Every convergent sequence is Cauchy

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Stub grade: C
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This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
I just did this to get the ball rolling. Page is of low grade due to ease of proof.

Statement

If a sequence [ilmath](a_n)_{n=1}^\infty[/ilmath] in a metric space [ilmath](X,d)[/ilmath] converges (to [ilmath]a[/ilmath]) then it is also a Cauchy sequence. Symbolically that is:

  • [ilmath]\Big(\forall\epsilon>0\ \exists N\in\mathbb{N}\ \forall n\in\mathbb{N}[n>N\implies d(a_{n},a)]\Big)\implies[/ilmath][ilmath]\Big(\forall\epsilon>0\ \exists N\in\mathbb{N}\ \forall n,m\in\mathbb{N}[n\ge m>N\implies d(x_n,x_m)<\epsilon]\Big)[/ilmath]

Proof

(Unknown grade)
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Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
The message provided is:
Easy proof, did it in my first year

See also


TODO: This too


References