MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 6 down vote accepted

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.

share|cite|improve this answer
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
I just needed to cite this result in a paper, and literally thought "I know that this result must be somewhere in Bourbaki, but don't want to go hunting; let me Google MO instead, and hope that someone has already done the hunting for me." Thank you for being that one! – L Spice Jan 5 at 5:37

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.