How is the topology on a vector space induced from its base field $\mathbb{R}$?

I don't remember where on this site, but I vaguely remember seeing that for a vector space with its base field being $\mathbb{R}$ (or more generally, a topological field?), there can be a natural (unique) topology induced from the base field to the vector space. I wonder how it is done? Thanks!

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A vector space is the direct sum of copies of the field, which can be seen as a subspace of the product, so we can give it the restriction of the product topology. –  Jose27 Apr 4 '12 at 2:26
@Jose27: Thanks! Why "a vector space is the direct sum of copies of the field"? How many copies are there? Some references? –  Tim Apr 4 '12 at 2:28
@Tim: This is the fact that any vector space has a basis. There are as many copies are there are vectors in a basis. –  Zhen Lin Apr 4 '12 at 8:35

Every vector space has a basis, this can be seen using a Zorn's lemma argument: $L=\{ X\subset V: X \text{ is linearly independent }\}$ is inductive and a maximal element is a basis.

So let $(v_\alpha)_{\alpha \in A}$ be a basis for $V$, and consider the linear transformation $T:V\to \oplus_A F_\alpha$ where $F_\alpha =F$ for all $\alpha$, such that ($x_\alpha \in F$)

$$\sum_{\alpha\in A} x_\alpha v_\alpha = x \mapsto (x_\alpha)_{\alpha\in A}$$

This is an isomorphism of vector spaces. Clearly we have

$$\bigoplus_A F_\alpha \subset \prod _A F_\alpha$$

To this last one we can give the product topology, and so we can give the direct sum the subspace topology. Now just declare $Y\subset V$ open if and only if $T(Y)$ is open in this last topology.

How big can $A$ be: as big as you want just build a $V$ via the direct sum above. This topology is useful (as far as I know) mostly for finite dimensional vector spaces though.

As for references, the basic linear algebra you can look up in Friedberg's "Linear Algebra", and the topology on, say, Kelley's or Munkres' books.

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Alternatively, one could embed $V$ in $c_0(A)$, the set of finitely supported sequences indexed by $A$, and endow this with the topology coming from some $\ell^p$ norm, for example. This would give $V$ a metrizable topology. However, the whole thing is completely non-constructive, thanks to Zorn's lemma, and it's hard to imagine such a topology being useful in practice. I'd hesitate to call it "natural". –  Nate Eldredge Apr 4 '12 at 4:22
@Nate: Thanks, I hadn't thought of that approach. A question on notation, isn't the space of finitely supported nets usually denoted $c_{00}(A)$ (I'm going by analogy with the case $A=\mathbb{N},\mathbb{Z}$). –  Jose27 Apr 4 '12 at 4:57
Thanks! WHat does "inductive" mean? –  Tim Apr 4 '12 at 11:42
@Tim: Basically, it means that Zorn's lemma can be applied, in other words it means that every chain in $L$ has an upper bound in $L$. –  Jose27 Apr 4 '12 at 14:52
@Jose27: You're right, I meant $c_{00}$. –  Nate Eldredge Apr 4 '12 at 14:59