Non-Closed Set of Real Numbers is not Compact

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
Consider the complement of $S$ in $\R$:
 * $S' = \complement_\R \left({S}\right) = \R \setminus S$

As $S$ is not closed, by definition $S'$ is not open.

Thus by definition there exists $x \in S'$ such that:


 * $\forall \epsilon \in \R_{>0}: B_\epsilon \left({x}\right) \notin S'$

where $B_\epsilon \left({x}\right)$ denotes the open $\epsilon$-ball of $x$.

That is:


 * $\forall \epsilon \in \R_{>0}: \exists y \in S: y \in B_\epsilon \left({x}\right)$

That is, for all positive $\epsilon$ there exists a $y$ within $\epsilon$ of $x$.

From Union of Open Sets of Metric Space:


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

is an open set of $\R$.

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 of $S$.

Let $\mathcal F \subseteq \mathcal C$ be a finite subset of $\mathcal C$.

Then:


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

for some $\epsilon' \in \R_{>0}$.

But, as has been seen, there exists $y \in S$ closer to $x$ than $\epsilon'$.

Thus $\displaystyle \bigcup \mathcal F$ does not cover $S$.

This construction demonstrates an open cover $\mathcal C$ of $S$ which has no finite subcover.

So, by definition, $S$ is not a compact subspace of $\R$.