Proving or disproving facts about Cauchy Sequences Exercise 2.6.2 in Understanding Analysis by Stephen Abbott
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I'd like someone to verify my solution, and check if my arguments are technically correct. Do I need to add a short proof for (c)?

Give an example of each of the following, or argue that such a request is impossible.


(a) A Cauchy sequence that is not monotone.


(b) A Cauchy sequence with an unbounded subsequence.


(c) A divergent monotone sequence with a Cauchy subsequence.


(d) An unbounded sequence containing a subsequence that is Cauchy.

Proof.
(a) By CC(Cauchy Criterion),
\begin{align*}
 \text{Cauchy sequences} \iff \text{Convergent sequences}
\end{align*}
A convergent sequence could be monotone or oscillating. Consider the sequence $(a_n)$ given by,
\begin{align*}
 a_n = \frac{1}{n} \sin \left(\frac{n\pi}{4}\right)
\end{align*}
is an example of a Cauchy sequence that is not monotone.
(b) This request is impossible. Cauchy sequences are bounded. All subsequences of a bounded sequence must be bounded.
(c) This request is impossible. By MCT, a monotone and bounded sequence is convergent. The contrapositive of this statement is; if a sequence is divergent, atleast one of the two possibilities must hold (1) the sequence is oscillating (2) the sequence is unbounded. As we are told, the sequence is monotone, it must be the case, that the sequence is unbounded. An unbounded monotone sequence $(a_n)$ cannot nest inside itself, a bounded subsequence $(a_{2n})$ or $(a_{f(n)})$.
(d) This request is impossible. An unbounded sequence cannot contain a bounded subsequence. And Cauchy sequences are always bounded.
 A: Besides (d) this is correct (see the comment). For (c) your proof makes sense but is a little unclear. A self contained proof would be as follows: If $x_n$ were increasing then of course $x_n \rightarrow \infty.$ Suppose there was a subsequence with $x_{n_k} \rightarrow x.$ (This is equivalent to $x_{n_k}$ being Cauchy). Then there exists $N \in \mathbb{N}$ such that $x_N > x.$ Further there exists $k \in \mathbb{N}$ such that $n_k > N.$ Using the monotoneness we have $x \geq x_{n_k} \geq x_N > x.$ You would then need to handle the case when $x_n$ were decreasing too but this is essentially the same argument with the inequality signs swapped.
A: Your example for (a) works, though there are certainly simpler ones, e.g., $a_n=\left(-\frac12\right)^n$.
(b) is fine.
In (c) the assertion that an unbounded monotone sequence cannot have a bounded subsequence requires proof. Without loss of generality suppose that the unbounded sequence $\sigma=\langle a_n:n\in\Bbb N\rangle$ is increasing and has a subsequence $\langle a_{n_k}:k\in\Bbb N\rangle$ bounded by $b$. Since $\sigma$ is unbounded, there is an $m\in\Bbb N$ such that $a_m>b$. But then there is a $k\in\Bbb N$ such that $n_k\ge m$ and hence $b\ge a_{n_k}\ge b_m>b$, which is absurd.
As has been pointed out in the comments, (d) is wrong. Any interlacing of an unbounded sequence with a Cauchy sequence will give you a counterexample; the one in the comments is about as simple as they come.
