Theorem 1.14. 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: (a) Suppose $U$ is a neighborhood of $0$ in $X$. Since scalar multiplication is continuous, there is a $\delta>0$ and there is a neighborhood $V$ of $0$ in $X$ such that $\alpha V\subset U$ whenever $|\alpha|<\delta$. Let $W$ be the union of all these sets $\alpha V$. Then $W$ is a neighborhood of $0$, $W$ is balanced, and $W\subset U$.
(b) Suppose $U$ is a convex neighborhood of $0$ in $X$. Let $A=\bigcap \alpha U$, where $\alpha$ ranges over the scalars of absolute value $1$. 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 $A^\circ$ of $A$ is a neighborhood of $0$. Clearly $A^\circ\subset U$. Being an intersection of convex sets, $A$ is convex; hence so is $A^\circ$. To prove that $A^\circ$ is a neighborhood with the desired properties, we have to show that $A^\circ$ balanced; for this is suffices 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 proof.
My question is:
- I don't understand "where $\alpha$ ranges over the scalars of absolute value $1$". Can it be interpreted as "$|a|\leq 1$"?
- Is it correct to understand that "$\alpha^{-1}=\frac{1}{\alpha}$"?
Can someone help me? Thanks.
