Closure of linear subspace in Topological vector space

Question

Let $X$ be a TVS, $x\in X$ and $M<X$ be a linear subspace. Does $x\in M+U$ for every open neighborhood $U$ of $0$ imply that $x$ is in the closure of $M$?

EDIT: This argument is used here: http://terrytao.wordpress.com/2011/05/24/locally-compact-topological-vector-spaces/ in the proof of Theorem 2. However, I think the existence of a bounded neighborhood is needed (which in the linked article is of course the case).

Answer 1

If for every open neighborhood $U$ we have $x\in M+U$ then $(x-U)\cap M\neq \emptyset $ hence $x\in\overline{M}.$

Answer 2

Actually it's an equivalence since $$ \cap\{M+U\mid U\in\mathscr{V}(0)\}=\overline{M}.$$ Here $\mathscr{V}(0)$ stands for the neighborhoods of $0$.