Editing Bolzano–Weierstrass theorem (section)

== Proof ==

First we prove the theorem for <math>\mathbb{R}</math> (set of all [[real number]]s), in which case the ordering on <math>\mathbb{R}</math> can be put to good use. Indeed, we have the following result:

'''Lemma''': Every infinite sequence <math>(x_n)</math> in <math>\mathbb{R}</math> has an infinite [[monotone sequence|monotone]] [[subsequence]] (a subsequence that is either [[non-decreasing]] or [[non-increasing]]).

'''Proof<ref>Bartle and Sherbert 2000, pp. 78-79.</ref>''': Let us call a positive integer-valued index <math>n</math> of a sequence a "peak" of the sequence when <math>x_m \leq x_n</math> for every <math>m > n</math>. Suppose first that the sequence has infinitely many peaks, which means there is a subsequence with the following indices <math>n_1<n_2<n_3<\dots<n_j<\dots</math> and the following terms <math>x_{n_1} \geq x_{n_2} \geq x_{n_3} \geq \dots \geq x_{n_j} \geq \dots</math>. So, the infinite sequence <math>(x_n)</math> in <math>\mathbb{R}</math> has a monotone (non-increasing) subsequence, which is <math>(x_{n_j})</math>. But suppose now that there are only finitely many peaks, let <math>N</math> be the final peak if one exists (let <math>N=0</math> otherwise) and let the first index of a new subsequence <math>(x_{n_j})</math> be set to <math>n_1=N+1</math>.  Then <math>n_1</math> is not a peak, since <math>n_1</math> comes after the final peak, which implies the existence of <math>n_2</math> with <math>n_1<n_2</math> and <math>x_{n_1} < x_{n_2}</math>. Again, <math>n_2</math> comes after the final peak, hence there is an <math>n_3</math> where <math>n_2<n_3</math> with <math>x_{n_2} \leq x_{n_3}</math>. Repeating this process leads to an infinite non-decreasing subsequence&nbsp; <math>x_{n_1} \leq x_{n_2} \leq x_{n_3} \leq \ldots</math>, thereby proving that every infinite sequence <math>(x_n)</math> in <math>\mathbb{R}</math> has a monotone subsequence.

Now suppose one has a [[bounded sequence]] in <math>\mathbb{R}^1</math>; by the lemma proven above [[there exists]] a monotone subsequence, likewise also bounded. It follows from the [[monotone convergence theorem]] that this subsequence converges.

The general case in <math>\mathbb{R}^n</math> can be proven using this lemma as follows. Firstly, we will acknowledge that any sequence <math>(x_m)_{m \in I}</math> in <math>\mathbb{R}^n</math> (where <math>I</math> denotes its [[index set]]) has a convergent subsequence if and only if there exists a countable set <math>K \subseteq I</math>  such that <math>(x_m)_{m \in K}</math> converges. Let <math>(x_m)_{m \in I}</math> be any bounded sequence in <math>\mathbb{R}^n</math>, then it can be expressed as an n-tuple of sequences in <math>\mathbb{R}</math> by writing <math>x_m = (x_{m1}, x_{m2}, \dots, x_{mn})</math>, where <math>(x_{mj})_{m \in I}</math> is a sequence for <math>j=1,2,\dots,n</math>. Since <math>(x_m)</math> is bounded, <math>(x_{mj})</math> is also bounded for <math>j=1,2,\dots, n</math>. It follows then by the lemma that <math>(x_{m1})</math> has a convergent subsequence and hence there exists a countable set <math>K_1 \subseteq I</math> such that <math>(x_{m1})_{m \in K_1}</math> converges. For the sequence <math>(x_{m2})</math>, by applying the lemma once again there exists a countable set <math>K_2 \subseteq K_1 \subseteq I</math> such that <math>(x_{m2})_{m \in K_2}</math> converges and hence <math>(x_{m2})</math> has a convergent subsequence. This reasoning may be applied until we obtain a countable set <math>K_n</math> for which <math>(x_{mj})_{m \in K_n}</math> converges for <math>j=1,2,\dots,n</math>. Hence, <math>(x_m)_{m \in K_n}</math> converges and therefore, since <math>(x_m)</math> was arbitrary, any bounded sequence in <math>\mathbb{R}^n</math> has a convergent subsequence.