When does the convergence of $f(\xi_n)$ imply the convergence of $\xi_n$? Let $(X_1,d_1)$ and $(X_2,d_2)$ be two (complete)metric spaces. If $(\xi_n)$ is a sequence in $X_1$ and $f:X_1\rightarrow X_2$ such that $(f(\xi_n))$ is convergent in $X_2$, then my question is under what conditions can we show that $(\xi_n)$ is convergent in $X_1$. I guess it is obviously true if $f$ is a homeomorphism, but can we have weaker requirements on $f$ or is it too general to say anything else? If so can we extend it to nets/filters in general topological spaces? 
In particular I am looking at the following problems.
Suppose that $(\xi_n)$ is sequence in $\mathbb{R}$, show that: 
i. If $\frac{\sin \xi_n}{\xi_n}\rightarrow 1$ then $\xi_n\rightarrow 0$
ii. If for all $t\in\mathbb{R}$, $(\exp[it\xi_n])$ converges in $\mathbb{C}$ then $(\xi_n)$ is convergent.
iii. If $(\exp[it\xi_n])$ converges in $\mathbb{C}$ for any $t$ in a set with positive Lebesgue measure then $(\xi_n)$ is convergent.
For i. It is easy to see $|\frac{\sin \xi_n}{\xi_n}-1|= 1-\frac{\sin \xi_n}{\xi_n}$, and is an even function. Then observing that for a $\epsilon>0$, $\inf_{t\geq\epsilon}\{1-\frac{\sin t}{t}\}:=\alpha_\epsilon>0$, so $|\frac{\sin \xi_n}{\xi_n}-1|<\alpha_\epsilon$ then implies that $|\xi_n|<\epsilon$. Any corrections, improvements or neater arguments to show this will be greatly appreciated. 
For ii. I was given the idea to consider $\lim_{n\rightarrow\infty}\lim_{m\rightarrow\infty}(\exp[it(\xi_n-\xi_m])=1$. Then using DCT we can integrate both sides over $[-1,1]$ which reduces the problem to $\lim_{n\rightarrow\infty}\lim_{m\rightarrow\infty}\frac{\sin(\xi_n-\xi_m)}{\xi_n-\xi_m}= 1$, so then we need to use i. to infer $\lim_{n\rightarrow\infty}\lim_{m\rightarrow\infty}|\xi_n-\xi_m|=0$.  However I am not to comfortable with this and think they must be a nicer approach.  Secondly don't we require the double limit $\lim_{n,m\rightarrow\infty}|\xi_n-\xi_m|=0$ to infer that the $(\xi_n)$ is a Cauchy sequence? If so how do we know that the iterated and double limit will coincide? So my confusion is with the iterated limits vs double limits. Any help again is most welcome, or an alternative argument will also be appreciated.
For iii. have no idea how to proceed as sets that are nowhere dense such as the Smith-Volterra-Cantor set, have to be considered.
I would appreciate any help with regards to any of these problems and any references (basic or advanced) will also be appreciated. Thanks in advance.
 A: (edited later to clarify some issue raised in comments)
For 1 I think your proof is fine - also note that you can replace $\xi_n$ with $|\xi_n|$ and assume $0<\xi_n<\frac{\pi}{2}, n \ge n_0$ by size considerations and then $\frac{\sin x}{x}$ is strictly decreasing on $[0, \frac{\pi}{2}]$ hence is a homeomorphism from that to $[\frac{2}{\pi},1]$ etc
I will skip 2 and present a proof for 3 directly: 
Since obviously $|e^{it\xi_n}|=1$, if $g(t)=\lim_{n} e^{it\xi_n}, t \in A$ of non zero measure, $|g(t)|=1, t \in A$, and $g$ is measurable so there is a measurable $f(t) \in (-\pi, \pi], t \in A$, s.t. $g(t)=e^{if(t)}, t \in A$ by taking the principal branch of the argument. 
First assume $\xi_n$ is bounded when we just need $A$ uncountable. Let $\xi$ any limit point of $\xi_n$. Then obviously $g(t)=e^{if(t)}=e^{it\xi}, t \in A$ so there is an integer $k(t,\xi)$ with $f(t)-t\xi=2k(t,\xi)\pi$. Assume $\zeta$ is another limit point distinct from $\xi$ it then follows $f(t)-t\zeta=2k(t,\zeta)\pi$, so:
$t(\xi-\zeta) \in 2\pi \mathbb{Z}, t \in A$ uncountable or if $\xi-\zeta=a \ne 0$, $t \in \frac{2\pi \mathbb{Z}}{a}$ which is a countable set and that is a contradiction. Hence there is only one limit point and the sequence converges.
Assume now $\xi_n$ unbounded and wlog (passing to a subsequence, and using that the conjugate $e^{-it\xi_n}$ also converges, so we can replace $\xi_n$ by $-\xi_n$ etc) we can assume $\xi_n \to \infty$. Since the Lebesgue measure is regular, for any set of positive measure $m(A)>0$ there are bounded measurable subsets $A_k$ of $A$ for which $m(A_k) \to m(A)$, so we can take one such bounded $A_k, m(A_k) >0$ and rename it $A$ for our set of convergence and (also for convenience and later use of the standard Riemann-Lebesgue lemma) we also can scale $A$ to be in $[-\pi, \pi]$ since we just replace $t\xi_n$ by $\frac{t}{C}(C\xi_n)$ for some constant $C>0$ that makes all elements in $A$ less than $\pi$ in absolute value, while $C\xi_n$ still goes to infinity and we rename it as $\xi_n$. 
Then let $\xi_n=m(n)+\alpha_n$ its decomposition into integer and fractional parts respectively, and we can assume wlog $m(n)$ strictly increasing to infinity and we also assume (by passing to a subsequence) that $\alpha_n \to \alpha$, so we reduce the problem to the case:
$e^{itm_n} \to e^{ih(t)}$ where $t \in A$ included in $[-\pi,\pi]$ of non zero measure and $m_n \to \infty$ integers, while $h$ is a measurable function with values in $[-2\pi,2\pi]$ since we can use $h(t)=f(t)-\alpha t$, and we have that $f(t) \in [-\pi,\pi], 0 \le \alpha \le 1, |t| \le \pi$. 
Since obviously by Riemann-Lebesgue for Fourier series applied to $\chi(A)e^{ikt}$ (and by conjugation which switches the sign of the exponent, keeping the limit still zero), we have $\int_{-\pi}^{\pi}\chi(A)e^{it(m_n-k)}dt \to 0$ for any integer $k$, and since $\chi(A)e^{it(m_n-k)} \to \chi(A)e^{i(h(t)-kt)}$ for all $t \in [-\pi,\pi]$, we get by DCT $\int_{-\pi}^{\pi}\chi(A)e^{i(h(t)-kt)}dt=0$ for any integer $k$ 
But now this means that the Fourier series of the integrable function $\chi(A)e^{ih(t)}$ is zero, hence the function is zero a.e. But $|\chi(A)e^{ih(t)}|=\chi(A)$, so we get $\chi(A)=0$ a.e hence $m(A)=0$ contradiction and we are done!
