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Let $X$ be a metrizable topological vector space (TVS) over $\mathbb R\text{ or }\mathbb C$, and consider the following two properties that $X$ might have.

  1. Local convexity
  2. Completeness (as a TVS, which implies completeness in any invariant compatible metric)

I would like to know which combinations of these properties are enough to guarantee what I will call the closed ball property (CBP), which says that for every compatible invariant metric d, the closure of the unit ball $B_{<1} = \{x\in X:d(0, x) < 1\}$ is the closed unit ball $B_{\leq 1} = \{x\in X : d(0, x) \leq 1\}$. This post gives a counterexample showing that metrizability alone does not guarantee CBP. The post also claims without proof that (1) + (2) together (which make $X$ a Fréchet space) does guarantee CBP. (Edit: I added the "for every compatible invariant metric" part above.)

This naturally raises the question: Does (1) or (2) alone suffice for CBP to hold? Please provide proofs or counterexamples showing which combinations of (1) and (2) guarantee CBP. If both (1) + (2) are both required, please give a proof of the (1) + (2) case, as I cannot find any.

I should add that this post discusses the same question on more general grounds, but it does not seem to contain any direct answers to the question I'm asking here.

WillG
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    Note that "metrizable space" and "metric space" is not the same thing, and that there's no concept of a ball in a metrizable space. So you need to clarify what you mean – Jakobian Apr 02 '25 at 20:58
  • @Jakobian Good point, I edited the post to clarify this. However, this makes me unclear what the OP in the linked question meant by saying that CBP is satisfied by Fréchet spaces. This might be wrong, since it seems the counterexample given there is in fact a Fréchet space. – WillG Apr 02 '25 at 21:11
  • No metrizable space with more than two points satisfies this CBP property of yours – Jakobian Apr 02 '25 at 21:14
  • @Jakobian So you are claiming that OP of the linked question is mistaken about Fréchet spaces? – WillG Apr 02 '25 at 21:20
  • @Jakobian I'm confused. You're claiming that every metrizable TVS has CBP (as currently described in my post)? What about the counterexample you gave and deleted earlier, also given here? – WillG Apr 02 '25 at 21:47
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    Oh sorry my bad. Yeah. You're right, no non-trivial metrizable topological vector space has this property. A metrizable topological vector space has an invariant metric $d$, and then the metric $d'(x, y) = \min(d(x, y), r)$ for small enough $r > 0$ serves as an example. – Jakobian Apr 02 '25 at 21:55

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