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Is the (algebraic) span a finite set of vectors in a Hausdorff topological vector space over a complete field always closed?

I suspect yes, but I can't come up with a proof, and it seems like locally convex might be needed to get this.

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Yes, this is true. I assigned it as a HW problem in a course last spring and a student solved it. I asked him to type it up, but apparently I don't have it. :( Anyway, sure, the proof that works over $\mathbb{R}$ (found e.g. in Rudin's Functional Analysis) goes over verbatim. –  Pete L. Clark Sep 6 '10 at 8:48

2 Answers 2

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This holds indeed for complete fields: see Theorem 2, Section I.2.3, of Bourbaki's "Espaces Vectoriels Topologiques".

Here is the argument.

Let $K$ be a (not necessarily commutative) field equipped with a complete nontrivial absolute value $x\mapsto|x|$, let $n$ be a positive integer, let $\tau$ be a Hausdorff vector space topology on $K^n$, and let $\pi$ be the product topology on $K^n$.

THEOREM $\tau=\pi$.

REMINDER A topological group $G$ is Hausdorff iff {1} is closed. [Proof: {1} is closed $\Rightarrow$ the diagonal of $G\times G$ is closed (because it's the inverse image of {1} under $(x,y)\mapsto xy^{-1}$) $\Rightarrow$ $G$ is Hausdorff.]

LEMMA The Theorem holds for $n=1$.

The Lemma implies the Theorem. We argue by induction on $n$. The continuity of the identity from $K^n_\pi$ to $K^n_\tau$ (obvious notation) is clear (and doesn't use the Lemma). To prove the continuity of the identity from $K^n_\tau$ to $K^n_\pi$, it suffices to prove the continuity of an arbitrary nonzero linear form $f$ from $K^n_\tau$ to $K_\pi$. By induction hypothesis, the kernel of $f$ is closed, and the Theorem follows from the Reminder and the Lemma.

Proof of the Lemma. We'll use several times the fact that $K^\times$ contains elements of arbitrary large and arbitrary small absolute value. As already observed, we have $\tau\subset\pi$. If $x$ is in $K^\times$, write $B_x$ for the open ball of radius $|x|$ and center 0 in $K$. Let $a$ be in $K^\times$, and let $\tau_0$ be the set of those $U$ such that $0\in U\in\tau$.

It suffices to check that $B_a$ contains some $U$ in $\tau_0$.

We can find a $b$ in $K^\times$ and a $V$ in $\tau_0$ such that $a$ is not in $B_bV$, and then a $c$ in $K$ with $|c|>1$ and a $W$ in $\tau_0$ such that $a$ is not in $B_cW$. Then $U:=c^{-1}W$ does the job.

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Umm, I wouldn't have known how to prove this result, but I don't see how it addresses my question, either. –  Ricky Demer Sep 8 '10 at 7:11
    
You asked if a finite dimensional space F in a Hausdorff topological vector space over a complete field is always closed. The result I prove (following Bourbaki) shows that F (equipped with the induced topology) is complete, and thus closed. –  Pierre-Yves Gaillard Sep 8 '10 at 7:32
    
How do you show that every complete field has an absolute value that induces its topology? –  Ricky Demer Sep 8 '10 at 8:30
    
I'm afraid I misunderstood your question. I took it for granted that you considered only fields complete with respect to a nontrivial absolute value. Sorry. [I know nothing about other kinds of complete fields.] –  Pierre-Yves Gaillard Sep 8 '10 at 9:01
    
Your answer is still more general (and so better) than Robin's. [If you don't know what another kind of complete field would be, see Definition, Completeness, and Example 3 at en.wikipedia.org/wiki/Uniform_space ] –  Ricky Demer Sep 8 '10 at 9:46

For real/complex vector spaces, this is Theorem 1.21 in Rudin's Functional Analysis (2nd ed.). I believe the proof works for any complete field, but haven't checked in detail.

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This holds indeed for complete fields: see Theorem 2, Section I.2.3, of Bourbaki's "Espaces Vectoriels Topologiques". –  Pierre-Yves Gaillard Sep 6 '10 at 8:47
    
Very interesting. Clearly Bourbaki was the right place! –  Pietro Majer Sep 6 '10 at 8:51

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