Bolzano-Weierstrass Theorem/Lemma 1

Lemma for Bolzano-Weierstrass Theorem
Let $S$ be a non-empty subset of the real numbers such that its supremum $\map \sup S$ exists.

Let $\map \sup S \notin S$.

Then $\map \sup S$ is a limit point of $S$.

Proof
$\map \sup S$ is not a limit point of $S$.

By the negation of the definition of a limit point, there is an $\epsilon \in \R_{>0}$ such that:
 * $\paren {\map {B_\epsilon} {\map \sup S} \setminus \set {\map \sup S} } \cap S = \O$

Since $\map \sup S \notin S$, adding back $\map \sup S$ to $\map {B_\epsilon} {\map \sup S} \setminus \set {\map \sup SS}$ still gives an empty intersection with $S$.

That is:
 * $\map {B_\epsilon} {\map \sup S} \cap S = \openint {\map \sup S - \epsilon} {\map \sup S + \epsilon} \cap S = \O$

So, since $\openint {\map \sup S - \epsilon} {\map \sup S} \subset \openint {\map \sup S - \epsilon} {\map \sup S + \epsilon}$, we also have:
 * $\openint {\map \sup S - \epsilon} {\map \sup S} \cap S = \O$

Now, because $\epsilon > 0$, $\openint {\map \sup S - \epsilon} {\map \sup S}$ is non-empty.

So, there is a real $r$ such that $\map \sup S - \epsilon < r < \map \sup S$.

This $r$ is an upper bound on $S$.

We show this as follows:

Note that for any $s \in S$:
 * $s < \map \sup S$

Suppose:
 * $\map \sup S - \epsilon < s < \map \sup S$

Then:
 * $s \in \openint {\map \sup S - \epsilon} {\map \sup S}$

This contradicts what we established earlier: that $\openint {\map \sup S - \epsilon} {\map \sup S}$ cannot have an element of $S$.

That is, $r$ is an upper bound on $S$.

Hence we finally have that:
 * $s \le \map \sup S - \epsilon < r < \map \sup S$

which makes $r$ a lower upper bound on $S$ than $\map \sup S$.

This contradicts the Continuum Property of $\map \sup S$.

Hence by Proof by Contradiction it must be the case that $\map \sup S$ is a limit point of $S$.