Isolated point of a separable F-space

Question

I am prooving a theorem where there is a statement saying if X is a separable F-space( I.e. a topological vector space with a complete translation invariant metric) it has no isolated point. I am unable to prove this statement so that I can proceed further. Kindly help me understanding the statement. Thank you for your time and effort

Answer 1

If $x$ is an isolated point then there exists $r>0$ such that $\{x\}=B(x,r)$. By translation invariance of the metric it follows that $\{y\}=y-x+\{x\}=y-x+B(x,r)=B(y,r)$ for any $y$. It follows that $\{y\}$ is open for every $y$. This contradicts separability. [ Write the whole space as a union of singletons. In a separable space any union of open sets is a countable union. This makes the entire space countable but no vector space other than $\{0\}$ is countable].

Answer 2

Well, this is false--the trivial vector space $\{0\}$ has an isolated point. But more generally, it is true that no nontrivial topological vector space $X$ has an isolated point (no need to assume it is a separable F-space). This is immediate from the continuity of scalar multiplication: for any $x\in X$, $f(t)=tx$ is a continuous map $[0,1]\to X$ which gives a path from $0$ to $x$, so $X$ is path-connected.