Difference between revisions of "Every convergent sequence is Cauchy"
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==Statement== | ==Statement== | ||
If a [[sequence]] {{M|1=(a_n)_{n=1}^\infty}} in a [[metric space]] {{M|(X,d)}} [[convergence (sequence)|converges]] (to {{M|a}}) then it is also a [[Cauchy sequence]]. | If a [[sequence]] {{M|1=(a_n)_{n=1}^\infty}} in a [[metric space]] {{M|(X,d)}} [[convergence (sequence)|converges]] (to {{M|a}}) then it is also a [[Cauchy sequence]]. | ||
Latest revision as of 21:23, 19 April 2016
Stub grade: C
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I just did this to get the ball rolling. Page is of low grade due to ease of proof.
Contents
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)
This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
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The message provided is:
The message provided is:
Easy proof, did it in my first year
See also
TODO: This too
References
Categories:
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