What if we remove some conditions in this statement, will it still be true?

Question

Let $(X,d)$ and $(Y,\rho)$ be metric spaces and $f:X\to Y$.

We can show that that if $f:X\to Y$ is uniformly continuous, then $f$ maps Cauchy sequences to Cauchy sequences, i.e., $$(x_n)\text{ is Cauchy in }X\implies(f(x_n))\text{ is Cauchy in }Y.$$

But what if $f$ is continuous but not uniformly continuous? Can we prove it in some ways or provide any counterexample?

What if $f$ is continuous and $(X,d)$ is complete? Any reason that we can tell?

What I think:

For the first question, I think if $f$ is continuous but not uniformly continuous the statement is false. But I cannot find any counterexample. Can anybody please give some help?

I'm also a bit not sure about the second question. If $(X,d)$ is complete, that means every Cauchy sequence has a limit in $X$. But what can we say about $Y$ ? Is it also complete? And is it necessary to know that $Y$ is complete in order to show the statement?

Thanks in advance for the help!

Answer 1

Consider $\tan$. It maps the Cauchy sequence $\frac{\pi}{2}-\frac{1}{n}$ to a non-converging sequence (in particular not Cauchy) in $\mathbb R$.

If $X$ is complete, the statement is of course true, since continous maps send converging sequences to converging sequences, in particular to Cauchy sequences.

But note that $Y$ can still be not complete. Consider $f: \mathbb R \to (0, \infty], x \mapsto e^{-x}$. $f$ is bijective and continuous,$\mathbb R$ is complete, but $(0,\infty]$ is not.

Answer 2

For the first part you can also consider the function $f(x)=\frac{1}{x}$ on $(0,1)$ to $\mathbb{R}$. Then take a sequence that convergences to $0$ say $\{\frac{1}{n}\}$. It is cauchy but $f(\frac{1}{n})$ is not.