Bounded, real-valued sequences with same limit points. I am stuck in the following exercise:
Let $\{\lambda_i\}$ and $\{\mu_j\}$ ($i,j\in \mathbb{N}$) be real, bounded sequences with the same limit points (I may initially only consider finitely many limit points). Show that there is a bijection $\sigma:\mathbb{N}\to\mathbb{N}$ such that $|\lambda_i-\mu_{\sigma(i)}|\to 0$ as $i\to\infty$. (This is exercise 2.9.4 in Analytic K-homology by Higson and Roe).
So far I have not even figured out where to start. Is it possible to explicitly construct this bijection $\sigma$?
 A: The case of finite limit points. Suppose $(\lambda_{i})$ and $(\mu_{i})$ are real bounded sequences, with limit points $r_{1},...,r_{n}$, we can find maps $f,g:\mathbb{N}\rightarrow\{1,...,n\}$ given by
$$f(k)=\min\{i=1,...,n:|r_{i}-\lambda_{k}|=\min_{1\leq j\leq n}|r_{j}-\lambda_{k}|\}$$
$$g(k)=\min\{i=1,...,n:|r_{i}-\mu_{k}|=\min_{1\leq j\leq n}|r_{j}-\mu_{k}|\}$$
Note that the subsequences 
$$(\lambda_{i})_{i\in\mathbb{N},f(i)=k}:=(\lambda_{k_{\lambda,j}})\text{ and }(\mu_{i})_{i\in\mathbb{N},g(i)=k}:=(\lambda_{k_{\mu,j}})$$
for all $k\in\{1,...,n\}$ are infinite and converge to $r_{k}$. We can now define $\sigma:\mathbb{N}\rightarrow\mathbb{N}$ by taking for $k\in\{1,...,n\}$ and $j\in \mathbb{N}$.
$$\sigma(k_{\lambda,j})=k_{\mu,j}.$$
Note that $\sigma$ is well-defined as for every $m\in\mathbb{N}$ there exists unique $k\in\{1,...,n\}$ and $j\in\mathbb{N}$ such that $m=k_{\lambda,j}$. Clearly for all $k\in\{1,...,n\}$ we have that $|\lambda_{k_{\lambda,i}}-\mu_{k_{\lambda,i}}|\rightarrow0$ as $i\rightarrow\infty$, so $|\lambda_{i}-\mu_{\sigma(i)}|$.
I'm not sure yet how to expand this to infinite limit points, I'll expand on this answer when I know more.
