Balanced Subset of Complex Plane is Bounded or Entire Space

Theorem

Consider $\C$ as a vector space over $\C$.

Let $E$ be a balanced subset of $\C$.

Then $E$ is bounded, or $E = \C$.

Proof

Suppose that $E$ is not bounded.

Then, for each $R > 0$ there exists some $z_R \in E$ such that $\size {z_R} > R$.

We show that:

$\map {B_R} 0 \subseteq E$ for each $R > 0$

where $\map {B_R} 0$ is the open ball centered at $0$ with radius $R$.

Let:

$w \in \map {B_R} 0$

Then:

$\ds \cmod {\frac w {z_R} } < 1$

So, since $E$ is balanced, we have:

$\ds z_R \cdot \paren {\frac w {z_R} } = w \in E$

giving:

$\map {B_R} 0 \subseteq E$ for each $R > 0$.

From Union of Subsets is Subset, we have:

$\ds \bigcup_{R > 0} \map {B_R} 0 \subseteq E$

giving:

$\C \subseteq E$

Since we also have $E \subseteq \C$, we obtain $E = \C$.

$\blacksquare$