I'm trying to understand. this theorem
Theorem 1.14. In a topological vector space ,
(a) every neighborhood of 0 contains a balanced neighborhood of 0.
The proof asserts that as scalar multiplication is continuous for every neighborhood of 0 let's call it U there exist $\delta > 0$ and another neighborhood V such that $\alpha V \subset U$ whenever $\left | \alpha \right | < \delta$.
My question: Why is this affirmation true? It is obvious when working on $R^n$ or even $C$ with a topology defined by a metric, however we do not have a norm, metric or even scalar product in the hypothesis.
Ok, I think I did realize the answer.
Given a topological vector space X and that $\cdot:F\times X \rightarrow X$ is continuous.
Given a neighborhood U of 0 there is a neighborhood $D \times V \subset F \times X$ such that $\cdot$ carries this neighborhood into U (this is because the continuity of $\cdot$).
Now 0 belongs to $D$ meaning that there is $\delta > 0$ such that $B(0,\delta) \subset D$ and for any $\alpha < \delta$ we have that $\cdot(\alpha, V) \subset U$