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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 :)

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    $\begingroup$ I'm convince by the first part. But not by the second one. Can you restate the definition of a Cauchy sequence? Also, I don't know why you mention Heine-Cantor theorem. Heine-Cantor theorem gives conditions implying that a map is uniform continuous. But here we're already supposing that it is uniform continuous. $\endgroup$ – mathcounterexamples.net May 30 '20 at 8:39
  • $\begingroup$ The second part follows from definition of uniform continuity and definition of a Cauchy sequence. I don't understand why you are quoting a theorem for this. $\endgroup$ – Kavi Rama Murthy May 30 '20 at 8:51
  • $\begingroup$ I thought that the cauchy sequence definition gives us, that a infinit amount of values lie in every epsilon region of our point of convergence. so there is this $\xi$ for which every $x_n$ with $N>n$ lies in any epsilon region around $\xi$. For any epsilon we might choose. And from the Heine Cantor theorem it follows that since the function is uniformly continuous our $(f(x_n))$ also has to lie in the close region of $f(\xi)$.. $\endgroup$ – CoffeeArabica May 30 '20 at 8:58
  • $\begingroup$ Cauchy sequence does not need to converge. $\endgroup$ – N.Quy May 30 '20 at 9:05
  • $\begingroup$ Ohh.. could you give me an example for that case? :) so I can ponder with that? $\endgroup$ – CoffeeArabica May 30 '20 at 9:13
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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.

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  • $\begingroup$ Sorry if i miss the point.. but wasnt i stating (especially) the second part in my post? $\endgroup$ – CoffeeArabica May 30 '20 at 9:01
  • $\begingroup$ @Coffee Arabic: Sorry,I don't see really your comment. I simply proved it by the definition of uniform continuity. I don't get that comment. $\endgroup$ – Aman Pandey May 30 '20 at 9:12

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