I'm reading Rudin's Functional Analysis and came upon this proof that every topological vector space has a balanced local base: a subset $S$ of a TVS $X$ is balanced if $\alpha S \subseteq S$ whenever $|\alpha| \leq 1$.
To do this, he shows that every neighborhood of $0$ contains a balanced neighborhood of $0$. However, I'm not clear on the first step of his proof below: why does continuity of scalar multiplication ensure the existence of such a $\delta > 0$ and an open neighborhood $V$ of $0$ such that $\alpha V \subseteq U$ whenever $|\alpha| < \delta$? Thank you so much.
Recall: the "continuity of scalar multiplication" means precisely that the function $\mu = (t,x) \mapsto tx : \mathbb{R} \times X \to X$ is continuous.
Since $U$ is a neighborhood of $0$, there is an open set $U_0 \subseteq U$ such that $0 \in U_0$. Now since $U_0$ is open, $\mu^{-1}(U_0)$ is an open subset of $\mathbb{R} \times X$. Moreover, $(0,0) \in \mu^{-1}(U_0)$.
So, $\mu^{-1}(U_0)$ is an open neighborhood of $(0,0)$ in $\mathbb{R} \times X$. By definition of the product topology, this means that there are open sets $A \subseteq \mathbb{R}$ and $V \subseteq X$ such that $(0,0) \in A \times V \subseteq \mu^{-1}(U_0)$. Note: in particular, $V$ is an open neighborhood of $0$!
Since $0 \in A$ and $A \subseteq \mathbb{R}$ is open, we have $(-\delta, \delta) \subseteq A$ for some $\delta > 0$. Thus, $(-\delta, \delta) \times V \subseteq A \times V \subseteq \mu^{-1}(U_0)$.
The claim now follows. Let $\alpha \in (-\delta, \delta)$ and $v \in V$ be arbitrary. Then $(\alpha, v) \in A \times V \subseteq \mu^{-1}(U_0)$, so $$\alpha v = \mu(\alpha, v) \in U_0.$$ Since $U_0 \subseteq U$, we have $\alpha v \in U$. Since $v$ was arbitrary, we have $\alpha V \subseteq U$. Since $\alpha$ was arbitrary, we are done. $\square$