Balanced Subset of Real Numbers is Bounded or Entire Space
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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 $E = \R$.
$\blacksquare$