Non-Closed Set of Real Numbers is not Compact/Proof 1

Proof
Consider the complement of $S$ in $\R$:
 * $S' = \relcomp \R S = \R \setminus S$

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

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


 * $\forall \epsilon \in \R_{>0}: \map {B_\epsilon} x \notin S'$

where $\map {B_\epsilon} x$ denotes the open $\epsilon$-ball of $x$.

That is:


 * $\forall \epsilon \in \R_{>0}: \exists y \in S: y \in \map {B_\epsilon} x$

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

From Union of Open Sets of Metric Space is Open:


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

is an open set of $\R$.

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 of $S$.

Let $\FF \subseteq \CC$ be a finite subset of $\CC$.

Then:


 * $\ds \bigcup \FF = \openint \gets {x - \epsilon'} \cup \openint {x + \epsilon'} \to$

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

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

Thus $\ds \bigcup \FF$ does not cover $S$.

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

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