Balanced Subset of Real Numbers is Bounded or Entire Space

Theorem
Consider $\R$ as a vector space over $\R$.

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

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

Proof
Suppose that $E$ is not bounded.

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

We show that:


 * $\closedint {-M} M \subseteq E$ for each $M > 0$.

Let:


 * $t \in \closedint {-M} M$

Then, we have:


 * $\ds \size {\frac t {x_M} } < 1$

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


 * $\ds x_M \cdot \paren {\frac t {x_M} } = t \in E$

So:


 * $\closedint {-M} M \subseteq E$ for each $M > 0$.

From Union of Subsets is Subset, we then have:


 * $\ds \bigcup_{M > 0} \closedint {-M} M \subseteq E$

So:


 * $\R \subseteq E$

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