Balanced set in $\mathbb{K}$

Question

Let $A$ be a balanced set in $\mathbb{K}$. Why do we have : if $A$ is not bounded then $A=\mathbb{K}$ ?

Answer 1

Let $\mathbb{K}$ be a field with an absolute value function $\lvert - \rvert : \mathbb{K} \to \mathbb{R}$. If $A$ is balanced, then for each $x \in A$ and each $\lambda \in \mathbb{K}$ such that $\lvert \lambda \rvert \le 1$ we have $\lambda x \in A$.

Hence obviously $0 \in A$.

For $x \in A \setminus \{ 0 \}$ we have $r = \lvert x \rvert > 0$. We show that $D(r) = \{ y \in \mathbb{K} \mid \lvert y \rvert \le r \} \subset A$. So let $y \in \mathbb{K} \setminus \{ 0 \}$ such that $\lvert y \rvert \le r$. Then $\lvert y/x \rvert = \lvert y \rvert/ \lvert x \rvert \le r/r = 1$, hence $y = y/x \cdot x \in A$.

Let $z \in \mathbb{K}$. Since $A$ is unbounded, there exists $x \in A$ such that $\lvert z \rvert \le \lvert x \rvert = r$. We conclude $z \in D(r) \subset A$.