Bolzano-Weierstrass Theorem/Lemma 0

Lemma for Bolzano-Weierstrass Theorem
Let $S' \subseteq S \subseteq \R$.

Then every limit point of $S'$ is a limit point of $S$.

Proof
Consider an arbitrary limit point $l$ of $S'$.

Fix $\epsilon \in \R_{>0}$.

By definition of limit point:
 * $\paren {\map {B_\epsilon} l \setminus \set l} \cap S' \ne \O$

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

Thus there exists a real $s_\epsilon$ in both $\map {B_\epsilon} l \setminus \set l$ and $S'$.

But since $S' \subseteq S$, $s_\epsilon \in S'$ implies $s_\epsilon \in S$.

So, in other words, $s_\epsilon$ is in both $\map {B_\epsilon} l \setminus \set l$ and $S$.

That is:
 * $\paren {\map {B_\epsilon} l \setminus \set l} \cap S \ne \O$

This is exactly what it means for $l$ to be a limit point of $S$.

As $l$ is arbitrary, this result is true for every limit point $l$ of $S'$.