Prove that if $x_{n}$ is a convergent sequence, then for all $\epsilon >0$, whenever $m,n \geq N$, it follows that $|x_{m}-x_{n}|< \epsilon$.

Asked Sep 17 '16 at 16:37

Active Sep 17 '16 at 17:20

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This is the original question: Prove the following characterization of convergence. If $\{x_{n}\}$ is a convergent sequence, then $\forall \varepsilon >0$ there is an $N(\epsilon) \in \mathbb{N}$ so that whenever $m,n\geq N(\epsilon)$, it follows that $$|x_{m}-x_{n}|<\varepsilon.$$ I know how to show something converges by using the definition of convergence for specific equations, but I have never written a proof for a general case. Any help on how to prove this would be very helpful!

edited Sep 17 '16 at 17:20

rtybase

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asked Sep 17 '16 at 16:37

Anne M.

1

Hint: If $x = \lim x_n$, $x-x=0$ and the triangle inequality. – Nigel Overmars Sep 17 '16 at 16:39
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Effectively, this asks to prove that every converging sequence is Cauchy (https://en.wikipedia.org/wiki/Cauchy_sequence) sequence as well? But this is a standard material for any analysis course ... – rtybase Sep 17 '16 at 16:41
E.g. $|x_n-x_m| = |x_n - x + x -x_m|<|x_n - x| + |x -x_m| < 2\epsilon$ – rtybase Sep 17 '16 at 16:43

Prove that if $x_{n}$ is a convergent sequence, then for all $\epsilon >0$, whenever $m,n \geq N$, it follows that $|x_{m}-x_{n}|< \epsilon$.

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