Let $f$ be continuous on $X$ and $(x_n)$ be a Cauchy sequence on $X$. Show that $(f(x_n))$ doesnt have to be a Cauchy sequence. And show that $(f(x_n))$ always has to be a Cauchy sequence when $f$ is uniformly continuous.
My attempt:
We take the Cauchy sequence $(x_n)=\frac{1}{n}$ and choose $f:\mathbb{R}^+\longrightarrow\mathbb{R}:x\mapsto\frac{1}{x}$ which is a continuous function on given domain.
Since $x_n \longrightarrow0\,\,\Longrightarrow (f(x_n))\longrightarrow+\infty$
So eventhough $(x_n)$ is a Cauchy sequence, $(f(x_n))$ on the continuous $f$ isnt!
$\exists f\in C^0(\mathbb{R}^+)$ where the requirement does not hold!
Now we show that from the fact that $f$ is uniformly continuous we can always conclude:
Is $(x_n)$ a Cauchy sequence $\Longrightarrow$ $(f(x_n))$ is a Cauchy sequence
A function $f$ is uniformly continuous on $X$ when the Heine-Cantor theorem holds:
$\forall \epsilon>0 \,\,\, \exists \delta>0 \,\,\,\forall x,y\in X:x\in \mathcal{U}_{\delta}(y)\Longrightarrow f(x)\in \mathcal{U}_{\epsilon}(f(y))$
So for a $N\in \mathbb{N}$ we know that for all $n>N:x_n\in \mathcal{U}_{\delta}(\xi)$
Where $\xi$ is the point $(x_n)$ is converging to.
Following the Heine-Cantor theorem we know that if $n>N:x_n\in \mathcal{U}_{\delta}(\xi) \Longrightarrow f(x_n)\in \mathcal{U}_{\epsilon}(f(\xi))$
So when $(x_n)$ is a Cauchy sequence, $(f(x_n))$ has to be one aswell!
Would be great if someone could check my reasoning and give me advice for improving 🙂
Best Answer
Consider $A = \{1, 1/2, 1/3, · · ·\}$ and $f(1/n) = \begin{cases}1, &if ~~n ~is~ odd\\−1,& if~~ n~ is ~even\end{cases}$.
Then $f$ is continuous but not uniformly continuous. The sequence $(x_n)= \frac{1}{n}$ in $A$ is Cauchy but the sequence $f(x_n) = (1, −1, 1, −1, · · ·)$ is not Cauchy.
So the Uniformely continuous is necessary condition for the statement you want to prove.
Now
If $f$ is uniformly continuous on $A$, then given $ε > 0$ there is $δ > 0$ such that if $x, y\in A$ and $|x−y| < δ$, then $|f(x) − f(y)| < ε$. Let $(x_n)$ be a Cauchy sequence in $A$. Then for given $δ > 0$ there is $M$ such that if $p, q > M$, then $|x_p − x_q| < δ$, and thus $|f(x_p) − f(x_q)| < ε$, implying that $(f(x_n))$ is a Cauchy sequence.