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 lemmata 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 and its related Euclidean topology.

Lemma $0$
The above trivial lemma is given purely for the sake of argument completeness.

It is assumed implicitly in all the proofs below.

Lemma $2$
Throughout the rest of the proof:
 * $\tilde s$ is used to mean $\map \sup S$
 * $\underline s$ is used to mean $\map \inf S$.

Lemma $3$
Lemma 3: (Bolzano-Weierstrass on $\R$) 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 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:


 * Case 1.0: $\tilde s_0 \notin S$: Then, by Lemma 1 above, $\tilde s_0$ is a limit point of $S$ and the proof is complete.


 * Case 2.0: $\tilde s_0 \in S$: Then, 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.

So, yet again, we have two cases:
 * Case 1.1: either $\tilde s_1 \notin S_1$, in which case we stop because we get a limit point of $S_1$ (and hence of $S$ as $S_1 \subset S$),
 * Case 2.1: or $\tilde s_1 \in S_1$, in which case we continue our analysis with $\tilde s_2 = \sup S_1 \setminus \set {\tilde s_1} = \sup S \setminus \set {\tilde s_0, \tilde s_1}$.

Continuing like this, we note that our analysis stops after a finite number of steps if and only if 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:


 * 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 claim easily follows.


 * $\tilde s_k > \tilde s_{k + 1}$.

To see this, note that $\tilde s_{k + 1} \in S_{k + 1} = S \setminus \set {\tilde s_0, \ldots, \tilde s_k} = S_k \setminus \set {\tilde s_k}$.

So, firstly, $\tilde s_{k + 1} \ne \tilde s_k$ and, secondly, because $\tilde s_k$ is by construction an upper bound on $S_k$ (and therefore on its subset $S_{k + 1}$), we have $\tilde s_k \ge \tilde s_{k + 1}$.

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 $S$.

Before moving onto the proof of the main theorem, I skim over the elementary concept of projection that will be used in the proof. Fix positive integers $m, n$ where $m \le n$. Then, for any set $X$,
 * There is a function $\pi_{1, \ldots, m}: X^n \to X^m$ such that $\map {\pi_{1, \ldots, m} } {x_1, \ldots, x_m, \ldots, x_n} = \tuple {x_1, \ldots, x_m}$. Essentially, $\pi_{1, \ldots, m}$ takes in a coordinate of $n$ elements of $X$ and simply outputs the first $m$ elements of that coordinate.
 * There is a function $\pi_m: X^n \to X$ such that $\map {\pi_m} {x_1, \ldots, x_m, \ldots, x_n} = x_m$. Essentially, $\pi_m$ takes in a coordinate of $n$ elements of $X$ and outputs just the $m$th element of that coordinate.
 * In general, for positive integers $i \le j < n$, there is a function $\pi_{i, \ldots, j}: X^n \to X^{j - i + 1}$ such that $\map {\pi_{i, \ldots, j} } {x_1, \ldots, x_i, \ldots, x_j, \ldots, x_n} = \tuple{x_i, \ldots, x_j}$.

We begin with an easy lemma:

Lemma $4$
Lemma 4: For positive integers $m < n$ and $S \subseteq X^n$, $S \subseteq \map {\pi_{1, \ldots, m} } S \times \map {\pi_{m + 1, \ldots, n} } S$.

Proof: Fix any $\tuple {x_1, \ldots, x_m, x_{m + 1}, \ldots, x_n} \in S$. Then, by the Definition:Image of Subset under Mapping, $\tuple {x_1, \ldots, x_m} \in \map {\pi_{1, \ldots, m} } S$ because $\map {\pi_{1, \ldots, m} } {x_1, \ldots, x_m, x_{m + 1}, \ldots, x_n} = \tuple {x_1, \ldots, x_m}$. Similarly, $\tuple {x_{m + 1}, \ldots, x_n} \in \map {\pi_{m + 1, \ldots, n} } S$.

So, by Definition:Cartesian Product, $\tuple {x_1, \ldots, x_m, x_{m + 1}, \ldots, x_n} \in \map {\pi_{1, \ldots, m} } S \times \map {\pi_{m + 1, \ldots, n} } S$.

Since $\tuple {x_1, \ldots, x_m, x_{m + 1}, \ldots, x_n}$ was an arbitrary element of $S$, this means that $S \subseteq \map {\pi_{1, \ldots, m} } S \times \map {\pi_{m + 1, \ldots, n} } S$.

Lemma $5$
Lemma 5: For positive integers $i \le j \le n$ and $S \subseteq \R^n$, if $S$ is a bounded space in $\R^n$, then so is $\map {\pi_{i, \ldots, j} } S$ in $\R^{j - i + 1}$.

Proof: For a contradiction, assume otherwise. So, by the negation of the definition of a bounded space, for every $K \in \R$, there are $x = \tuple {x_i, \ldots, x_j}$ and $y = \tuple {y_i, \ldots, y_j}$ in $\map {\pi_{i, \ldots, j} } S$ such that $\map d {x, y} = \size {x - y} = \sqrt {\ds \sum_{s \mathop = i}^j \paren {x_s - y_s}^2} > K$ where we get the formula $\size {x - y} = \sqrt {\ds \sum_{s \mathop = i}^j \paren {x_s - y_s}^2}$ because we are working with the euclidean metric on all real spaces (after a suitable change of variables in the summation). Now, by definition of the image set $\map {\pi_{i, \ldots, j} } S$, there are points $x' = \tuple {x_1, \ldots, x_i, \ldots, x_j, \ldots, x_n}$ and $y' = \tuple {y_1, \ldots, y_i, \ldots, y_j, \ldots, y_n}$ in $S$ from which $x$ and $y$ originated as coordinate components. Therefore, $\map d {x', y'} = \size {x' - y} = \sqrt {\ds \sum_{s \mathop = 1}^n \paren {x_s - y_s}^2} \ge \sqrt {\ds \sum_{s \mathop = i}^j \paren {x_s - y_s}^2} > K$ contradicting the fact that $S$ is a bounded space.

Lemma $6$
Lemma 6: For any function $f: X \to Y$ and subset $S \subseteq X$, if $S$ is infinite and $\map f S$ is finite, then there exists some $y \in \map f S$ such that $\map {f^{-1} } y \cap S$ is infinite. Here, $\map {f^{-1} } y$ is the preimage of the element $y$.

Proof: If there weren't such an element in $\map f S$, then for all $y \in \map f S$, $\map {f^{-1} } y \cap S$ would be finite. Also, since $\map f S$ is finite, we may list its elements: $y_1, \ldots, y_n$ (there must be at least one image element as $S$ is non-empty).

Then, by repeated applications of Union of Finite Sets is Finite, we get that:
 * $\ds \bigcup_{y \mathop \in \map f S} \paren {\map {f^{-1} } y \cap S} = \paren {\map {f^{-1} } {y_1} \cap S} \cup \cdots \cup \paren {\map {f^{-1} } {y_n} \cap S}$

must be finite.

But notice that:

This contradicts the fact that $S$ is infinite.

Recall the statement of the theorem:

We proceed by the Principle of Mathematical 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 $\R^{n + 1}$.

Consider the image of $S$ under the projection functions $\pi_{1, \ldots, n}$ and $\pi_{n + 1}$: $S_{1, \ldots, n} = \map {\pi_{1, \ldots, n} } S$ and $S_{n + 1} = \map {\pi_{n + 1} } S$. Then,
 * Because $S$ is a bounded space of $\R^{n + 1}$, $S_{1, \ldots, n}$ and $S_{n + 1}$ must be bounded spaces of $\R^n$ and $\R$ respectively by Lemma 5.
 * Also, $S \subseteq S_{1, \ldots, n} \times S_{n + 1}$ by Lemma 4. So, by the fact that $S$ is infinite and Subset of Finite Set is Finite, $S_{1, \ldots, n} \times S_{n + 1}$ is infinite. But then, by Product of Finite Sets is Finite, either $S_{1, \ldots, n}$ or $S_{n + 1}$ must be infinite.

Let us analyze the case that $S_{1, \ldots, n}$ is infinite first. Then, $S_{1, \ldots, n}$ is an infinite bounded space of $\R^n$. So, by the induction hypothesis, it has a limit point $l_{1, \ldots, n} = \tuple {l_1, \ldots, l_n}$.

By definition of limit point, for every $\epsilon > 0$, there is an $s_\epsilon \in \paren {\map {B_\epsilon} l \setminus \set l} \cap S_{1, \ldots, n}$. To this $s_\epsilon$, which is in $S_{1, \ldots, n}$, there corresponds the set of all $\paren {n + 1}$th coordinates of $S$-elements that have $s_\epsilon = \tuple {s_{\epsilon, 1}, \ldots, s_{\epsilon, n} }$ as their first $n$ coordinates: $\tilde S_{\epsilon, n + 1} = \map {\pi_{n + 1} } {\map {\pi_{1, \ldots, n}^{-1} } {s_\epsilon} \cap S} \subseteq S_{n + 1}$ and collect every element of such sets in one set: $\tilde S_{n + 1} = \ds \bigcup_{\epsilon \mathop > 0} \tilde S_{\epsilon, n + 1} = \map {\pi_{n + 1} } {\paren {\bigcup_{\epsilon \mathop > 0} \map {\pi_{1, \ldots, n}^{-1} } {s_\epsilon} } \cap S} \subseteq S_{n + 1}$.

Now, if $\tilde S_{n + 1}$ is