Proving that $f(x_{n})$ is a Cauchy sequence when $(x_{n})$ is Cauchy and $f$ is continuous.

Question

I am aware of this question but I want to request feedback on my proofs specifically. I am required to prove the following two conditionals:

1. If $f:(0,1) \rightarrow R$ is uniformly continuous and $(x_{n})$ is a Cauchy sequence, then $f(x_{n})$ is a Cauchy sequence.

Proof: Suppose $(x_{n})$ is a Cauchy sequence. Let $\epsilon > 0$. Then, by definition, there exists $N \in N$ s.t. for all $m, n \geq N$, $| x_{n}-x_{m}| < \epsilon$. Now, pick $\delta = \epsilon$. Then, suppose $f$ is uniformly continuous on $(0,1)$. Then, $| x_{n}-x_{m}| < \delta$ implies $|f(x_{n}) - f(x_{m})| < \epsilon$ which shows that $f(x_{n})$ is a Cauchy sequence. I am a little apprehensive about the "which shows that $f(x_{n})$ is a Cauchy sequence." Is this proof accurate?

Secondly, I am required to prove:

2. If $f:[0,\infty) \rightarrow R$ is continuous and $(x_{n})$ is a Cauchy sequence, then $f(x_{n})$ is a Cauchy sequence.

Proof: Suppose $(x_{n})$ is a Cauchy sequence. Then, $(x_{n})$ converges, that is, $\lim (x_{n}) = x$ exists. Note that $[0,\infty)$ is closed since its complement is open. So, $x \in [0,\infty)$. Suppose$f:[0,\infty) \rightarrow R$ is continuous. Then, $f(x_{n})$ must be convergent. Since all convergent sequences are Cauchy, $f(x_{n})$ must be Cauchy. Am I missing something in this proof? Thank you!

Answer 1

In your first proof, you use $\epsilon$ twice, but they are not the same epsilon.

I think I would begin with the $f(x)$ is uniformly continuous.

$\forall\epsilon >0, \exists \delta > 0,\forall x,y \in (0,1): |x-y|<\delta \implies |f(x)-f(y)| < \epsilon$

This establishes our epsilon. Now, we can work to keep the differences less than epsilon.

$\{x_n\}$ is Cauchy tells us there exists an $N >0$ such that $n,m>N \implies |x_n-x_m| < \delta$

Note that I have used delta instead of the traditional epsilon. This is the same delta from the line above.

Therefore $n,m>N \implies |f(x_n) - f(x_m)| < \epsilon.$

This is the same epsilon, and the proof flows more directly from givens to the conclusion.

The second proof still has some holes.

You state that $[0,\infty)$ is closed, but not why that is relevant. You state that $f(x_n)$ is convergent, without any justification. Invoke the definition of limits to show convergence.

$\forall\epsilon >0, |x_n-x|<\delta \implies |f(x_n) - f(x)|<\epsilon$

Answer 2

Admittedly (assuming I had to proof read this -- which I have not done for years) I'd ask for a more stringent presentation, at least in case of i).

For i) The statement you have to show is: for $\varepsilon >0 $ there is $N$ such that $|f(x_n)- f(x_m)|< \varepsilon$ for every $n,m >N$. So you should start by looking at this particular situation, take some $\varepsilon$ and look at the difference $|f(x_n)- f(x_m)|$ , and then adapt the choice of the other quantities accordingly, using what you know by hypothesis. It's a litte nitpicking, admittedly, but if claims get more complicated it is really helpful if you know how to formally and cleanly write this down. The idea is fine.

ii) is also fine, in principle. It would be just nice if you'd say explicitly why $f(x_n) $ converges and what the limit is.