V$ be a vector space over the real numbers $\mathbf{R}$. (You can replace $\mathbf{R}$ by any topological field in everything that follows, although for the properties at the end you'll need your topological field to be complete of characteristic zero and Hausdorff.)

Suppose that $V$ is finite-dimensional. Any isomorphism $V\cong \mathbf{R}^n$ endows $V$ with a topology compatible with the vector space structure, where we endow $\mathbf{R}^n$ with the product topology. (Such an isomorphism boils down to the choice of a basis for $V$.) It is easy to see that this topology is independent of the chosen isomorphism.

Now, in general, if $V$ is not finite-dimensional, you can endow $V$ with the "inductive limit topology". This topology is characterized by the following property:

A map $f:V\to A$ is continuous if and only if its restriction $f|_{W}:W\to A$ to every finite-dimensional subspace $W$ of $V$ is continuous.

It is a direct consequence of the definitions that the inclusion $W\to V$ of a finite-dimensional subspace $W$ of $V$ is continuous. More generally, any $\mathbf{R}$-linear map $V_1\to V_2$ of vector spaces endowed with the inductive limit topology is continuous.

Properties:

  1. Any vector space of $V$ is closed.

  2. The topology is Hausdorff.

  3. Let $K\subset V$ be compact. Then $K$ is contained in a finite-dimensional subspace of $V$.

The first two properties are not so hard to prove. In fact, to prove that $V$ is Hausdorff proceed as follows. Let $x$ and $y$ be in $V$. Consider the subspace $W$ generated by $x$ and $y$. Choosing a basis for $V$ gives you a projection $p:V\to W$. This projection is continuous. Let $U_x$ be an open of $W$ containing $x$ and $U_y$ an open of $W$ containing $y$ such that $U_x\cap U_y$ is empty. (Here you use that the base field is Hausdorff.) Then $p^{-1}(U_x)$ and $p^{-1}(U_y)$ separate $x$ and $y$ in $V$.

The third one is a bit trickier. You'll need to use that the base field is complete and of characteristic zero.

Harry
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