The Set of All Subsequential Limits Given $\{a_n\}_{n=0}^\infty$ and $\{b_n\}_{n=0}^\infty$ bounded sequences; show that if $\lim \limits_{n\to \infty}a_n-b_n=0$ then both sequences have the same subsequential limits.
My attempt to prove this begins with: Let $E_A=\{L|L$ subsequential limit of $a_n$}
and $E_B=\{L|L$ subsequential limit of $b_n$}. We need to show that $E_A=E_B$.
Given bounded sequence $a_n$ and $b_n$ we know from B.W that each sequence has a subsequence that converges, therefore both $E_A$ and $E_B$ are not empty; 
Let $L\in E_A$. 
How can I show that $L\in E_B$?
Thank you very much.
 A: You can approach it very directly. Let $L\in E_A$. Then there is a subsequence $\langle a_{n_k}:k\in\Bbb N\rangle$ converging to $L$; now use the hypothesis that $\langle a_n-b_n:n\in\Bbb N\rangle\to 0$ to show that $\langle b_{n_k}:k\in\Bbb N\rangle$ converges to $L$.
Note that you really only have to prove that $E_A\subseteq E_B$: if $\langle a_n-b_n:n\in\Bbb N\rangle\to 0$, then clearly also $\langle b_n-a_n:n\in\Bbb N\rangle\to 0$, and you can appeal to the first half to conclude that $E_B\subseteq E_A$.
A: I think you may want to prove that:

*

*Consider two sequences $\{x_n\}$ and $\{y_n\}$ such that $x_n-y_n \to l$ and $y_n \to y$, then, $$x_n \to l+y$$


*Given a convergent sequence $\{x_n\}$ that converges to $x$, all its subsequences converge to the same limit, $x$.
Do you see how that would pay here?


*

*Let $r \in E_B$. That is, $r$ is a limit point of the sequence $\{b_n\}$. So, there is a subsequence of $\{b_n\}$, say $\{b_{n_k}\}$ that converges to $r$.


*Now, consider the same subsequence of $\{a_n-b_n\}$, namely $\{a_{n_k}-b_{n_k}\}$. Since this is a subsequence of a convergent subsequence, $\{a_{n_k}-b_{n_k}\}$ converges to $0$, $a_{n_k}-b_{n_k} \to 0$ by $(2)$.
Now putting together the two claims, by $(1)$. you have that $a_{n_k} \to r$. That is, $r \in E_A$. This proves one inclusion, $E_B \subseteq E_A$. The proof of the other inclusion is similar.
For the other inclusion, as Brian observes, note that $a_n -b_n \to 0$ implies $b_n-a_n \to 0$. Now appeal to the previous part, to see that $E_A \subseteq E_B$.

