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Note that $X$ is a metric space and $f:(X,\rho)\rightarrow (X,\sigma)$.

Disproving: Suppose $(x_n)$ is a Cauchy sequence, $f$ is continuous and $(f(x_n))$ is not a Cauchy sequence, then $\exists\epsilon'>0$ such that $\forall M\in\Bbb{N}$, there exists $i,j\geq M$ where $\sigma(f(x_i),f(x_j))>\epsilon'$. Also, knowing that $f$ is continuous, then we know that $\exists\delta'>0$ such that if $\rho(x_i,y)<\delta'$, such that $y\in X$, then $\sigma(f(x_i),f(y))<\epsilon'$. However, $y$ is not necessarily a term of the Cauchy sequence $(x_n)$. Hence, it does not necessarily mean that $(f(x_n))$ is also a Cauchy sequence.

Now, I just noticed after typing that this may get tagged as duplicate since the same question has been asked here: If $X = \{x_n:n \in \mathbb N\}$ is a cauchy sequence in a metric space $S$ and $f : S \rightarrow T$ is continuous , is $f(x_n)$ a cauchy sequence?, but would my argument be valid in general? Thank you!

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    $\begingroup$ How do you choose your $y$ for it to work? And it should rather be $\sigma (f(x_i),f(y))<\epsilon '$ right? $\endgroup$ – Suzet Jul 18 '18 at 13:58
  • $\begingroup$ What do you mean? I am not trying to make it work because I'm arguing it would not always work. $\endgroup$ – TheLast Cipher Jul 18 '18 at 14:00
  • $\begingroup$ oh right. typo! thanks $\endgroup$ – TheLast Cipher Jul 18 '18 at 14:01
  • $\begingroup$ I've corrected it now. $\endgroup$ – TheLast Cipher Jul 18 '18 at 14:01
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    $\begingroup$ Oh, actually the statement is untrue, I didn't get it. Well, the only way to show that it is false is to give a counter example for it. Trying to explain why a proof of it wouldn't work is good, but you do not get a definitive conclusion before you give a counter example. $\endgroup$ – Suzet Jul 18 '18 at 14:03
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Another counterexample that goes into a slightly different direction.

Choose $X=\mathbb Q$,

$$f(x)= \left\{ \begin{matrix} 0 & \text{, if } x < \sqrt{2}\\ 1 & \text{, if } x > \sqrt{2}. \\ \end{matrix} \right. $$

This is a continuous function $\mathbb Q \to \mathbb Q$.

If you choose a sequence $\{x_n\}$ of rationals that tends to $\sqrt{2}$ from both sides (infinitely many terms both above and below $\sqrt{2}$), then $\{x_n\}$ is Cauchy, but $\{f(x_n)\}$ is not, as it will contain infinitely many 0's and 1's.

The main idea is to realize that your proposition becomes true when $X$ is a complete metric space, so both my example and that of Salahamam Fatima use a non-complete space.

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  • $\begingroup$ Interesting. I've proven the result when $X$ is a complete metric space, but somehow I still lacked intuition and struggle with finding counterexamples. Thank you for your input. This is informative. $\endgroup$ – TheLast Cipher Jul 19 '18 at 7:25
  • $\begingroup$ That's why I posted my solution in addition to the already existing ones. Intuition about how things can 'break' from what one would expect from the 'normal' case comes from having seen lots of examples. $\endgroup$ – Ingix Jul 19 '18 at 7:49
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Take $$X=(0,+\infty)$$

$$x_n=\frac 1 n$$

$$f:x\mapsto \frac{1}{x}$$

$$(x_n) \text{ is Cauchy}$$

$$f \text{ is continuous}$$

$$f(x_n)=n \text{ is not Cauchy}$$

Your statement will be true if $f$ is UNIFORMLY CONTINUOUS.

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    $\begingroup$ This does not provide any comment on the proof attempt, (like if it is correct or not). $\endgroup$ – supinf Jul 18 '18 at 16:00
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    $\begingroup$ If you read his question, then you will notice he tried to disprove it too... $\endgroup$ – supinf Jul 18 '18 at 16:04
  • $\begingroup$ @supinf The OP had to write $\delta'_i$ instead of $\delta'$ since it depends on $x_i$. $\endgroup$ – hamam_Abdallah Jul 18 '18 at 16:09
  • $\begingroup$ @Salahamam_Fatima: I never wrote $\delta_{i}'$. What do you mean it depends on $x_i$? $\endgroup$ – TheLast Cipher Jul 19 '18 at 7:20
  • $\begingroup$ @TheLastCipher Yes. {}{}{}{}{} $\endgroup$ – hamam_Abdallah Jul 19 '18 at 8:26
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No, it is not correct. You start talking about $y$ without telling what $y$ is. But, above all, your approach only tells us why you think that the sequence $\bigl(f(x_n)\bigr)_{n\in\mathbb N}$ might not be a Cauchy sequence. The best approach, in this case, consists in providing a counterexample.

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  • $\begingroup$ I don't understand why arguing that something is not necessarily true is insufficient proof that it is not always true :? I have a hunch that this follow up falls under logic and philosophy? Does it? $\endgroup$ – TheLast Cipher Jul 18 '18 at 14:34
  • $\begingroup$ What you said in your post was that the sequence $\bigl(f(x_n)\bigr)_{n\in\mathbb N}$ might not be a Cauchy sequence. That doesn't prove that there are cases in which it isn't a Cauchy sequence, which is what you were supposed to prove. $\endgroup$ – José Carlos Santos Jul 18 '18 at 14:36

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