Theorem: Bolzano-Weierstrass Theorem/Lemma 3

Every bounded, infinite subset $S$ of $\R$ has at least one limit point.

Proof:
As $S$ is bounded, it is certainly bounded above.
Also, since {{hypothesis}} $S$ is infinite, it is of course non-empty.
Hence, by the completeness axiom of the real numbers, $\tilde s_0 = \sup S$ exists as a real.
There are two cases:
;Case $1.0$ -- $\tilde s_0 \notin S$:
By '''Lemma 1''', $\tilde s_0$ is a limit point of $S$ and the proof is complete.
;Case $2.0$ -- $\tilde s_0 \in S$:
Because $S$ is infinite, $S_1 = S \setminus \set {\tilde s_0}$ is non-empty.
Of course, as $S_1 \subset S$, it is still bounded above because $S$ is.
Hence, $\tilde s_1 = \sup S_1$ exists, again, by the completeness axiom of the reals.
Once again, we have two cases:
;Case $1.1$ -- $\tilde s_1 \notin S_1$
In this case we have a limit point of $S_1$.
Hence we have a limit point of $S$ as $S_1 \subset S$.
;Case $2.1$ -- $\tilde s_1 \in S_1$
In this case we continue our analysis withP:
:$\tilde s_2 = \sup S_1 \setminus \set {\tilde s_1} = \sup S \setminus \set {\tilde s_0, \tilde s_1}$
Continuing like this, we note that our analysis stops after a finite number of steps {{iff}} we ever reach a case of the form '''Case $1.k$''' for some $k \in \N$.
In this case:
:$\tilde s_k = \sup S_k \notin S_k$
and we use '''Lemma 1''' to show that $\tilde s_k$ is a limit point of $S_k$ and, therefore, of $S$.
Otherwise, the proof continues indefinitely if we keep getting cases of the form '''Case $2.k$''' for all $k \in \N$.
In that case:
:$\tilde s_k \in S_k$
and we get a sequence $\tilde S = \sequence {\tilde s_k}_{k \mathop \in \N}$ of reals with the following properties:
:$(1): \quad$ Each $\tilde s_k$ is in $S$.
This is because, as remarked earlier, the only way we get our sequence is if $\tilde s_k \in S_k$.
But $S_k$ is either $S$ when $k = 0$ or $S \setminus \set {\tilde s_0, \ldots, \tilde s_{k - 1} }$ when $k \ge 1$.
In both cases, $S_k$ is a subset of $S$.
From this fact, the result easily follows.
:$(1): \quad \tilde s_k > \tilde s_{k + 1}$
To see this, note that:
:$\tilde s_{k + 1} \in S_{k + 1} = S \setminus \set {\tilde s_0, \ldots , \tilde s_k} = S_k \setminus \set {\tilde s_k}$
So, first:
:$\tilde s_{k + 1} \ne \tilde s_k$
and secondly, because $\tilde s_k$ is by construction an upper bound on $S_k$, and therefore on its subset $S_{k + 1}$:
:$\tilde s_k \ge \tilde s_{k + 1}$
Combining both these facts gives our present claim.
Now, the first property says that the set of all the $\tilde s_k$'s, which is $\tilde S$, is a subset of $S$.
So, it is bounded because $S$ is.
Then, certainly, it is also bounded below.
Also, $\tilde S$ is obviously non-empty because it is infinite.
Hence, one final application of the completeness axiom of the reals gives that $\underline s = \inf \tilde S$ exists as a real.
Note that $\underline s \notin \tilde S$.
Otherwise, if $\underline s = \tilde s_k$ for some $k \in \N$, by the second property of our sequence, we would have $\underline s > \tilde s_{k + 1}$.
This would contradict the fact that $\underline s$ is a lower bound on $\tilde S$.
But then, by '''Lemma $2$''' above, $\underline s$ is a limit point of the set $\tilde S$ and, therefore, of its superset $S$.
{{qed|lemma}}