Cauchy sequence question

Question

Let $f$ be a continuous function in $ (0,1)$.

Determine of the following is right or wrong:

If $ (x_n) $ is a Cauchy sequence such that $0< (x_n) <1$ then $f(x_n) $ is a Cauchy sequence.

Actually... I don't know where to start. How to approach this question?

Thanks!

Answer 1

Look at $f(x) = 1/x$, and the sequence $x_n = 1/n$, which is evidently a Cauchy sequence.

Letting $y_n = f(x_n)$, you get that $y_n = 1/(1/n) = n$, which is not a Cauchy sequence.

As for your real question, which is "where do I start?", my answer is this:

A Cauchy sequence is one where the terms eventually stay very near each other, but a continuous function can stretch things apart...so if we can just find a continuous function that does this enough to make our sequence non-Cauchy, we're done. Now comes the hard part: finding a function that pulls apart nearby items, no matter how nearby they are. Such a function, if differentiable, would have to have infinite derivative at the limit point. That doesn't make sense for continuous functions, so it must be that the limit point for the sequence is not actually at a point where $f$ is well-defined. And then if the derivative of $f$ approaches infinity as we head to that limit, everything will work out.

Well, a Cauchy sequence in $\mathbb R$ has a limit, so the only hope we've got is to pick a sequence in $(0, 1)$ whose limit falls outside...and an easy choice is something like $1/n$.

And now we have to find a function whose derivative goes to infinity as $x \to 0$. Everyone's favorite is $f(x) = 1/x$. Combine, and it happens to work out really nicely.

Answer 2

Idea: Since every uniformly continuous function is Cauchy-continuous (and what you want is an example of continuous function on $(0,1)$ contradicting Cauchy continuity), a natural start is to consider a continuous function $f$ that is not uniformly continuous. In particular, it'll have to have a "problem" at one of the endpoints, $0$ or $1$; which in turn implies that $(x_n)_n$ should converge (being Cauchy, it'll have to converge) to either $0$ or $1$ for the "bad behavior" of $f$ to manifest itself.

Hint: now, take your favorite continuous-but-not-uniformly-continuous function $f$ on $(0,1)$, say with a uniform continuity issue at $0$ (there is a good chance it'll work, i.e. will not be Cauchy continuous), and choose accordingly a sequence $(x_n)_n \in (0,1)^{\mathbb{N}}$ converging to $0$. See what happens.