I am self-learning Real Analysis from the text Understanding Analysis by Stephen Abbott. I'd like someone to verify if my proofs/counterexamples to below exercise are rigorous and correct.
[Abbott, 4.4.6] Give an example of each of the following, or state that such a request is impossible. For any that are impossible, supply a short explanation of why this is the case.
(a) A continuous function $\displaystyle f:( 0,1)\rightarrow \mathbf{R}$ and a Cauchy sequence $\displaystyle ( x_{n})$ such that $\displaystyle f( x_{n})$ is not a Cauchy sequence.
(b) A uniformly continuous function $\displaystyle f:( 0,1)\rightarrow \mathbf{R}$ and a Cauchy sequence $\displaystyle ( x_{n})$ such that $\displaystyle f( x_{n})$ is not Cauchy.
(c) A continuous function $\displaystyle f:[ 0,\infty )\rightarrow \mathbf{R}$ and a Cauchy sequence $\displaystyle ( x_{n})$ such that $\displaystyle f( x_{n})$ is not a Cauchy sequence.
Proof.
(a) Consider $\displaystyle f( x) =\frac{1}{x}$ defined and continuous on $\displaystyle ( 0,1)$.
Let $\displaystyle ( x_{n}) =\frac{1}{n}$ be a Cauchy sequence in $\displaystyle ( 0,1)$, where $\displaystyle ( x_{n})\rightarrow 0$.
$\displaystyle f( x_{n})$ is an unbounded sequence, and so it's not Cauchy.
(b) This request is impossible.
Suppose $\displaystyle ( x_{n})$ is a Cauchy sequence in $\displaystyle ( 0,1)$. Then, for all $\displaystyle \delta >0$, there exists $\displaystyle N\in \mathbf{N}$, $\displaystyle | x_{n} -x_{m}| < \delta $ for all $\displaystyle n >m\geq N$.
Since $\displaystyle f$ is uniformly continuous, for all $\displaystyle \epsilon >0$, there exists $\displaystyle \delta >0$, such that for all points $\displaystyle x,y$ satisfying $\displaystyle | x-y| < \delta $, we have $\displaystyle | f( x) -f( y)| < \epsilon $.
Consequently, for all $\displaystyle \epsilon >0$, there exists $\displaystyle N\in \mathbf{N}$, such that $\displaystyle | f( x_{n}) -f( x_{m})| < \epsilon $ for all $\displaystyle n >m\geq N$.
So, $\displaystyle f( x_{n})$ is Cauchy.
(c) This request is impossible.
$\displaystyle [ 0,\infty )$ is a closed set. For all Cauchy sequences $\displaystyle ( x_{n}) \subseteq [ 0,\infty )$ such that $\displaystyle ( x_{n})\rightarrow c$, the image sequence $\displaystyle f( x_{n})\rightarrow f( c)$. So, $\displaystyle f( x_{n})$ is Cauchy.
Best Answer
Your answers are correct. However, the justification for the third answer is not well written. I would put it as follows:
If $(x_n)_{n\in\Bbb N}$ is a Cauchy sequence of elements of $[0,\infty)$, then it converges to some $c\in\Bbb R$ and, since $[0,\infty)$ is a closed subset of $\Bbb R$, $c\in[0,\infty)$. So, since $f$ is continuous, $\lim_{n\to\infty}f(x_n)=f(c)$ amd the fact that $\bigl(f(x_n)\bigr)_{n\in\Bbb N}$ converges implies that it is a Cauchy sequence.
Another possibility consists in saying that if $(x_n)_{n\in\Bbb N}$ is a Cauchy sequence, then it is bounded. So, for some $M\in[0,\infty)$, $(\forall n\in\Bbb N):x_n\in[0,M]$. But the restriction of $f$ to $[0,M]$ is uniformly continuous, and uniformly continuous functions map Cauchy sequences into Cauchy sequences.