1
$\begingroup$

It is fairly straightforward to prove that over a normed space $ (V,\| \cdot \|)$ the existence of a Schauder basis $ \{ e_n\}_{n=1}^\infty$ implies the separability of the space.

I was however wondering if this implication also held for metric linear spaces with a metric that is not translation invariant.

To recall a metric linear space is a vector spaces with a metric such that both the addition and multiplication functions are continuous where the topology of the field $\mathbb{K} \in \{\mathbb{R}, \mathbb{C} \}$ is given by the standard absolute value I think. (This last part I am not so sure about)

I've tried to solve using the same proof as with the normed case but without any luck so far. I've only managed to prove it for the case of a translation invariant metric.

Does anyone know if the proposition is true ? And if not does someone know of a counter-example ?

EDIT: I added how I think we define the "distance function" on the field.

$\endgroup$
  • $\begingroup$ If you want $V$ to be a topological vector space then you should take $\mathbb{K}$ to have its usual topology. $\endgroup$ – Rhys Steele May 10 at 14:13
1
$\begingroup$

The result is true. It's possible that there is a straightforward proof but here is a way to "nuke" the problem.

Since you assume your topological vector space is metrisable, it is first-countable and Hausdorff. Hence it satisfies the hypotheses of the Birkhoff-Kakutani Theorem and as a result the topology is induced by a translation invariant metric (see Theorem 1 and Remark 2 here). This reduces the problem to the case of translation invariant metrics which you have already solved.

$\endgroup$
1
$\begingroup$

In the meantime I came up with the following elementary proof. We assume $\mathbb{K} = \mathbb{R}$ with the standard absolute value (the case $\mathbb{K} = \mathbb{C}$ is similar).

Let $\{e_n \}_{n=1}^\infty$ be a Schauder basis and define the following set

$$S= \{\sum_{k=1}^n a_n e_n | n \in \mathbb{N} \phantom{0} a_n \in \mathbb{Q} \} $$

Pick $ x = \sum_{n=1}^\infty c_n e_n \in V$. Then for $n$ big enough we have: $$d(x,\sum_{k=1}^n c_k e_k ) < \epsilon$$

Take a sequence $b^m \in \mathbb{Q}^n$ st $ b^m \rightarrow a= \{a_1, \ldots, a_n \}$ This is possible by the fact that $\mathbb{Q} \subset \mathbb{R} $ is dense. So since we know multiplication and addition to be continuous we get.

$$d(\sum_{k=1}^n c_k e_k,\sum_{k=1}^n b_k^m e_k ) \rightarrow 0$$

So for some big enough $m$ we get the result.

$$d(x,\sum_{k=1}^n b_k^m e_k) \leq d(x,\sum_{k=1}^n c_k e_k) + d(\sum_{k=1}^n c_k e_k,\sum_{k=1}^n b_k^m e_k)\leq 2\epsilon $$

It should however be noted that if we do not assume $\mathbb{K} \in \{\mathbb{R}, \mathbb{C} \}$ to have the standard topology the result can fail for example (a commentator posted this but deleted his comment in the meantime and I don't know how to find his name).

Take $\mathbb{C}$ as a $\mathbb{C}$-VS with the metric.

$$ d(x,y) = 1\text{ if } x=y \text{ and } d(x,y) = 0 \text{ if } x=y$$

for both the field and the vector space. Then this space is finite dimensional which implies that it has a Schauder basis. Moreover it has the discrete topology so addition and multiplication are continuous.

However is cannot be separable since it has the discrete topology.

$\endgroup$
  • $\begingroup$ (+1) I know that you can't accept your own answer yet but I think you should mark this as the accepted answer in two days instead of my answer. My answer is somewhat silly (though correct) and this is what my answer should have been. It shows that the usual argument really works. $\endgroup$ – Rhys Steele May 10 at 15:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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