This is a follow-up to [this answer](https://mathoverflow.net/a/355402/120251).


If $E$ is a (real or complex) topological vector space, we say that a sequence $\{x\_n\}\_{n=1}^\infty$ in $E$ *can be made to converge to zero* if there exists a sequence $\{\alpha\_n\}\_{n=1}^\infty$ of strictly positive real scalars such that $\lim\_{n\to\infty} \alpha\_n x\_n = 0$.


In the aforementioned answer, I compared this with two related notions, and I gave a generic example of a space where every sequence can be made to converge to zero: take a metrizable topological vector space and pass to a weaker (i.e. coarser) topology, possibly with a different dual. My question is whether every example is of this form. In other words:


**Question.** *If $E$ is a topological vector space where every sequence can be made to converge to zero, is there a finer linear topology on $E$ that is metrizable?*


Maybe it is false in general, but true in certain special cases? (E.g. if $E$ is locally convex/complete/separable, or if $E$ the strong dual of a Fréchet space?)



Metrizability of a topological vector space where every sequence can be made to converge to zero
For the counterexample below we need a little lemma about barrelled spaces (i.e., every barrel = closed, absolutely convex, and absorbing set is a $0$-neighbourhood):


*If $(E,\mathcal T)$ is a barrelled locally convex space which has a finer metrizable vector space topology $\mathcal S$ then $(E,\mathcal T)$ is metrizable.*


Indeed, let $(V\_n)\_{n\in\mathbb N}$ be a basis of the $0$-neighbourhoods in $(E,\mathcal S)$ und $U\_n =\overline{\Gamma(V\_n)}^{\mathcal T}$ the closed absolutely convex hulls of $V\_n$. These are barrels in in $(E,\mathcal T)$ and hence $\mathcal T$-neighbourhoods of $0$. On the other hand, every $\mathcal T$-neighbourhood of $0$ contains a closed absolutely convex one which itself contains some $V\_n$ (because $\mathcal S$ is finer) and hence some $U\_n$. This means that $(U\_n)\_{n\in\mathbb N}$ is a countable basis of the $0$-neighbourhoods of $(E,\mathcal T)$ which is thus metrizable.


Now the counterexample: Let $I$ be an uncountable index set and
$$E=\{(x\_i)\_{i\in I}\in \mathbb R^I: \{i\in I: x\_i\neq 0\} \text{ is countable}\}$$
endowed with the relative topology of the product topology on $\mathbb R^I$. According to proposition 4.2.5(ii) in the book *Barrelled Locally Convex Spaces* of Bonet and Perez-Carreras, $E$ is barrelled. Since every $0$-neighbourhood only gives conditions on finitely many coordinates there is no countable basis of all $0$-neighbourhoods, i.e., $E$ is not metrizable and hence, by the lemma, it does not admit a finer metrizable vector space topology. On the other hand, for every sequence $x^n\in E$ the union $J$ of the supports of the $x^n$ is countable, so that $\{x^n:n\in\mathbb N\}$ is contained in the subspace $F=\mathbb R^J\times \{0\}^{I\setminus J}$ which is metrizable. Therefore, there is a sequence of $t\_n>0$ such that $t\_nx\_n$ converges to $0$ in $F$ and hence also in $E$.




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As you also asked for the particular case of strong duals of Frechet spaces $X$: This is not only covered by the answer of Tomasz Kania but it is rather trivial: If $X$ is not isomorphic to a Banach space there is a fundamental sequence of semi-norms $p\_n$ which are pairwise non-equivalent. Then $X'=\bigcup\_{n\in\mathbb N} X\_n'$ where $X\_n'$ is the space of functionals which are continuous with respect to $p\_n$. Choosing $\phi\_n\in X\_n'\setminus X\_{n-1}'$ there is no sequence of $t\_n>0$ such that $t\_n\phi\_n\to 0$ because every bounded subset of $X'$ is contained in some $X\_n'$.




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EDIT. Another counterexample is the space of absolutely summable families $$\ell^1(I)=\{(x\_i)\_{i\in I}: \sum\_{i\in I} |x\_i|<\infty\}.$$
Every absolutely summable family has only countably many non-zero terms and hence a somehow natural topology (of course, besides the Banach space topology of the $\ell^1$-norm) is that of the inductive limit $\ell^1(I)=\lim\limits\_\to \ell^1(J)$ for all countable subsets $J\subseteq I$. This finer topology is barrelled but not metrizable (hence it does not admit a finer metrizable vector space topology) but every sequence can be made bounded.


