Non-Closed Set of Real Numbers is not Compact/Proof 3

Theorem

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

Let $S \subseteq \R$ be non-closed 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 closed.

Consider the relative complement of $S$ in $\R$:

$T = \relcomp \R S = \R \setminus S$

It remains to be shown that $T$ is open in $\R$.

Let $x \in T$.

For all strictly positive real numbers $\epsilon \in \R_{>0}$, it follows from Union of Open Sets of Metric Space is Open that:

$O_\epsilon := \openint \gets {x - \epsilon} \cup \openint {x + \epsilon} \to$

is open in $\R$.

Let $\CC = \set {O_\epsilon: \epsilon \in \R_{>0} }$.

Then:

$\ds \bigcup \CC = \R \setminus \set x \supseteq S$

as $x \notin S$.

So $\CC$ is an open cover for $S$.

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

Then:

$\ds \bigcup \GF = \openint \gets {x - \delta} \cup \openint {x + \delta} \to \supseteq S$

for some strictly positive real number $\delta \in \R_{>0}$.

By the rule of transposition, it follows that:

$\openint {x - \delta} {x + \delta} \subseteq T$

Hence, $T$ is open in $\R$.

$\blacksquare$