A continuous function $f : (0, 1)\to\Bbb R$ and a Cauchy sequence $x_n$ such that $f(x_n)$ is not a Cauchy sequence 
Provide an example or show that it is impossible. A continuous function $f : (0, 1) → R$ and a Cauchy sequence $(x_n)$ such that $f(x_n)$ is not a Cauchy sequence.

I saw several examples that shows that it is possible. However, I want to understand why. It seems to me that the statement contradicts the definition of continuity of a function.
Say $x_n$ is Cauchy then $|x_{n}-x_{n-1}| < \sigma$, if $f(x_n)$ is not Cauchy then  $|f(x_n)-f(x_{n-1})| \ge \epsilon$, but that says that $f(x)$ is not convergent.  
What is wrong with the argument?
 A: The issue is the following : $f$ continuous, means that if $x_n$ is a convergent sequence to $x$, then so is $f(x_n)$ , converging to $f(x)$.
Note that every convergent sequence is Cauchy. However, the converse is true(in $\mathbb R$), only in a closed set. Note that $(0,1)$ is not a closed set. Therefore, there are Cauchy sequences in $(0,1)$ that are not convergent, and the problem is that the continuity of $f$ does not imply anything at all, since it comes into play only for convergent sequences.
The best example of such a function is $f(x) = \frac 1x$ on $(0,1)$ , with the sequence $\frac 1n$ : clearly, $x_n$ is Cauchy while $f(x_n) = n$ is not. What is happening here, is that $x_n$ is not convergent, so the continuity of $f$ does not help.

The "theory" part of things is this :
Suppose $f$ is continuous. Now, let $x_n$ be any Cauchy sequence in $(0,1)$. Note that Cauchy sequences in a subset of $\mathbb R$ converge to a point in the closure of that set, in this case $(0,1)$. Therefore, two things can happen :


*

*Suppose that $x_n$ converges to $x$, where $x \in (0,1)$. Then by the definition of continuity, since $x_n$ is a convergent sequence, we have that $f(x_n)$ converges to $f(x)$.



But let us look more deeper, and see "why". We are actually using continuity at the point $x$ for this, because : given $\epsilon > 0$, first we know there exists $\delta > 0$  such that $|y - x| < \delta$ implies $|f(y) - f(x)| < \delta$. Now we can find $N$ such that $$n > N \implies |x_n - x| < \delta \implies |f(x_n) - f(x)| < \epsilon$$, so $f(x_n) \to f(x)$ by the definition of convergence of a sequence of points.



*

*Suppose that $x_n$ converges to either $0$ or $1$. Then $x_n$ is only Cauchy, not convergent in $(0,1)$. By the previous post, we saw that $f(x_n)$, if it were a convergent sequence, must go to $f(x)$, and the proof depended crucially on continuity at the point $x$, which was the limit of the $x_n$. But here, the limit $x$ does not belong to the set $(0,1)$, and therefore $f$ is not even defined on $x$. If it were possible to define $f$ there, we could speak of extension, which could be continuous if $f$ was continuous at $x$. 



Now for the more subtler point : continuity in $(0,1)$, is equivalent to continuity at each point of $(0,1)$. That is, continuity in $(0,1)$ can be written as follows :

Given any point $x \in (0,1)$, and given any $\epsilon > 0$, there exists a $\delta > 0$, depending on $x$ and $\epsilon$, such that $|y-x| < \delta \implies |f(y) - f(x)| < \epsilon$. 

To put this in plain English : if we fix a point, then if another point is sufficiently close to that point, the function values of the two points are sufficiently close.
Now, let us write down what it means for $f$ to take Cauchy sequences to Cauchy sequences : it means that if a sequence of points get eventually close to each other, so must the function values of the points. There is no fixing of a point here, rather a sequence is fixed,so the definition of pointwise continuity cannot be used.
Why would we intuitively think this is true if $f$ is continuous? Let's put it this  way : if points are getting close, then fixing any two of them which are "sufficiently" close to each other, their function values are "sufficiently" close, by using continuity at either point. 

Alas, this is not true. Why? Because, given $\epsilon > 0$, the $\delta$ depends on $x$, that is why. Suppose I fix an $\epsilon > 0$ for trying to show that $f(x_n)$ is Cauchy. I do know that there exists $N$ such that $m,n > N \implies |x_m - x_n| < \epsilon$, because $x_n$ is Cauchy. Now, let us fix $P,Q > N$, and consider the points $x_P$ and $x_Q$. By continuity at $x_P$, there is $\delta_P > 0$ depending on $x_P$ such that if $|y - x_P| < \delta_P$ then $|f(y) - f(x_P)| < \epsilon$. Similarly $\delta_Q$ for $x_Q$.
The inevitable question arises : what if $|x_P - x_Q|$ is  greater than both $\delta_P$ and $\delta_Q$? Then, $|f(x_Q) - f(x_P)|$ cannot be controlled, because we require the points to be close enough to apply the continuity criterion.
Thus, the complete answer to the question is this :

Fix $\epsilon > 0$. Even if the points $x_n$ are getting closer and closer to each other, the $\delta$ corresponding to each point of the sequence and the given $\epsilon$, could possibly reduce more rapidly than the distances between the points. Thus, even though the distances between points is decreasing, for example $|x_P - x_Q|$ is really small, yet since $|x_P - x_Q| > \delta_P$ and $\delta_Q$, neither of these points fall into the other's "range of continuity" (i.e where the definition of "continuity at a point" can applied for that point) for the given $\epsilon$, and since continuity says nothing about what happens if $|x-y| > \delta$, this allows the function to behave unwantedly at these points i.e. $|f(x_P) - f(x_Q)|$ may be very large even though $|x_P - x_Q|$ is not.

Back to our example : what serves as $\delta_x$ for a point $x$? See that $ \left|\frac{1}{y} - \frac 1x\right| = \frac{|x-y|}{|xy|}$, so given $\epsilon > 0$ , you can check that $\delta = \frac{\epsilon x^2}{1 + \epsilon x}$ does the job (and no higher $\delta_x$ does the job at each $x$). Now, $|\frac 1n - \frac 1m| = \frac{|m-n|}{mn}$, and $\delta_{\frac 1l} = \frac{\epsilon}{l(l+\epsilon)}$. So, you can check that if $\epsilon = \frac 12$ then $\delta _{\frac 1n},\delta_{\frac 1m}$ both fall short of $|\frac 1n -\frac 1m|$. So here, for a fixed $\epsilon > 0$, the $\delta$s are decreasing much faster than $\frac 1n - \frac 1m$, and continuity of $f$ would help only if $\frac 1n - \frac 1m$ were small enough to lie in one of the $\delta$ neighbourhoods.

How do we tackle this situation i.e. what condition do we put on $f$ so that $x_n$ Cauchy implies $f(x_n)$ Cauchy? There are a few ways around it.


*

*If $x_n$ is a convergent sequence to $x$, then even though $f(x_n)$ and $f(x_m)$ cannot directly be shown to be close to each other, we can show they are both close to $f(x)$ via continuity, and then the triangle inequality shows they are close to each other. This is the condition that the domain be closed, or that the domain of $f$ be continuously extendable to a closed domain.

*Don't let the $\delta$s fall so fast. Maybe, they should depend only on $\epsilon$. Such a function is called a uniformly continuous function. For example, $f(x) = x$ on $(0,1)$ is uniformly continuous. (Can you find the $\delta$ for each $\epsilon$? It should not be difficult!)
This should be a comprehensive discussion on the desired topic.
A: In complete metric spaces, Cauchy sequences are the same thing as convergent sequences, so in complete spaces, the statement contradicts the definition of continuity. 
However, in your example, the space $(0,1)$ that your function is defined on is not complete. Hence there are Cauchy sequences that do not converge to a limit in $(0,1)$, for example, $x_n=\frac1n$. [And then choosing $f(x)=\frac1x$ maps this into the non-Cauchy sequence $y_n=n$). 
