Don't understand proof that convergence implies Cauchy

Question

So we are given that $x_n \rightarrow x$, so we can let $\epsilon = \epsilon'/2$ and there definitely exists an $N$ such that for all $n \geq N$, $|x_n - x| < \epsilon'/2$. Also by the triangle inequality, it suffices to show that $|x_n - x| + |x_m - x| < \epsilon'$. We know that $|x_n - x| < \epsilon'/2$, but we are given no information about $|x_m - x|$. It is possible that $\{x_m\}$ does not even converge, so I don't get how you can just claim that $\{x_m\}$ does converge without a separate proof/lemma.

Answer 1

The idea here is that you choose a pair of numbers, $n$ and $m$ such that $n\geq N$ for the appropriate $N$, implying you have both that $$|x_n-x|<\varepsilon'/2$$ $$|x_m-x|<\varepsilon'/2.$$ From which point the proof follows by the triangle equality you set up. We're not saying $|x_n-x|<\varepsilon'/2$ because $n$ is special - we're saying that's true because $n$ is big (and so is $m$). That is, we're trying to prove something about the "tail-end" of the sequence using only that the sequence stays within arbitrarily small bounds. So, $n$ and $m$ are ranging over the same sequence - so it makes little sense to think $\{x_m\}$ is different from $\{x_n\}$, since we're trying to consider the values of a single sequence, at different indices. The reason we choose $n$ and $m$ arbitrarily, only restricted in that they must be at least $N$ is because to prove a sequence is Cauchy, we need to show that, for any $\varepsilon$, there is a big enough $N$ such that for any $n$ and $m$ larger than $N$, the difference $|x_n-x_m|<\varepsilon$.