# In the first countable TVS, if every Cauchy sequence convergence then every Cauchy net convergent

Let $X$ be a topological vector space with the first countable topology(that is, every point has a countable neighborhood basis).If every Cauchy sequence convergence, we want to show that every Cauchy net converges.

For every given Cauchy net $\langle x_i\rangle$, I think I can inductively find a subnet $\langle x_{k_n}\rangle$of $\langle x_i\rangle$ which is actually a Cauchy sequence. By assumption, we know $\langle x_{k_n}\rangle$ converges to a point $x$ in $X$, I need to show $\langle x_i\rangle \rightarrow x$.

If $X$ is a metric space, then the argument may be easier since we can apply "$2\epsilon$ argument" to $x_i-x_{k_n}+x_{k_n}-x$. But how to do this in a general 1st countable topological vector space? Plus, if $\{U_n\}$ is a countable neighborhood basis at $0$, is $\{U_n+U_n\}$ again a neighborhood basis at $0$?

if $\{U_n\}$ is a countable neighborhood basis at $0$, is $\{U_n+U_n\}$ again a neighborhood basis at $0$?
Yes. Let $U$ be an arbitrary neighborhood of the zero. There exists a neighborhood $V$ of the zero such that $V+V\subset U$, and there exists a base neighborhood $U_n$ of the zero such that $U_n\subset V$. This should allow you to apply "2$\epsilon$ argument"