Let $E$ be a TVS over $\mathbb R$ or $\mathbb C$ such that $\overline{\lbrace 0 \rbrace} = E$. Does $E$ carry the trivial topology?

Question

I would very much be interested in seeing a nontrivial example of a TVS over the real or complex numbers such that the closure of the origin is the entire space. Is this possible?

Answer 1

Since translation by any element $v\in E$ is a homeomorphism, $\overline{\{v\}}=E$ as well. So $E$ is the only closed subset of $E$ which contains any point. That is, the only nonempty closed subset of $E$ is $E$, which means it has the trivial topology.