This is a question I've stumbled upon and the question asks me to prove something which I don't think is true.


Let $f(x)$ be a continuous function in $\mathbb{R}$, and let $(a_n)$ be a Cauchy sequence. Prove that $f(a_n)$ is a Cauchy sequence.


Since the reals covers $[-\infty,\infty]$, I thought I might as well look for a simple interval to prove that it isn't true. Let $f(x) : (0,1) \to \mathbb{R}$. Since they asked for a continuous function, let $f(x)=\frac{1}{(x)}$, which is continuous. Looking at a Cauchy sequence $(\frac{1}{(n)})$, I see that

$f(\frac{1}{(n)})=n$, which is not Cauchy.

Am I doing something wrong or is the question improperly stated?

Attempt #2:

Let $f(x)$ be a continuous function $\in \mathbb{R}$. (As Chris said) I'll lock $(a_n)$ to a compact interval and then I can work with $f$ there. Let $\epsilon>0$ be given. Since $f$ is continuous on a compact set, it is also uniformly continuous. Since $f$ is uniformly continuous there exists $\delta>0$ such that for all $x,y\in \mathbb{R}$ such that $|x-y|<\delta$, $|f(x)-f(y)|<\epsilon$. And since $(a_n)$ is Cauchy and $\delta>0$, there exists a $N\in \mathbb{N}$ such that for all $n,m \geq N$ we have $|x_n - x_m|<\delta$. Thus $|f(x)-f(y)|<\epsilon$, $\epsilon>0$. Thus $f((x_n))$ is Cauchy.

  • 2
    $\begingroup$ $\frac{1}{x}$ is not continuous on all of $\mathbb{R}$ (it has a discontinuity at $0$). As a hint, observe that any Cauchy sequence lies in a compact set and that any continuous function on a compact set is uniformly continuous. For this problem it matters a lot that the domain is all of $\mathbb{R}$. $\endgroup$ – Chris Janjigian Dec 8 '13 at 2:59
  • $\begingroup$ $\frac{1}{x}$ is just a counterexample I used. I'm trying to show that for a subset of $\mathbb{R}$ it doesn't hold, so it won't hold for all of $\mathbb{R}$. $\endgroup$ – sillyme Dec 8 '13 at 3:02
  • $\begingroup$ The result is true as stated. It matters that $f$ is assumed to be continuous on all of $\mathbb{R}$. $\endgroup$ – Chris Janjigian Dec 8 '13 at 3:04
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    $\begingroup$ @sillyme: Right. And what Chris is trying to point out is that there is no way to transform a function on a subset of $\mathbb{R}$ that fails this theorem to a function on all of $\mathbb{R}$ that fails it as well. $\endgroup$ – Eric Stucky Dec 8 '13 at 3:05
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    $\begingroup$ @sillyme $\mathbb{R}$ is not compact. $\endgroup$ – Chris Janjigian Dec 8 '13 at 3:14

Do you have to do it as an epsilon-delta proof? I would have just used the fact that, since $f(x)$ is continuous and ${a_n}$ converges, $f(a_n)$ converges. And since any convergent sequence is Cauchy (in the reals), you're done.

  • $\begingroup$ I honestly wrote my attempt 2 for practice in TeX. $\endgroup$ – sillyme Dec 8 '13 at 3:58
  1. The first attempt fails because -though are homeomorphic,- the metric is deformed so much between $\Bbb R$ and $(0,1)$ such that sequences which tend to either infinity (which are not Cauchy sequences) correspond to sequences convergent to either $0$ or $1$.
  2. You're almost there, but $\Bbb R$ is not compact. However, as Chris suggested in a comment, we can enclose our Cauchy sequence to a compact interval and restrict $f$ there.

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