I'm reading through the Banach-Steinhaus, there are few bits I don't get.
The proof starts as
Pick balanced neighborhoods $W,U$ of $0$ in $Y$ such that $\bar{U} + \bar{U} \subset W$.
I can pick balanced neighborhoods because of theorem 1.14 (any neighborhood contains a balanced neighborhood). But why I can choose $U$ such that $\bar{U} + \bar{U} \subset W$?
Other bit I struggle is the following
If $x \in B$ then $\Gamma(x) \subset nU$ for some $n$, so that $x \in nE$ consequently $B \subset \bigcup_{n=1}^{\infty} nE$.
Here $E$ is defined as
$$ E = \bigcap_{\Lambda \in \Gamma} \Lambda^{-1}(\bar{U}) $$
Now.. $x \in B$ means $\Gamma(x)$ is bounded, which implies in can scale $U$ with an integer $n$ in such a way that $\Gamma(x) \in nU$ (It's almost the definition of boundness but there's the bit "why I can use $U$ to bound $\Gamma(x)$m since $U$ is balanced"). Also why does it follows that $x \in nE$?
The map $(x,y) \to x+y$ is continuous and the point $(0,0)$ is mapped to $0$ under this map. Since $W$ is a neighborhood $o$ there exists an open set $V$ containing $(0,0)$ such that $(x,y) \in V$ implies $x+y \in W$. By the definition of product topology there exists a neighborhood $S$ of $0$ such that $S \times S \subset V$, It follows now that $S+S \subset W$. Now you can choose an open set $U$ containing $0$ such that $\overset {-} U \subset S$. [This entire sequence of steps is there in the book; perhaps you are jumping to this theorem before reading the earlier theorems]. This gives $\overset {-} U+\overset {-} U\subset W$. Second question: $\Gamma (x)\subset nU$ implies $\Lambda (x) \in nU$ for any $\Lambda \in \Gamma$ so $\Lambda (\frac x n) \in U$. This implies $\frac x n \in E$ so $x \in nE$.