# If the difference of two sequence converge to zero, then both sequences must have same cluster points.

Suppose $\{x_n\}$ and $\{y_n\}$ are two sequences of real numbers which satisfy $x_n-y_n\rightarrow 0$ as $n\rightarrow 0$. I need to show that both sequences must have same cluster points.

I can see the result to make intuitive sense, it's very straight forward however I'm having a hard time trying to rigorously prove this. Here is how I thinking about it.

Let $z_n=x_n-y_n$, $z_n\rightarrow0$ means that $x_n$ and $y_n$ are getting closer and closer to each other as $n\rightarrow\infty$. $z_n\rightarrow0$ implies that for any $\delta>0$ there exists $N\in\mathbb{N}$ such that for all $n\geq N$ we have,

$$x_n-y_n\in(-\delta,\delta)$$

I don't know how to proceed from here. Any ideas?

Hint If $a$ is a cluster point of $x_n$, then you can find some $x_{k_n} \to a$.
Then $$y_{k_n}= x_{k_n}-(x_{k_n}-y_{k_n})\to a$$
and hence $a$ is a cluster point for $y$ too.
Next, do the same proof with $x,y$ interchanged.
Denote $A, B$ be the sets of cluster points of the sequences $\{x_n\}, \{y_n\}$ respectively. You wish to show $A = B$. Thus if $a \in A\implies (a-\epsilon, a+\epsilon)$ contains those $x_k$'s with $k \ge N_0$. But $|y_n-a| \le |y_n-x_n|+|x_n-a|< \epsilon/2+\epsilon/2= \epsilon$, for $n \ge N_1\ge N_0\implies (a-\epsilon,a+\epsilon)$ contains those $y_k$'s with $k \ge N_1\implies a \in B\implies A \subseteq B$. Similarly, $B \subseteq A$. Thus $A = B$.