Let $E$ be a topological vector space and $U$ be an arbitrary neighborhood of $0$. I would like to know if $V=U \cap -U$ is balanced, that is $\lambda V \subset V$ for all $\lambda \in \mathbb{C}$ such that $|\lambda|\leq 1$.
I used this neighborhood to proof that every connected topological group $G$ can be written as $G=\bigcup_{n=1}^{\infty} U^n$ where $U$ is a negihborhood of $e$ in $G$. Now I was trying to see if this set has more properties when we are working with topological vector spaces.
But I was unable to prove that fact. If $\lambda \in \mathbb{C}$ such that $|\lambda|\leq 1$ and $x \in V$ then $x \in U$ and $x \in -U$. Thus, $x=u_0$ and $x=-u_1$ with $u_0,u_1 \in U$. I can't see a way to continue.
No: take $E=\Bbb C$ and $U$ an open square centred at zero. Then $U=-U$ but $V=U$ is not invariant under multiplication by $\frac15(3+4i)$ say.