Let $X$ be a totally bounded metric space. If $f$ is a uniformly continuous mapping from $X$ to a metric space $Y$, show that $f(X)$ is totally bounded. Is the same true if $f$ is only required to be continuous?
I got the first part, but now I am stuck on the second part. I think the answer is no and I think $f(x) = \frac{1}{x}$ with $X = (0,1]$ would be a counterexample, but I'm struggling showing that its not totally bounded
EDIT: Similarly, if $X$ is complete and $f$ continuous, is it true that $f(X)$ is complete? Again, I think the answer is no, but I'm having trouble coming up with any example here
Take $\iota\colon[1,+\infty)\longrightarrow\mathbb R$ defined by $\iota(x)=\frac1x$. The space $[1,+\infty)$ is complete, $\iota$ is continuous, but $\iota\bigl([1,+\infty)\bigr)=(0,1]$, which is not complete.