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## Solution to Understanding Analysis Exercise 2.6

##### Exercise 2.6.1

Let $\varepsilon > 0$ be arbitrary. Since $(x_n)$ converges to $x$, there exits $N\in\mathbf N$ such that \begin{equation}\label{eq2.6.1.1}|x_n-x|<\frac{\varepsilon}{2}\end{equation} for all $n\geqslant N$.
Whenever $n,m\geqslant N$, we have\begin{align*}|x_n-x_m|\leqslant &~|x_n-x|+|x-x_m|\\ \text{by }\eqref{eq2.6.1.1}\quad< &~ \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.\end{align*}Hence $(x_n)$ is a Cauchy sequence.

##### Exercise 2.6.2

(a) Consider the sequence $a_n$ $a_n=(-1)^n\frac{1}{n}.$It is clear convergent to $0$ and hence a Cauchy sequence.

(b) Impossible by Lemma 2.6.3.

(c) Impossible. If a monotone sequence has a Cauchy subsequence, then such a sequence has a convergent sequence. By Exercise 2.5.2 (d) this sequence converges.

(d) Consider the sequence $(a_n)$$(1,0,2,0,3,0,4,0,\cdots).$It is clear this sequence is unbounded. Moreover the subsequence $(a_{2n})$ is Cauchy.

##### Exercise 2.6.3

(a) Let $\varepsilon>0$ be arbitrary. Since $(x_n)$ and $(y_n)$ are Cauchy sequences, there exist $N_1,N_2\in\mathbf N$ such that$|x_n-x_m|<\frac{\varepsilon}{2}$for all $n,m> N_1$ and $|y_n-y_m|<\frac{\varepsilon}{2}$for all $n,m> N_2$.

Let $N=\max \{N_1,N_2\}$. For all $n,m > N$, then by triangle inequality we have \begin{align*}|(x_n+y_n)-(x_m+y_m)|=&~|(x_n-x_m)+(y_n-y_m)|\\ \leqslant &~ |x_n-x_m|+|y_n-y_m| \\ < &~ \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.\end{align*}Hence $(x_n+y_n)$ is a Cauchy sequence.

(b) Let $\varepsilon>0$ be arbitrary. Since $(x_n)$ and $(y_n)$ are Cauchy sequences, they are bounded by Lemma 2.6.3. Hence there exist $M>0$ such that $|x_n|< M$ and $|y_n|< M$ fro all $n\in\mathbf N$. Since $(x_n)$ and $(y_n)$ are Cauchy sequences, there exist $N_1,N_2\in\mathbf N$ such that$|x_n-x_m|<\frac{\varepsilon}{2M}$for all $n,m> N_1$ and $|y_n-y_m|<\frac{\varepsilon}{2M}$for all $n,m> N_2$.

Let $N=\max \{N_1,N_2\}$. For all $n,m > N$, then by triangle inequality we have \begin{align*}|x_ny_n-x_my_m|=&~|x_ny_n-x_my_n+x_my_n-x_my_m|\\ \leqslant &~ |x_ny_n-x_my_n|+|x_my_n-x_my_m| \\ =&~ |y_n||x_n-x_m|+|x_m||y_n-y_m|\\ < &~ M\cdot\frac{\varepsilon}{2M}+M\cdot\frac{\varepsilon}{2M}=\varepsilon.\end{align*}Hence $(x_ny_n)$ is a Cauchy sequence.

##### Exercise 2.6.4

(a) True. Since $(a_n)$ and $(b_n)$ are Cauchy, they are convergent. Hence by Algebraic Limit Theorem, the sequence $(a_n-b_n)$ is convergent. By Exercise 2.3.10 (b), the sequence $(|a_n-b_n|)$ is convergent as well. Therefore by Cauchy Criterion, $(|a_n-b_n|)$ is Cauchy.

(b) False. Counterexample: Let $a_n\equiv 1$ for all $n\in\mathbf N$.

(c) False. Counterexample: Let $a_n=(-1)^n\dfrac{1}{n+1}$, then it is easy to see that $c_n=\begin{cases}0, &\text{if }n \text{ is even},\\ -1,&\text{if }n \text{ is odd}.\end{cases}$Therefore $c_n$ is divergent and hence not Cauchy.

##### Exercise 2.6.5

(a) False. Take the harmonic series$s_n=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}.$Let $\varepsilon>0$ be arbitrary. Take $N> \dfrac{1}{\varepsilon}$. Then for all $n>N$, we have$|s_{n+1}-s_n|=\frac{1}{n+1}<\frac{1}{N}<\varepsilon.$Hence $(s_n)$ is pseudo-Cauchy and not bounded.

(b) True. It follows from the triangle inequality. Let $\varepsilon>0$ be arbitrary. Then there exist $N_1,N_2\in\mathbf N$ such that $|x_{n+1}-x_n|<\frac{\varepsilon}{2}$for all $n>N_1$ and $|y_{n+1}-y_n|<\frac{\varepsilon}{2}$for all $n>N_2$.

Let $N=\max\{N_1,N_2\}$. For all $n> N$, we have by triangle inequality that\begin{align*}|(x_{n+1}+y_{n+1})-(x_n+y_n)| \leqslant &~ |x_{n+1}-x_n|+|y_{n+2}-y_n|\\ <& ~\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.\end{align*}Hence $(x_n+y_n)$ is pseudo-Cauchy.