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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!

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marked as duplicate by Math1000, Arpit Kansal, Joey Zou, JonMark Perry, Shailesh Sep 18 '16 at 1:59

This question was marked as an exact duplicate of an existing question.

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

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