Topological vector space, and balanced sets.

Question

Wikipedia states:

Every topological vector space has a local base of absorbing and balanced sets.

I'm not a big expert in the subject. But I'd like to see the proof of such statement. Any clue?

I've attempted to prove it using the definition of basis for a topology, and the continuity of these operations in t.v.s.

Answer 1

Let $U$ be a neighborhood of $0$. Since $0*x=0$ and by the continuity of the scalar multiplication, we get $ \delta >0$ such that $tx \in U$ for all $|t| \le \delta$. Hence: $x \in \lambda U$ for $|\lambda| \ge 1/ \delta$. This shows that U is absorbing.

Since $0*0=0$ and by the continuity of the scalar multiplication,we get $ \delta >0$ and a neighborhood $V$ of $0$ such that $tV \subset U$ for $|t| \le \delta$. Hence $ \delta V \subset \lambda U$ for $|\lambda| \ge 1$.

It's your turn to show that $U$ contains a balanced neighborhood of $0$.