Does sequential completeness imply completeness in a metrizable topological vector space? Let a metrizable topological vector space $(X, \tau)$ be given (i.e., its topology can be described by a metric). If it sequentially complete (i.e., every Cauchy sequence is convergent), does it follow that it is complete (i.e., every Cauchy net is convergent)?
If it is true, I would be grateful if someone would be able to provide a source with a citation, since I need to use this fact in my thesis.
 A: More generally, even equivalence is true in any metrizable space $X$ (at least assuming the axiom of countable choice, which is weaker than the axiom of choice), as follows:
Let first $\mathcal F$ be a Cauchy filter. We choose a metric $d$ on $X$ which generates the topology of $X$. A fundamental system of entourages is given by the sets
$$
V_n := \left\{(x, y) \in X \times X \middle| d(x,y) < \frac{1}{2^n} \right\}, n \in \mathbb N.
$$
Let $n \in \mathbb N$ be given. Since $\mathcal F$ is a Cauchy filter, we find $A_n \in \mathcal F$ such that for $x, y \in A_n$ we have $(x, y) \in V_n$. The set
$$
B_n := \bigcap_{k \le n} A_k
$$
will be in $\mathcal F$. Furthermore, using countable choice we pick a sequence $x_n \in B_n$ and observe that it is a Cauchy sequence, which will hence converge to an $x \in X$ by completeness in the metric sense. We claim that $\mathcal F$ converges to $x$. This is because it contains the neighbourhood filter of $x$, since the $V_n$ are a fundamental system of entourages.
Let now $(x_n)_{n \in \mathbb N}$ be a Cauchy sequence (in the metric sense). The co-finite subsets of $\{x_n \mid n \in \mathbb N\}$ are a Cauchy filter, which hence converges. The limit will be the desired limit in the metric sense.
Google Books found the following reference:
https://books.google.de/books?id=mX3rBwAAQBAJ&pg=PA120&dq=uniform+completeness+coincides&hl=de&sa=X&ved=0ahUKEwi8hvLm2fzSAhVHXiwKHR06DwAQ6AEIKDAF#v=onepage&q=uniform%20completeness%20coincides&f=false
