One of the most important epistemic reasons we study sequences in real analysis is that they provide us with a way to construct the real numbers from the rationals. This was alluded to in our very first conversation this semester about the decimal \(0.999\ldots\text{.}\)
Here, the idea was that the real numbers are the limits of convergent sequences of rational numbers. This situates convergence (rather than completeness) as the tool capable of "filling in the gaps" between the rationals. The problem is that in order to say that a sequence of rational numbers converges, we need to specify its limit ahead of time: namely, in the definition
\begin{align*} \forall \epsilon > 0 \amp\amp \exists N \in \mathbb{N} : \amp\amp \forall n \geq N \amp\amp |s_n-L|\lt \epsilon. \end{align*}where do we get our limit \(L\text{?}\) If our known universe includes only rational numbers, then \(L\) may only be rational. But if we want convergence to produce for us a whole new set of numbers, it would be circular reasoning to somehow "select" our \(L\) from that as-yet unknown set!
We resolve this tension by creating a new type of sequence in which the terms are not said to grow arbitrarily close to a fixed limit, but instead arbitrarily close to each other. These are the Cauchy sequences.
Let \(s_n\) be a sequence. We say that it is a Cauchy sequence if, for all \(\epsilon \gt 0\text{,}\) there exists an \(N\in\mathbb{N}\) such that, for all \(m,n \geq N\text{,}\) we have
\begin{gather*} \bigl|s_n-s_m\bigr| \lt \epsilon. \end{gather*}Written in logical notation, a sequence \(s_n\) is Cauchy if
\begin{align} \forall\epsilon\gt 0 \amp\amp \exists N\in\mathbb{N} : \amp\amp \forall n,m \geq N \amp\amp \bigl|s_n-s_m|\lt \epsilon.\tag{3.1} \end{align}It's not immediately clear that we're defining something different from convergence here. One of the reasons for that lack of clarity is our intuition that if a sequence converges (grows arbitrarily close to a limit) then of course it must be Cauchy (grows arbitrarily close to "itself").
Indeed, it is always the case that convergent sequences are Cauchy:
Let \(s_n\) be a convergent sequence. Then \(s_n\) is a Cauchy sequence.
Denote by \(L = \lim_{n\to\infty} s_n\) the limit of the sequence. We must show that \(s_n\) meets the definition of Cauchy sequence, so we'll use the logical form (3.1) to structure the argument that follows.
Let \(\epsilon \gt 0\) be arbitrarily chosen.
By definition of convergence, there exists a natural number \(N\in\mathbb{N}\) such that
\begin{align} \forall n \geq N \amp\amp \bigl|s_n-L\bigr|\lt \frac{\epsilon}{2}.\tag{3.2} \end{align}Let \(n,m \geq N\) both be arbitrarily chosen.
Then we can use an add-subtract trick to relate the distance of the terms \(s_n, s_m\) from each other to their respective distances to \(L\) as follows.
\begin{align*} |s_n-s_m| \amp= |s_n - L + L - s_m| \amp \\ \amp \leq \bigl|s_n-L\bigr| + \bigl|L-s_m\bigr| \amp \text{(triangle inequality)}\\ \amp = |s_n-L| + |s_m-L| \amp\\ \amp \lt \frac{\epsilon}{2} + \frac{\epsilon}{2} \amp \text{because n,m satisfy }\knowl{./knowl/mdn-7.html}{\text{(3.2)}}\\ \amp = \epsilon. \end{align*}This completes the proof.
What's not clear, and which is the "big reveal" of this chapter, is that the converse of this theorem is also true for sequences of rational numbers. Namely, that any Cauchy sequence of rational numbers will also be convergent (though, of course, its limit might not be rational).
Let \(s_n\) be a Cauchy sequence of rational numbers. Then \(s_n\) is a convergent sequence, and there exists a real number \(L \in\mathbb{R}\) such that \(L=\lim_{n\to\infty} s_n\text{.}\)
(Note, here \(L\) is a real number in the sense of the completeness axiom, that is, it is the least upper bound of some set of rational numbers.)
In order to prove that \(s_n\) converges, we will have to produce (from somewhere) a real number \(L\) to which to apply the definition of convergence. We'll get that by identifying a subsequence of \(s_n\) which we can show is convergent. This will take four steps:
Step 1. According to Lemma 5 below, every sequence has a monotonic subsequence. Applying this lemma we obtain a monotonic subsequence \(s_{n_1},s_{n_2},s_{n_3},\ldots\text{,}\) though we can't know whether this sequence is increasing or whether it is decreasing.
Step 2. Is this subsequence bounded? Yes, since in fact every Cauchy sequence is (fully) bounded, by Lemma 4 below, we can conclude that this subsequence is in particular also a bounded sequence of rational numbers.
Step 3. By the monotone convergence theorem, since it is monotonic and bounded, the subsequence \(s_{n_1}, s_{n_2},s_{n_3},\ldots\) is convergent. In particular, if we denote
\begin{align} L = \lim_{k\to\infty} s_{n_k} = \sup\{ s_{n_k}\} \amp\amp \text{using }\knowl{./knowl/monotone-convergence.html}{\text{Theorem 2.2}}\tag{3.3} \end{align}we cast \(L\) as the supremum of a set of rational numbers. This means that \(L \in\mathbb{R}\) according to the completeness axiom of the real numbers.
Step 4. We need to show that the whole sequence, not just this subsequence, also converges to \(L\text{.}\) So applying the definition of convergence to structure our proof, we:
Let \(\epsilon\gt 0\) be arbitrarily chosen.
Since \(s_n\) is a Cauchy sequence, there exists a natural number \(N_1 \in\mathbb{N}\) such that
\begin{align} \forall n,m \geq N_1 \amp\amp |s_n-s_m| \lt \epsilon/2.\tag{3.4} \end{align}Meanwhile, since \(s_{n_k}\) converges to \(L\text{,}\) there exists a natural number \(N_2 \in\mathbb{N}\) such that
\begin{align} \forall k \geq N_2 \amp\amp |s_{n_k}-L| \lt \epsilon/2.\tag{3.5} \end{align}Define \(N = \max \{ N_1,N_2 \}.\)
Now, let \(n \geq N\) be arbitrarily chosen.
We can use an add-subtract trick to transform what we know about how close the terms are to one another (3.4) as well as how close the subsequence is to its limit (3.5) into a statement about how close all the tail terms are to the limit. Let's choose \(k\) so that \(n_k \geq N\text{,}\) i.e., choose any term in the subsequence that we know is also in the tail of our full sequence. Then we have
\begin{align*} |s_n - L| \amp = \bigl| s_n - s_{n_k} + s_{n_k} - L \bigr|\\ \amp \leq |s_n-s_{n_k}| + |s_{n_k}-L| \amp \text{(triangle inequality)}\\ \amp \lt |s_n-s_{n_k}| + \frac{\epsilon}{2} \amp \text{by }\knowl{./knowl/mdn-10.html}{\text{(3.5)}}\\ \amp \lt \frac{\epsilon}{2} + \frac{\epsilon}{2} \amp \text{by }\knowl{./knowl/mdn-9.html}{\text{(3.4)}}\\ \amp = \epsilon. \end{align*}This completes the proof.
This proof relied on the following two lemmas.
Let \(s_n\) be a Cauchy sequence of real numbers. Then \(s_n\) is a bounded sequence.
As in our proof of Theorem 1.2, we begin by applying the definition of Cauchy to split our sequence into a finite head and an infinite tail.
Choose \(\epsilon = 1\text{.}\) Then since \(s_n\) is a Cauchy sequence, there exists a natural number \(n\in\mathbb{N}\) such that
\begin{align} \forall n,m \geq N \amp\amp |s_n-s_m| \lt \epsilon.\tag{3.6} \end{align}The head of the sequence \(\{s_1,s_2,\ldots,s_N\}\) being a finite set, we are on firm ground to define \(M_1 = \max \{|s_1|,|s_2|,\ldots,|s_N|\}\text{.}\)
Why is the tail of the sequence \(\{s_N, s_{N+1}, \ldots\}\) bounded? By the Cauchy property (3.6) we know that all terms in this tail are within \(\epsilon\) of one another. So in particular, all of them are within \(\epsilon\) of the first tail term, \(s_N\text{.}\) Thus
\begin{align*} \forall m \geq N \amp\amp \amp\amp |s_N-s_m| \lt \epsilon.\\ \amp\amp\amp s_N-\epsilon \amp \lt s_m \amp\lt s_N+\epsilon\\ \amp\amp\amp\amp |s_m| \amp\lt |s_N| + \epsilon. \end{align*}So all of the tail terms are bounded by the quantity \(|s_N|+\epsilon\) and we can now choose the larger of the head bound and tail bound as the bound for our entire sequence,
\begin{gather} M = \max \{ M_1, |s_N|+\epsilon\}.\tag{3.7} \end{gather}Now, for any arbitrary natural number \(n\in\mathbb{N}\) we will have
and this proves that the entire sequence is bounded by \(M\text{.}\)
Let \(s_n\) be a sequence of real numbers. Then there exists a sequence of natural numbers
\begin{equation*} n_1 \lt n_2 \lt n_3 \lt \cdots \end{equation*}such that the sequence
\begin{equation*} s_{n_1}, s_{n_2}, s_{n_3}, \ldots \end{equation*}is a monotonic sequence.
(Note, we will later call this an example of a subsequence of \(s_n\text{.}\))
A proof of this result can be found here.