A Proof of Bolzano–Weierstrass

A Proof of Bolzano–Weierstrass

Definition

For any infinite sequence $\{a_n\}, n \in \mathbb{N}$ that is bounded within the closed real interval $[x, y]$, there is a convergent subsequence.

Proof

Recall the definition of a convergent sequence:

Aa sequence $\{a_n\}, n \in \mathbb{N}$, converges to $L$ when:

$\forall_{\varepsilon > 0}$ $\exists_M$ $m \geq M \Rightarrow |a_m - L| < \varepsilon$

Given an infinite sequence $\{a_n\}$, let $s_1$ be the supremum (resp infimum) of $\{a_n\}$. Then note that either $s_1 \in \{a_n\}$ or $s_1 \not\in \{a_n\}$.

1. If $s_1 \in \{a_n\}$, note that it is the maximum of $\{a_n\}$. We will call this $m_1$.

   a. If there are an infinite number of maxima all equal to $m_1$, we are done. In the sequence $(m_1, m_1, m_1, \dots)$, $\forall_{\varepsilon \gt 0}$ $\forall_{m \geq 1}$, $|m_m - m_m| \lt \varepsilon$.

   b. If there are finite number of maxima, then for our existing maxima $a_i = m_1$, there is a next greatest element, $a_j = m_2$ such that $m_1 > m_2$ and $j > i$

We will start with a lemma:

Lemma:

If the supremum (resp infimum) of an infinite bounded sequence is not a maximum (resp minimum) then there exists a sequence that converges to said supremum (resp infimum).

Without loss of generality, we will prove this for the supremum. The proof for the infimum a direct analogue.

Let $\{s_i\}$ be a sequence with $S$ as its supremum, but not its maximum. Note that $S \notin \{s_i\}$.

Then $\forall \epsilon \gt 0$, the open interval $(S-\epsilon, S)$ includes infinitely many elements of $\{s_i\}$. (Otherwise there would be a last element, ie: a maximum, but we don't have a maximum.)

Since this holds for all $\epsilon \gt 0$, we can pick $s_{N-1} = \inf (S-\epsilon, S)$, noting that the subequent element in the sequence, $s_N \in (S-\epsilon, S)$, meaning $s_N \gt S-\epsilon$.

Re-arranging, we have $S - s_n \lt \epsilon$

Since we can do this for any $\epsilon \gt 0$, we have a convergent sequence.

Rest of Proof

Now we will prove that any bounded infinite sequence $\{a_n\}, n \in \mathbb{N}$ has a convergent subsequence.

First, we create two sequences from members of $\{a_n\}$. One sequence based on suprema, $\{a_{Si}\}_{i=1}^{n}$ and one based on infima, $\{a_{Ii}\}_{i=1}^{n}$. Let N be the first index where $\{a_{SN}\}$ is a non-maximal supremum. (Note, we can also use a sequence of infima and thr proof will work the same way.)

We then have three possibilities.

1. $N$ is infinite and we have an infinite sequence of equal suprema that are maxima. That is, for some $n$, $\{a_{Sn}\} = \{a_{Sn+1}\} = \{a_{Sn+2}\} = ...$. In such case, we have a convergent subsequence with $\{a_{Sn}\}$ as its limit.
2. $N$ is infinite and we have an infinite sequence of decreasing suprema that are maxima. That is $\{a_{Sn}\} \geq \{a_{Sn+1}\} \geq \{a_{Sn+2}\} \geq ...$. Then we have a convergent subsequence with $\inf(a_{Sn})$ as its limit.
3. Finally, if $N$ is finite, then, we have a non-maximal supremum $\{a_{SN}\}$ to which a subsequence will converge.

By noting that a supremum that is not a maximum of a sequence, must be the value to which it converges, we have shown that for any sequence we can either:

1. Find a non-maximum supremum (resp non-minimum infimum) to which a subsequence converges in a finite number of steps or:
2. We end up demonstrating convergence by the process of finding the infimum of an infinite sequence of suprema.

Thus we conclude that $\forall_{x,y \in \mathbb{R}}$, any infinite sequence $\{a_n\}, n \in \mathbb{N}$ that is bounded within the closed interval $[x, y]$, there is a convergent subsequence.