Bolzano-Weierstrass Theorem/Lemma 3

Theorem
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 $S$ 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 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 $S$.