Definition 2.4.1.
We call \(X=(x_n)_{n\in\mathbb{N}}\) a Cauchy sequence if
\begin{equation*}
\forall\epsilon\gt 0,\exists K\in\mathbb{N}:m,n\gt K\Rightarrow \left\lvert x_n-x_m\right\rvert\lt \epsilon\ \ .
\end{equation*}
Section 2.4 Cauchy sequences
Theorem 2.3.3 and Corollary 2.3.13 do give us partial converses to Theorem 2.1.14, but they do leave something to be desired. Theorem 2.3.3 puts far too strong a condition on the sequence; Corollary 2.3.13 only gives us subconvergence. Can we do better? That is, can we come up with a condition that guarantees a sequence converges, without already knowing the limit of that sequence?
Aside
The informal notion of \(X\to L\) is that "the terms of \(X\) are bunching up around \(L\text{.}\)" If we think about the problem of the existence of \(\sqrt{2}\text{,}\) part of that issue is: we can straightforwardly write down a sequence of rational numbers which "bunches up" around \(\sqrt{2}\text{,}\) but doesn’t actually converge to anything. So whether "bunched up" sequences must have limits ought to be pretty intimately connected to completeness.
Subsection 2.4.1 The Cauchy property
To make this precise, here’s a formal definition of what it means for a sequence to "bunch up".
Aside
Aside
Example 2.4.3.
Consider the sequence \(x_n=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\text{.}\) This sequence is not Cauchy.
Proof.
Consider the difference
\begin{equation*}
\left\lvert x_n-x_m\right\rvert=\frac{1}{n+1}+\cdots+\frac{1}{m}
\end{equation*}
(here, without loss of generality, we’ve assumed \(n\lt m\text{.}\))
We will show that we can make this difference greater than \(\frac{1}{2}\text{,}\) for some choice of \(m,n\text{,}\) no matter how large we require both \(m\) and \(n\) to be.
Given \(K\in\mathbb{N}\text{,}\) note that \(2^K\gt K\text{,}\) so choose \(n=2^K\) and \(m=2^{K+1}\text{.}\) Notice that there are \(2^{K}\) terms between \(n\) and \(m\text{.}\) Therefore
\begin{align*}
\left\lvert x_n-x_m\right\rvert&=\frac{1}{2^K+1}+\cdots+\frac{1}{2^{K+1}}\\
&\gt \frac{1}{2^{K+1}}+\cdots+\frac{1}{2^{K+1}}\\
&=\frac{2^K}{2^{K+1}}=\frac{1}{2}
\end{align*}
Remark 2.4.4.
Sit with Example 2.4.3 for a moment. Notice that subsequent differences are indeed very small (\(\lvert x_n-x_{n+1}\rvert=\frac{1}{n+1}\)). But the definition of Cauchy requires us to consider pairs of terms which are arbitrarily far apart.
Cauchiness is related to other properties a sequence might or might not have:
Proposition 2.4.5.
Proposition 2.4.6.
Every Cauchy sequence is bounded.
A bounded sequence which diverges must do so because it has at least two distinct subsequential limits, that is, subsequences \(\tilde{X}_1,\tilde{X}_2\text{,}\) so that \(\tilde{X}_1\to L_1\text{,}\) \(\tilde{X}_2\to L_2\text{,}\) with \(L_1\neq L_2\text{.}\)
Proposition 2.4.7.
If \(X\) is Cauchy and there is a convergent subsequence \(\tilde{X}\to L\text{,}\) then \(X\to L\text{.}\)
Proof.
Call the subsequence \(\tilde{X}=\left(x_{n_k}\right)_{k\in\mathbb{N}}\text{.}\)
Given \(\epsilon\gt 0\text{,}\) there is \(I\in\mathbb{N}\) so that \(k\geq I\) guarantees \(\left\lvert x_{n_k}-L\right\rvert\lt \frac{1}{2}\epsilon\text{.}\)
There is also \(J\in\mathbb{N}\) so that \(n,m\geq J\) guarantees \(\left\lvert x_n-x_m\right\rvert\leq \frac{1}{2}\epsilon\text{.}\)
Choose \(K=J+n_I\text{.}\) Then \(n_K\geq n_{n_I}\geq n_I\text{,}\) so by the convergence of \(\tilde{X}\text{,}\) \(\left\lvert x_{n_K}-L\right\rvert\lt \frac{1}{2}\epsilon\text{.}\) Furthermore, if \(n\geq K\text{,}\) then \(n\geq J\text{,}\) so \(\left\lvert x_n-x_{n_K}\right\rvert\lt\frac{1}{2}\epsilon\text{.}\)
Thus, for any \(n\geq K\text{,}\)
\begin{align*}
\left\lvert x_n-L\right\rvert&\leq \left\lvert x_n-x_{n_K}\right\rvert+\left\lvert x_{n_k}-L\right\rvert\\
&\lt \frac{1}{2}\epsilon+\frac{1}{2}\epsilon=\epsilon\ \ .
\end{align*}
Theorem 2.4.8.
Every Cauchy sequence of real numbers converges.
Subsection 2.4.2 Decimal representations
In the "real world" we normally interact with real numbers via their decimal representations. But what is a decimal representation?
Definition 2.4.9.
The real number represented by \(d_0.d_1d_2d_3\ldots\) is the limit of the sequence
\begin{equation*}
x_n=d_0+d_1\frac{1}{10}+d_2\frac{1}{100}+\cdots+d_n\frac{1}{10^n}
\end{equation*}
Proposition 2.4.10.
If the numbers \(d_1,d_2,\ldots\) each have \(d_k\in\left\{0,1,2,3,4,5,6,7,8,9\right\}\text{,}\) then the sequence \(x_n\) in Definition 2.4.9 converges.
Proof.
This is your job. There are two approaches (and you should flesh out both):
-
explain why \((x_n)_{n\in\mathbb{N}}\) is bounded and monotone;or
-
explain why \((x_n)_{n\in\mathbb{N}}\) is Cauchy.
Proposition 2.4.11.
Every real number \(x\in \mathbb{R}\) can be written as the limit of some decimal representation.
Proof.
To make the exposition easier, let’s assume \(x\gt 0\text{.}\)
If \(x\leq 1\text{,}\) set \(d_0=0\text{.}\) Otherwise, the set \(\left\{n\in\mathbb{N}\middle\vert n\lt x\right\}\) is nonempty. Clearly this set is bounded above, so has a supremum. Since this is a set of natural numbers, that supremum is a maximum. Set \(d_0=\sup\left\{n\in\mathbb{N}\middle\vert n\lt x\right\}\text{.}\)
Now, consider the set \(\left\{n\in\mathbb{N}\cup\{0\}\middle\vert d_0+n\frac{1}{10}\lt x\right\}\text{.}\) This set is bounded above by \(9\) (why?) and nonempty (why?) so it has a supremum. Because it’s a set of natural numbers, that supremum is a maximum. Set \(d_1=\sup\left\{n\in\mathbb{N}\cup\{0\}\middle\vert d_0+n\frac{1}{10}\lt x\right\}\) and \(x_1=d_0+\frac{1}{10}d_1\text{.}\)
For the same reason, we can define \(d_2=\sup\left\{n\in\mathbb{N}\cup\{0\}\middle\vert d_0+d_1\frac{1}{10}+n\frac{1}{100}\lt x\right\}\text{,}\) and \(x_2=d_0+\frac{1}{10}d_1+\frac{1}{100}d_2\) and so on.
It remains to show that \(x_n\to x\text{.}\)
Notice that by construction, each \(x_n\) has \(x-\frac{1}{10^n}\lt x_n\lt x\text{.}\) By Proposition 2.2.7, this means \(x_n\to x\text{.}\)
Checkpoint 2.4.12.
Apply the algorithm in the proof of Proposition 2.4.11 to these real numbers to obtain their decimal representations:
-
\(\displaystyle x=\frac{1}{3}\)
-
\(\displaystyle x=2\)
-
\(\displaystyle x=\sqrt{7}\)
Subsection 2.4.3 Cauchy-completeness
Notice that the definition of Cauchiness makes sense in any ordered field.
Definition 2.4.13.
We call an ordered field Cauchy-complete if every Cauchy sequence (in that ordered field) converges to a limit (in that ordered field).
We already showed that a complete ordered field must be Cauchy-complete; the converse is true as well:
Theorem 2.4.14.
A Cauchy-complete ordered field must be complete.
We’ll talk more about this idea later.
