Skip to main content

Section 2.3 monotone sequences and the Bolzano-Weierstraß Theorem

One of the difficulties in proving a sequence converges is: we sort of need to know what the limit is before we can get started on the proof. Wouldn’t it be nice to be able to say that a sequence converged, without having to know in advance what the limit was?
Naively, one might look at Theorem 2.1.14 and hope that its converse held, too. That is, we might try and prove
But unfortunately this is just plain false.

Example 2.3.1.

These sequences are bounded but do not converge:
  1. \(\displaystyle \left((-1)^n\right)_{n\in\mathbb{N}}\)
  2. \(\displaystyle \left(\sin(n)\right)_{n\in\mathbb{N}}\)
But! We will be able to say something sort of like the converse of Theorem 2.1.14. In fact, there are two sort-of converses. Let’s start with a definition.

Definition 2.3.2.

We call a sequence \(X=(x_n)_{n\in\mathbb{N}}\) nondecreasing if \(\forall n, x_{n}\leq x_{n+1}\text{.}\)
We call a sequence \(X=(x_n)_{n\in\mathbb{N}}\) nonincreasing if \(\forall n, x_{n}\geq x_{n+1}\text{.}\)
We call a sequence \(X=(x_n)_{n\in\mathbb{N}}\) increasing if \(\forall n, x_{n}\lt x_{n+1}\text{.}\)
We call a sequence \(X=(x_n)_{n\in\mathbb{N}}\) decreasing if \(\forall n, x_{n}\gt x_{n+1}\text{.}\)
If \(X\) is either increasing or decreasing, we call \(X\) strictly monotone. If \(X\) is any of these, we say \(X\) is monotone.

Proof.

Observe that one direction is Theorem 2.1.14, so we’ll only prove the interesting direction, namely that bounded monotone sequences converge.
Here’s the proof for a nondecreasing sequence.
Set \(L=\sup\left\{x_n\middle\vert n\in\mathbb{N}\right\}\text{.}\) That is, \(L\) is the supremum of the image of \(X\text{.}\) We’ll show that \(x_n\to L\text{.}\)
To this end, we let \(\epsilon\gt 0\) be given. By the \(\epsilon\)-characterization of supremum, there is \(x_K\) so that \(L-x_K\lt \epsilon\text{.}\) Use this \(K\in\mathbb{N}.\)
To see that this \(K\) works, consider any \(n\geq K\text{.}\) Because \(X\) is nondecreasing, \(x_n\geq x_K\text{.}\) Therefore
\begin{gather*} \lvert L-x_n\rvert=L-x_n\leq L-x_K\lt \epsilon \end{gather*}
which is what we need.

Remark 2.3.4.

The proof of Theorem 2.3.3 features a supremum very prominently. In fact, we can think of the conclusion of Theorem 2.3.3 as a property that an ordered field might or might not have; that is, we say an ordered field has the Monotone Sequence Property if every monotone, bounded sequence in that ordered field converges to a limit in that ordered field.

Checkpoint 2.3.5.

The proof of Theorem 2.3.3 shows that if an ordered field has the least upper bound property, then it has the monotone sequence property. Prove the converse: that an ordered field with the monotone sequence property must be complete.
Now, you might be concerned that calling this a partial inverse to Theorem 2.1.14 is a bit of an oversell. After all, most sequences aren’t monotone! Theorem 2.3.3 doesn’t say anything about lots of sequences which we might want to prove converge. That’s a fair critique. We get around it by means of another definition.

Definition 2.3.6.

Let \(X=(x_n)_{n\in\mathbb{N}}\text{.}\) A subsequence of \(X\) is a sequence obtained by composing with a strictly increasing function \(n:\mathbb{N}\to\mathbb{N}\text{.}\)
What a damned mess of a definition! Let’s unpack it some. First of all, we normall write, instead of function notation \(n(3)\text{,}\) something like \(n_3\text{.}\) The idea is to pick a sequence of indices. Here’s an example.

Example 2.3.7.

Let \(X=\left(\frac{1}{n}\right)\text{.}\) Define \(n_k=k^2.\text{.}\) Then corresponding subsequence is \(X\circ n=\left(x_{n_k}\right)_{k\in\mathbb{N}}\text{,}\) which is better known as
\begin{equation*} \left(\frac{1}{k^2}\right)_{k\in\mathbb{N}}\ \ . \end{equation*}

Checkpoint 2.3.8.

The \(K\)-tail \(X_{(K)}\) of a sequence \(X\) (see Definition 2.2.4) is a subsequence. What is the correspoding index-selection function \(n:\mathbb{N}\to\mathbb{N}\text{?}\)

Checkpoint 2.3.9.

Which of these is a subsequence of \(\left(\frac{1}{n}\right)_{n\in\mathbb{N}}\text{?}\)
  1. \(\displaystyle \left(1,\frac{1}{3},\frac{1}{9},\frac{1}{27},\frac{1}{81},\ldots\right)\)
  2. \(\displaystyle \left(\frac{1}{3},\frac{1}{2},\frac{1}{5},\frac{1}{4},\ldots\right)\)
  3. \(\displaystyle \left(1,\frac{1}{3},\frac{1}{5},\frac{1}{7},\frac{1}{9},\ldots\right)\)
  4. \(\displaystyle \left(\frac{1}{2},\frac{1}{6},\frac{1}{8},\frac{1}{10},\ldots\right)\)
  5. \(\displaystyle \left(-\frac{1}{2},\frac{1}{3},-\frac{1}{4},\frac{1}{5},-\frac{1}{6},\ldots\right)\)
Hint.
If you think something is a subsequence, it’s your job to say what the index-selection function \(n:\mathbb{N}\to\mathbb{N}\) is.

Example 2.3.10.

The even subsequence is what we get from choosing \(n(k)=2k\text{.}\)
The odd subsequence is what we get from choosing \(n(k)=2k-1\text{.}\)

Remark 2.3.11.

Checkpoint 2.3.9 shows that there’s an significant distinction between the notions of subset and subsequence.

Proof.

Proof.

You ought to be able to put this together.

Aside