This is a question I've stumbled upon and the question asks me to prove something which I don't think is true.
Question:
Let $f(x)$ be a continuous function in $\mathbb{R}$, and let $(a_n)$ be a Cauchy sequence. Prove that $f(a_n)$ is a Cauchy sequence.
Attempt:
Since the reals covers $[-\infty,\infty]$, I thought I might as well look for a simple interval to prove that it isn't true. Let $f(x) : (0,1) \to \mathbb{R}$. Since they asked for a continuous function, let $f(x)=\frac{1}{(x)}$, which is continuous. Looking at a Cauchy sequence $(\frac{1}{(n)})$, I see that
$f(\frac{1}{(n)})=n$, which is not Cauchy.
Am I doing something wrong or is the question improperly stated?
Attempt #2:
Let $f(x)$ be a continuous function $\in \mathbb{R}$. (As Chris said) I'll lock $(a_n)$ to a compact interval and then I can work with $f$ there. Let $\epsilon>0$ be given. Since $f$ is continuous on a compact set, it is also uniformly continuous. Since $f$ is uniformly continuous there exists $\delta>0$ such that for all $x,y\in \mathbb{R}$ such that $|x-y|<\delta$, $|f(x)-f(y)|<\epsilon$. And since $(a_n)$ is Cauchy and $\delta>0$, there exists a $N\in \mathbb{N}$ such that for all $n,m \geq N$ we have $|x_n - x_m|<\delta$. Thus $|f(x)-f(y)|<\epsilon$, $\epsilon>0$. Thus $f((x_n))$ is Cauchy.
