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

Question

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?

Answer 1

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.

Answer 2

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 $y_n=n$).