# Showing that a set is balanced but its interior is not

I'm having some difficulty visualizing a particular set for a functional analysis homework problem. That is, let $B= \{(z_1, z_2) \in \mathbb{C}^2: |z_1| \leq |z_2|\}$. I'm asked to show that $B$ is balanced, i.e. for a scalar $\alpha$ with $|\alpha| \leq 1$ we have $\alpha B \subset B$, and that the interior of $B$ is not. I've never worked in complex $2$-space before, so it's hard for me to picture the set $B$. My guess is that $B$ is some sort of annular region, so that would make it clear that it is a balanced set, I think.

I'm not so sure about the interior, but perhaps that follows from the origin somehow not being in the interior of $B$? This is an exercise from Rudin's $\textit{Functional Analysis}$ on p. $37$.

• You are right: $(0,0)$ is not an interior point of $B$. In any neighborhood of $(0,0)$ you can find points of the form $(2a,a)$ for some $a>0$, for instance. – njguliyev Sep 3 '13 at 22:29

## 1 Answer

You don't need to visualize $B$ to show that it's balanced. Just check the definition: if $(z_1,z_2)\in B$ and $|\alpha|\le 1$, then $(\alpha z_1,\alpha z_2)\in B$.

For any $\epsilon>0$, the point $(\epsilon,0)$ is not in $B$. Therefore, $(0,0)$ is not in the interior point of $B$, which implies the latter is not a balanced set.

• Why is it necessary that $|\alpha|\leq 1$ to have that $(\alpha z_1, \alpha z_2)\in B$? Doesn't it just always hold for any $(z_1, z_2)\in B$? – PPR Jun 7 '15 at 15:37