Suppose that $\lim (x_n) = x$ and that $|x_n-y_n|=1$ for all $n$ exists in $\mathbb N$.

Question

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?

Answer 1

Because $\{x_n\}$ converges, it is bounded by some $B>0$. And because $|y_n|\leq|y_n-x_n|+|x_n|$, the sequence $\{y_n\}$ is bounded (by $B+1$). So it follows from the Bolzano-Weierstrass Theorem that $\{y_n\}$ admits a convergent subsequence $\{y_{n_k}\}$ that converges to some $R$.

Now, $x_{n_k}\to x$ so $(x_{n_k}-y_{n_k})^2\to (x-R)^2$. Yet $\{(x_{n_k}-y_{n_k})^2\}_k$ is the constant sequence in which every entry equals $1$. Thus, we must have $$ 1=(x-R)^2\implies R=x+1\text{ or }R=x-1. $$