$X$ is a TVS and $O$ is an open neighbourhood of $0$. Is there an open neighbourhood $U$ of $0$ such that $U-U\subseteq O$?

Question

Assume $X$ is Hausdorff.

This is true for locally convex TVS, since every open set contains an open ball in respect to some seminorm.

Is it still true when $X$ is not locally convex?

Answer 1

Hint:

Yes, choose a balanced neighborhood $V$ of $0$ such that $V+V \subset O$.

Since $(x,y) \mapsto x+y$ is continuous, for all neighborrhood $O$ of $0$ in $X$ there exists nbhd's $W, Z \subset X$ of $0$ such that $W+Z \subset O$. Define $V'=W\cap Z$. Thus, $V'+V' \subset W+Z \subset O$. Since $X$ is a TVS, for all neighborhood $U$ of $0$, there exists a neighborhood $V$ of $0$ which is balanced and $V \subset U$. In particular, for $V'$, there exists a balanced neighborhood $V$ of $0$ such that $V\subset V'$. Therefore, $V+V \subset V'+V' \subset O$. Since $V$ is balanced $V=-V$. Can you conclude from here?