Theorem 1.14 (b) Rudin functional analysis, few clarifications.

Question

Going through such theorem which states:

In a topological vector space $X$ : (a) every neighborhood of $0$ contains a balanced neighborhood of $0$ and (b) every convex neighborhood of $0$ contains a balanced convex neighborhood of $0$

Proof of (a):

We pick a neighborhood $U$ of $0$ because of continuity of multiplication there's a $\delta > 0$ and a neighborhood $V$ such that $\alpha V \subset U$ when $|\alpha| < \delta$. Now we pick

$$ W = \bigcup_{|\alpha| <\delta} \alpha V $$

And $W$ is a neighborhood of $0$ (this is because is obtained as arbitrary union of open sets that are neighborhoods of $0$ right?). It is also balanced because if we pick $0 \leq |\beta| \leq 1$ we have

$$ \beta W = \beta \bigcup _{|\alpha| < \delta} \alpha V = \bigcup _{|\alpha| < \delta} \beta \alpha V \subset W $$

is this right?

Proof of (b): This is a bit confusing to me, all over the proof, I'll just try to highlight what I don't understand.

1. Is $W$ chosen as balanced neighborhood subset of $U$ (convex neighborhood of $0$), is it because of (a)?
2. Why $\alpha^{-1}W = W$? This should be consequence of the fact that $W$ is balanced, but I really don't see why.
3. Why $A^o \subset U$?
4. Given $A$ is convex so is $A^o$, why? I guess this because of theorem 1.13 (d), right?
5. It might be me but I really get confused with the proof that $A$ is balanced, can you clarify?

Proof of (b):

Suppose $U$ is a convex neighborhood of $0$ in $X$. Let $A = \bigcap \alpha U$ where $| \alpha | = 1$, choose $W$ as in part (a). Since $W$ is balanced, $\alpha^{-1} W = W$ when $|\alpha| = 1$; hence $W \subset \alpha U$. Thus $W \subset A$, which implies that the interior $A^o$ of $A$ is a neighborhood of $0$. Clearly $A^o \subset U$. Being an intersection of convex sets, $A$ is convex; hence so is $A^{o}$. To prove that $A^o$ is a neighborhood with the desired properties we have to show that $A^o$ is a neighborhood with the desired properties, we have to show that $A^o$ is balanced; for this it suffeces to prove that $A$ is balanced. Choose $r$ and $\beta$ so that $0 \leq r \leq 1, |\beta| = 1$. Then $$ r\beta A = \bigcap_{|\alpha| = 1} r\beta \alpha U = \bigcap_{|\alpha| = 1} r\alpha U, $$ Since $\alpha U$ is a convex set that contains $0$, we have $r \alpha U \subset \alpha U$. Thus $r\beta A \subset A$ which completes the proof.

Answer 1

1. Yes, $W$ is chosen to be the balanced open neighbourhood that was proven to exist in part (a).
2. If $|\alpha| = 1$ then $| \alpha^{-1} | = 1$ also. So if $W$ is balanced then $\alpha^{-1} W \subseteq W$ and $\alpha W \subseteq W$ whenever $|\alpha | = 1$. But $\alpha W \subseteq W$ implies that $W \subset \alpha^{-1} W$ so we must have $W = \alpha^{-1}W$.
3. By taking $\alpha = 1$ we see that $A \subseteq U$ so certainly $A^o \subseteq U$.
4. You are right, this is part of the contents of Theorem 1.13 (d).

For 5. let me elaborate a bit more. To show that $A$ is balanced, we must show that whenever $|\gamma| \leq 1$ we have that $\gamma A \subseteq A$. Equivalently, we want to show that if $|\beta| = 1$ and $0 \leq r \leq 1$ then $r \beta A \subseteq A$ (take $\gamma = r \beta$).

Then we have $$r \beta A = r \beta \bigcap_\limits{|\alpha| = 1} \alpha U = \bigcap_\limits{|\alpha| = 1} r \beta \alpha U = \bigcap_\limits{|\alpha| = 1} r \alpha U$$ where the last equality follows from the fact that, for $|\beta| = 1$, $\{ \alpha : |\alpha| = 1\} = \{ \alpha \beta : |\alpha| = 1\}$.

Now if $x \in \alpha U$, then since $r \in [0,1]$ and $0 \in \alpha U$, we have that $rx = rx + (1-r)0 \in \alpha U$ (using convexity of $\alpha U$ here). This implies that $r \alpha U \subseteq \alpha U$. Hence $$r \beta A = \bigcap _\limits{|\alpha| = 1} r \alpha U \subseteq \bigcap _\limits{|\alpha| = 1} \alpha U = A$$ so that $A$ is a balanced set.