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While reading over the definition of a Cauchy sequence in $\mathbb{R}$, I conjectured that the following condition

$\forall \epsilon>0, \exists N\in\mathbb{N}: \forall n>N, |a_n-a_N|<\epsilon$

is necessary and sufficient for a sequence in $\mathbb{R}$ to converge.

Is this true? I haven't been able to find any counterexamples or proofs of this conjecture on this site.

If this is indeed true, I'm guessing that a proof would involve subsequences and/or the Cauchy sequence condition.

FabrizzioMuzz
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    In $\mathbb{R}$ every sequence is indeed a Cauchy sequence. Convergent sequences are clearly Cauchy. You can prove the converse using the fact that Cauchy sequences are bounded and each bounded sequence has a convergent subsequence. – Student Apr 04 '23 at 19:59
  • I'll try that. Thanks for the insight. – FabrizzioMuzz Apr 04 '23 at 20:00
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    In case you are stuck: https://math.stackexchange.com/questions/2132993/prove-that-a-cauchy-sequence-is-convergent, but it's a nice exercise to try and prove it yourself :) – Student Apr 04 '23 at 20:02
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    The condition you write is equivalent to being Cauchy (even more generally in metric spaces). Then, in $\Bbb{R}$, Cauchyness is equivalent to convergence, so putting the two equivalences together, your condition is equivalent to convergence. – peek-a-boo Apr 04 '23 at 20:02
  • @Student By $\mathbb{R}$ every sequence is indeed a Cauchy sequence do you mean that every sequence satisfying the condition written out in the post is Cauchy? – FabrizzioMuzz Apr 04 '23 at 20:08

1 Answers1

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You don't even have to discuss convergence - the condition you specify is equivalent to being a Cauchy sequence in any metric space. Just use the triangle inequality

$$|x_m-x_n|\le |x_m -x_N|+|x_n-x_N|$$

and realize that if you can make both of the terms on the right hand side small then you can make the term on the left hand side small.

(You obviously should make the argument more formal. If you've seen a standard $\epsilon/2$ argument before it won't be hard. And if you don't recognize such an argument, here's some good practice!)

JonathanZ
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