Convergence of sequences, General Topology Prove: Show that if $x_{n}\rightarrow x$ and $|x_{n} - y_{n}| \rightarrow 0$ then $y_{n}\rightarrow x$.
proof: Let $\epsilon > 0$, since $x_{n}$ converges to $x$, then there exists a positive integer $n_{0}$ such that $$n > n_{0} \Rightarrow |x_{n} - x| < \epsilon$$ Also, since $|x_{n} - y_{n}|$ converges to $0$ then there also exists a positive integer $n_{1}$ such that $$n > n_{1} \Rightarrow |x_{n} - y_{n}| < \epsilon$$. So, now we can choose an $N\in \mathbb{N}$ such that $|x_{n} - x| < 2\epsilon $ and $|x_{n} - y_{n}| < \epsilon \ \  \forall n\geq N$. Hence, $\forall n\geq N$ we have $$|y_{n} - x| = ||x_{n} - y_{n}| - |y_{n} - x|| = |x_{n} - x| - |x_{n} - y_{n}| < 2\epsilon - \epsilon = \epsilon$$ Therefore, $y_{n} \rightarrow x$
I am not sure if I am right, any suggestions would be greatly appreciated.
 A: The step
$$
|y_{n} - x| = ||x_{n} - y_{n}| - |y_{n} - x|| \color{red}{=} |x_{n} - x| - |x_{n} - y_{n}| \color{red}{<} 2\epsilon - \epsilon = \epsilon
$$
is incorrect. However, a quick fix is: for $n\geq\max\{n_0,n_1\}$
$$
|y_n-x|=|y_n-x_n+x_n-x|\leq|y_n-x_n|+|x_n-x|<\epsilon+\epsilon=2\epsilon.
$$
As $\epsilon>0$ is arbitrary, this is enough for convergence.
A: You can't guarantee that $|x_n-y_n|\geq |x_n-x|$... However, your are on the good track. Instead of taking $N$ so that 
$|x_{n} - x| < 2\epsilon $ and $|x_{n} - y_{n}| < \epsilon$ for all $n>N$ 
which doesn't guarantee that $|x_{n} - x| \leq |x_{n} - y_{n}|$, you'd better take $N$ such that
$|x_{n} - x| < \epsilon/2 $ and $|x_{n} - y_{n}| < \epsilon/2$ for all $n>N$
plug that in your last equation and use the classic triangle inequality to conclude.
A: You should try to abuse the triangle inequality:
There exists $N$ such that if $n\geq N$, then $|x_n-x|<\epsilon /2$ and $|x_n-y_n|<\epsilon/2$. Try to use the triangle inequality for $|y_n-x|$.
