About a proof that in a topological vector space every convex neighbourhood of $0$ contains a balanced convex neighbourhood of $0$

Question

In page 192 of the following notes

http://www.ma.huji.ac.il/~razk/iWeb/My_Site/Teaching_files/TVS.pdf

The balanced convex "neighbourhood" of $0$ was constructed by $A=\bigcap_{|\alpha|=1}\alpha U$ from the given convex neighbourhood $U$ of $0$. But it seems like it didn't show that $A$ is in fact a neighbourhood. How do one show this? Or equivalently how do we know the interior of $A$ is non-empty and contains $0$?

Answer 1

One shows first that each neighbourhood of $0$ does contain a balanced neighbourhood. Let $V$ be a neighbourhood of $0$. By continuity of multiplication, there is $\delta>0$ and an open neighbourhood $W$ of $0$ such that $tw\in V$ for all $v\in V$ and $t$ with $|t|<\delta$. The union of all $tV$ with $|t|<\delta$ will be a balanced open neighbourhood of $0$ contained in $V$.

In your case let $W$ be a balanced open neighbourhood of $0$ contained in $U$. Then $W\subseteq A$.