Unbounded Set of Real Numbers is not Compact/Proof 1

Theorem

Let $\R$ be the set of real numbers considered as a Euclidean space.

Let $S \subseteq \R$ be unbounded in $\R$.

Then $S$ is not a compact subspace of $\R$.

Proof

By the rule of transposition, it suffices to show that if $S$ is a compact subspace of $\R$, then $S$ is bounded.

Let $\CC$ be the set of all open $\epsilon$-balls of $0$ in $\R$:

$\CC = \set {\map {B_\epsilon} 0: \epsilon \in \R_{>0}}$

We have that:

$\ds \bigcup \CC = \R \supseteq S$

From Open Ball of Metric Space is Open Set, it follows that $\CC$ is an open cover for $S$.

Let $\FF$ be a finite subcover of $\CC$ for $S$.

Then $\ds \bigcup \FF$ is the largest open $\epsilon$-ball in $\FF$.

Thus:

$S \subseteq \ds \bigcup \FF \in \CC$

Hence, $S$ is bounded.

$\blacksquare$

Sources

• 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{III}$: Metric Spaces: Compactness