Suppose that $\lim (x_n) = x$ and that $|x_n-y_n|=1$ for all n exists in N. Prove that there exists a subsequence of $(y_n)$ that converges to either x+1 or x-1.
I understand the logic behind this, I'm just having an issue formulating my argument.
I know properties of limits say we can do addition such that $\lim (x_n) + \lim (y_n)=\lim ((x_n)+(y_n))$.
Can we say that since $\lim (x_n)=x$, $|x_n-y_n| = |x - \lim(y_n)| = 1$, which could be broken into
$-1 < x - \lim(y_n) < 1$, thus
$-x - 1 < - \lim(y_n) < 1 - x$ thus
$x + 1 > \lim (y_n) > x -1$
This shows that it is bounded by these, but does that prove there is a subsequence that converges to these?