Convergence - continuous mapping theorem, the other implication Let $f\colon \mathbb{R} \rightarrow \mathbb{R}$ be an injective continuous function and let $(x_n)$ be a sequence of real numbers. 
I'm wondering if the following are true


*

*Assume that $(x_n)$ is bounded and $f(x_n)$ converges to $f(x)$ for
some $x \in \mathbb{R}$, is it true that $x_n \to x$?

*Assume that $f(x_n)$ converges to some $y\in \mathbb{R}$, is $(x_n)$ bounded?

*If 1. holds, would it still hold if we replace $\mathbb{R}$ with $\mathbb{C}$ and real sequence with a complex sequence?


Clearly, point 1 does not hold if $f$ is not injective. To generalise the question, do we know the class of functions for which the opposite implication to the one in the continuous mapping theorem (https://en.wikipedia.org/wiki/Continuous_mapping_theorem) 
holds. 
 A: I think 1) and 3) hold. In fact for every subsequence $(x_{\phi(n)})$ of $(x_n)$ which converges to $a$ for example we have $(f(x_{\phi(n)}))$ converges also to $f(a)$. Since $f$ is injective this implies $x=a$
A: I want to explain the idea by Youem a bit more in detail.
If $(x_n)_n$ is bounded, there exists a convergent subsequence. Let $(x_{n_k})_k$ be such a sequence converging to some $x'$. Now $f(x_{n_k})$ converges to $f(x')$ (by continuity) and to $f(x)$ (by assumption). Using injectivity, it follows that $(x_{n_k})_k$ converges to $x$. 
We now know that every convergent subsequence of $(x_n)_n$ has limit $x$. We still have to prove that $(x_n)_n$ is convergent. So let us assume that $(x_n)_n$ were not convergent. Then there exists a subsequence $(x_{n_k})_k$ such that $|{x_{n_k}} - x| > \varepsilon$ for some $\varepsilon > 0$ and all $k$. Since $(x_{n_k})_k$ is bounded, this sequence has a convergent subsequence and by what we have proven so far, it has to converge to $x$. However, we also have 
$|{x_{n_k}} - x| > \varepsilon$ for all $k$ which gives the desired contradiction.
Note that the target of $f$ did not really matter. It could be $\Bbb R$, $\Bbb C$ or any other topological space in which limits of sequences are unique. The domain of $f$ can be $\Bbb R$ or $\Bbb C$ or any other metric space in which bounded sequences have convergent subsequences. Surely, this is not the most general type of space, but I just wanted to note that this is not a special property of $\Bbb R$ or anything like this. Anyway the proof above works for 1. as well as for 3.
A: *

*I think Youem sniped me as I was typing.

*No - consider $f(x) = e^{-x}$. Take $x_n = n$. Clearly $f(x_n) \to 0$ as $n \to \infty$, but $x_n$ is unbounded.
A: Hint:
for $ 1,3$
$x_n$ is bounded so it is contained in compact set, and we know continuous . one-to-one (invertible) functions defined on compact sets has continuous inverse.
$2$: does not hold take $f(x) = e^{-x}$  
