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

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 $\R$:
 * $T = \complement_\R \left({S}\right) = \R \setminus S$

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:


 * $O_\epsilon := \left({-\infty \,.\,.\, x - \epsilon}\right) \cup \left({x + \epsilon \,.\,.\, {+\infty}}\right)$

is open in $\R$.

Let $\mathcal C = \left\{{O_\epsilon: \epsilon \in \R_{>0}}\right\}$.

Then:


 * $\displaystyle \bigcup \mathcal C = \R \setminus \left\{{x}\right\} \supseteq S$

as $x \notin S$.

So $\mathcal C$ is an open cover for $S$.

Let $\mathcal F$ be a finite subcover of $\mathcal C$ for $S$.

Then:


 * $\displaystyle \bigcup \mathcal F = \left({-\infty \,.\,.\, x - \delta}\right) \cup \left({x + \delta \,.\,.\, {+\infty}}\right) \supseteq S$

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

By the rule of transposition, it follows that:
 * $\left({x - \delta \,.\,.\, x + \delta}\right) \subseteq T$

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