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?
Actually, for most purposes we'll put sequence to use for, subconvergence will turn out to be good enough.
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 \(\sqrt\text\) part of that issue is: we can straightforwardly write down a sequence of rational numbers which "bunches up" around \(\sqrt\text\) but doesn't actually converge to anything. So whether "bunched up" sequences must have limits ought to be pretty intimately connected to completeness.
To make this precise, here's a formal definition of what it means for a sequence to "bunch up".
\begin \forall\epsilon\gt 0,\exists K\in\mathbb:m,n\gt K\Rightarrow \left\lvert x_n-x_m\right\rvert\lt \epsilon\ \ . \end
This idea is named for , who did most of his work at the French military engineering college École Polytechnique. Broadly speaking, most of the ideas in these notes were first hinted at in Cauchy's works (during the first half of the 19th century) and later refined and rigorized by Weierstraß.
What does it mean for a sequence to not be Cauchy?
It's a bit unfortunate that we don't have an adjectival word for Cauchy sequences. In French, for example, one says a sequence is de Cauchy, that is, "Cauchy's".
Consider the sequence \(x_n=1+\frac+\frac+\cdots+\frac\text<.>\) This sequence is not Cauchy.
Consider the difference
\begin \left\lvert x_n-x_m\right\rvert=\frac+\cdots+\frac \end(here, without loss of generality, we've assumed \(n\lt m\text<.>\) )
We will show that we can make this difference greater than \(\frac\text\) for some choice of \(m,n\text\) no matter how large we require both \(m\) and \(n\) to be.
Given \(K\in\mathbb\text\) note that \(2^K\gt K\text\) so choose \(n=2^K\) and \(m=2^\text<.>\) Notice that there are \(2^\) terms between \(n\) and \(m\text<.>\) Therefore
Sit with Example 2.4.3 for a moment. Notice that subsequent differences are indeed very small ( \(\lvert x_n-x_\rvert=\frac\) ). 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:
If \(X\) converges, then \(X\) is Cauchy.
Every Cauchy sequence is bounded.
A bounded sequence which diverges must do so because it has at least two distinct , that is, subsequences \(\tilde_1,\tilde_2\text\) so that \(\tilde_1\to L_1\text\) \(\tilde_2\to L_2\text\) with \(L_1\neq L_2\text<.>\)
If \(X\) is Cauchy and there is a convergent subsequence \(\tilde\to L\text\) then \(X\to L\text<.>\)
Call the subsequence \(\tilde=\left(x_\right)_
Given \(\epsilon\gt 0\text\) there is \(I\in\mathbb\) so that \(k\geq I\) guarantees \(\left\lvert x_-L\right\rvert\lt \frac\epsilon\text<.>\)
There is also \(J\in\mathbb\) so that \(n,m\geq J\) guarantees \(\left\lvert x_n-x_m\right\rvert\leq \frac\epsilon\text<.>\)
Choose \(K=J+n_I\text<.>\) Then \(n_K\geq n_\geq n_I\text\) so by the convergence of \(\tilde\text\) \(\left\lvert x_-L\right\rvert\lt \frac\epsilon\text<.>\) Furthermore, if \(n\geq K\text\) then \(n\geq J\text\) so \(\left\lvert x_n-x_\right\rvert\lt\frac\epsilon\text<.>\)
Thus, for any \(n\geq K\text\)
\begin \left\lvert x_n-L\right\rvert&\leq \left\lvert x_n-x_\right\rvert+\left\lvert x_-L\right\rvert\\ &\lt \frac\epsilon+\frac\epsilon=\epsilon\ \ . \end
Every Cauchy sequence of real numbers converges.
In the "real world" we normally interact with real numbers via their decimal representations. But what is a decimal representation?
The real number represented by \(d_0.d_1d_2d_3\ldots\) is the limit of the sequence
\begin x_n=d_0+d_1\frac+d_2\frac+\cdots+d_n\frac \endIf the numbers \(d_1,d_2,\ldots\) each have \(d_k\in\left\\text\) then the sequence \(x_n\) in Definition 2.4.9 converges.
This is your job. There are two approaches (and you should flesh out both):
Every real number \(x\in \mathbb\) can be written as the limit of some decimal representation.
Given \(x\in\mathbb\text\) we need to produce the decimal digits \(d_0,d_1,\ldots\text<.>\)
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\
Now, consider the set \(\left\
For the same reason, we can define \(d_2=\sup\left\
It remains to show that \(x_n\to x\text<.>\)
\(x_n\) converges by Proposition 2.4.10; the trouble is to show that the limit is \(x\text<.>\)
Notice that by construction, each \(x_n\) has \(x-\frac\lt x_n\lt x\text<.>\) By Proposition 2.2.7, this means \(x_n\to x\text<.>\)
Apply the algorithm in the proof of Proposition 2.4.11 to these real numbers to obtain their decimal representations:
Notice that the definition of Cauchiness makes sense in any ordered field.
We call an ordered field 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:
A Cauchy-complete ordered field must be complete.
We'll talk more about this idea later.