For the two scenarios below, either give an example if such a request is possible, or argue why such an request is impossible. I think the first is possible and second is impossible. However, I can't seem to find an example to support 1. and for 2. I am not sure if my argument is correct.
A continuous function $f:(0,1) \rightarrow [-1, 1]$ and a sequence $(x_n) \subset (0,1)$ which converges to $0$ such that the sequence $(f(x_n))$ does not converge.
A continuous function $f:(0,1) \rightarrow [-1, 1]$ and a sequence $(x_n) \subset (0,1)$ such that no subsequence of $(f(x_n))$ converges in $\mathbb{R}$.
My reasoning:
I think 1. is possible because $c=0$ is not part of the domain of $f$, hence even if $(x_n) \rightarrow 0$ (with $x_n \in (0,1)$), this does not mean $f(x_n)$ has to converge to $f(c)$. But I am struggling to find an example of this. What I've done so far is let $f(x) = \sin(\pi x)$ be the continuous function on $(0,1)$ and pick $(x_n) = \frac{1}{n+1}$, but $f(x_n)$ does converge... Any examples would be good.
I think 2. is impossible. Since $(x_n)$ is any sequence in $(0,1)$ then the sequence $(f(x_n))$ must be bounded between $[-1,1]$. then by the Bolzano-Weierstrass Theorem, this sequence must contain a convergent subsequence. Is this argument correct? How come I didn't need to use the assumption that $f$ is continuous in my argument at all? Then does this mean that this statement is false for ANY generic function $f:(0,1) \rightarrow [-1,1]$?
The key thing to realise with question $1$ is that if $f$ could be extended to a continuous function on $[0,1]$, then the result would be true for this extension, and so also true in the original context. So you have to find some example where no matter what you chose for $f(0)$, the resulting extension would not be continuous.
Your argument in question $2$ appears to be correct.