Bolzano-Weierstrass Theorem/General Form

Theorem
Every infinite bounded space in a real Euclidean space has at least one limit point.

Proof
The proof of this theorem will be given as a series of lemmas that culminate in the actual theorem in the end. Unless otherwise stated, all real spaces occurring in the proofs are equipped with the euclidean metric/topology.

Lemma 0: Suppose $S' \subseteq S \subseteq \mathbb{R}$. Then, any limit point of $S'$ is a limit point of $S$.

Proof: Consider any limit point $l$ of $S'$ and fix an $\epsilon > 0$. Then, by definition, $(B_\epsilon(l) \setminus \{l\}) \cap S' \neq \emptyset$. Thus, there is a real $s_\epsilon$ in both $B_\epsilon(l) \setminus \{l\}$ 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 $B_\epsilon(l) \setminus \{l\}$ and $S$ i.e. $(B_\epsilon(l) \setminus \{l\}) \cap S \neq \emptyset$. This is exactly what it means for $l$ to be a limit point of $S$.

This trivial lemma is given purely for the sake of argumentative completeness. It is assumed implicitly in all the proofs below.

$\square$

Lemma 1: Suppose $S$ is a non-empty subset of the reals such that its supremum, $\tilde{s}$, exists. If $\tilde{s} \notin S$, then $\tilde{s}$ is a limit point of $S$.

Proof: For a contradiction, suppose $\tilde{s}$ were not a limit point of $S$. So, by the negation of the definition of a limit point, there is an $\epsilon > 0$ such that $(B_\epsilon(\tilde{s}) \setminus \{\tilde{s}\}) \cap S = \emptyset$.

Since $\tilde{s} \notin S$, adding back $\tilde{s}$ to $B_\epsilon(\tilde{s}) \setminus \{\tilde{s}\}$ still gives an empty intersection with $S$. That is, $B_\epsilon(\tilde{s}) \cap S = (\tilde{s}-\epsilon, \tilde{s}+\epsilon) \cap S = \emptyset$. So, since $(\tilde{s}-\epsilon, \tilde{s}) \subset (\tilde{s}-\epsilon, \tilde{s}+\epsilon)$, we also have $(\tilde{s}-\epsilon, \tilde{s}) \cap S = \emptyset$.

Now, because $\epsilon > 0$, $(\tilde{s}-\epsilon, \tilde{s})$ is non-empty. So, there is a real $r$ such that $\tilde{s}-\epsilon < r < \tilde{s}$.

This $r$ is an upper bound on $S$. To see this, note that for any $s \in S$, $s < \tilde{s}$. Indeed, $s \leq \tilde{s}-\epsilon$ because otherwise, $\tilde{s}-\epsilon < s < \tilde{s}$ and $s$ would be in $(\tilde{s}-\epsilon, \tilde{s})$ contradicting what we established earlier: that $(\tilde{s}-\epsilon, \tilde{s})$ cannot have an element of $S$.

Hence, we finally have $s \leq \tilde{s}-\epsilon < r < \tilde{s}$, making $r$ a lower upper bound on $S$ than $\tilde{s}$. This contradicts the supremum property of $\tilde{s}$.

$\square$

Lemma 2: Suppose $S$ is a non-empty subset of the reals such that its infimum, $\underline{s}$, exists. If $\underline{s} \notin S$, then $\underline{s}$ is a limit point of $S$.

Proof: The proof is entirely analogous to that of the first lemma.

$\square$

Lemma 3: (Bolzano-Weierstrass on $\mathbb{R}$) Every bounded, infinite subset $S$ of $\mathbb{R}$ has at least one limit point.

Proof: As $S$ is bounded, it is certainly bounded above. Also, since it is infinite by hypothesis, it is of course non-empty. Hence, by the completeness axiom of the reals, $\tilde{s}_0 = \sup S$ exists as a real. Now, there are two cases:


 * $\tilde{s}_0 \notin S$: Then, by Lemma 1 above, $\tilde{s}_0$ is a limit point of $S$ and we are done.


 * $\tilde{s}_0 \in S$: Then, because $S$ is infinite, $S_1 = S \setminus \{\tilde{s}_0\}$ 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. So, yet again, we have two cases: either $\tilde{s}_1 \notin S_1$, in which case we again get a limit point of $S_1$ (and hence, of $S$ as $S_1 \subset S$), or $\tilde{s}_1 \in S_1$, in which case we have to restart our analysis with $\tilde{s}_2 = \sup S_1 \setminus \{\tilde{s}_1\} = \sup S \setminus \{\tilde{s}_0, \tilde{s}_1\}$.

Continuing in this manner, we note that the only way the proof stops after a finite number of steps is if we ever find an $\tilde{s}_k = \sup S_k \notin S_k$. In that case, 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 $\tilde{s}_k \in S_k$ (possible because $S$ is infinite). In this case, we get a sequence $\tilde{S} = \{\tilde{s}_k\}_{k \in \mathbb{N}}$ of reals with the following properties:


 * 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 $S \setminus \{\tilde{s}_0, \ldots, \tilde{s}_{k-1}\}$ when $k \geq 1$. In both cases, $S_k$ is a subset of $S$. From this fact, the claim easily follows.


 * $\tilde{s}_k > \tilde{s}_{k+1}$. To see this, note that $\tilde{s}_{k+1} \in S_{k+1} = S \setminus \{\tilde{s}_0, \ldots, \tilde{s}_k\} = S_k \setminus \{\tilde{s}_k\}$. So, firstly, $ \tilde{s}_{k+1} \neq \tilde{s}_k$ and, secondly, because $ \tilde{s}_k$ is by construction an upper bound on $S_k$ (and therefore on its subset $S_{k+1}$), we have $ \tilde{s}_k \geq \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 $k \in \mathbb{N}$, 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 3 above, $\underline{s}$ is a limit point of the set $\tilde{S}$ and, therefore, of its superset $S$. Thus, in every possible case, a limit point of $S$ exists.

$\square$

Main Theorem: (Bolzano-Weierstrass in $\mathbb{R}^n$) Every infinite, bounded subspace $S$ of $\mathbb{R}^n$ has at least one limit point.

Proof: We proceed by induction on the positive integer $n$:

(Base Case) When $n = 1$, the theorem is just Lemma 3 above which has been adequately proven.

(Inductive Step) Suppose that the theorem is true for some positive integer $n$. We must show that it is also true for the positive integer $n+1$. So, fix any infinite, bounded subset $S$ of $\mathbb{R}^{n+1}$.

Now, notice that $\mathbb{R}^{n+1} = \mathbb{R}^n \times \mathbb{R}$. So, there are two functions:
 * One maps any $(x_1, \ldots, x_n, x_{n+1}) \in \mathbb{R}^n \times \mathbb{R}$ to $(x_1, \ldots , x_n) \in \mathbb{R}^n$. Call this function $\pi_{1, \ldots , n}$.
 * The other maps any $(x_1, \ldots, x_n, x_{n+1}) \in \mathbb{R}^n \times \mathbb{R}$ to $x_{n+1} \in \mathbb{R}$. Call this function $\pi_{n+1}$.

Consider the image of $S$ under each of these functions: $S_{1, \ldots, n} = \pi_{1, \ldots , n}(S) = \{(x_1, \ldots , x_n) \in \mathbb{R}^n:\ (x_1, \ldots , x_n, x_{n+1}) \in S\}$ and $S_{n+1} = \pi_{n+1}(S) = \{x_{n+1} \in \mathbb{R}:\ (x_1, \ldots , x_n, x_{n+1}) \in S\}$.

Note that $S \subseteq S_{1, \ldots, n} \times S_{n+1}$ (to see this, simply fix any $(x_1, \ldots , x_n, x_{n+1}) \in S$; then consider where its $\pi_{1, \ldots , n}$ and $\pi_{n+1}$ images land). Thus,
 * Because $S$ is a bounded space of $\mathbb{R}^{n+1}$, $S_{1, \ldots, n}$ must be a bounded space of $\mathbb{R}^n$. To see this, assume otherwise. So, by the negation of the definition of a bounded space, for every $K \in \mathbb{R}$, there are $x=(x_1, \ldots , x_n)$ and $y=(y_1, \ldots , y_n) $ in $S_{1, \ldots , n}$ such that $$d(x,y) = |x - y| = \sqrt{\sum\limits_{i=1}^{n}(x_i -y_i)^2} > K$$ where we get the formula $|x - y| = \sqrt{\sum\limits_{i=1}^{n}(x_i -y_i)^2}$ because we are working with the euclidean metric on all real spaces. Now, by definition of $S_{1, \ldots , n}$ there are points $x' = (x_1, \ldots , x_n, x_{n+1})$ and $y' = (y_1, \ldots , y_n, y_{n+1})$ in $S$ from which $x$ and $y$ originated. Therefore, $$d(x',y') = |x'-y'| = \sqrt{\sum\limits_{i=1}^{n+1}(x_i -y_i)^2} \geq \sqrt{\sum\limits_{i=1}^{n}(x_i -y_i)^2} > K$$ contradicting the fact that $S$ is a bounded space. Similarly, for the same reason, $S_{n+1}$ must be bounded in $\mathbb{R}$.
 * Also, by the fact that $S$ is infinite and the contrapositive of Subset of Finite Set is Finite, $S_{1, \ldots, n} \times S_{n+1}$ must be infinite. But then, by the contrapositive of Product of Finite Sets is Finite, either $S_{1, \ldots , n}$ or $S_{n+1}$ must be infinite.

$\blacksquare$

Also known as
Some sources refer to this result as the Weierstrass-Bolzano theorem.