Proof of Cauchy Sequence Convergence with Subsequence

[tarheelborn]

Homework Statement

If {s_n} is a Cauchy sequence of real numbers which has a subsequence converging to L, prove that {s_n} itself converges to L.

Homework Equations

The Attempt at a Solution

I know that all Cauchy sequences are convergent, and I know that any subsequences of a convergent sequence are convergent to the same limit as the sequence, but I am not sure if I can turn the second part of the statement around to say that if a subsequence is convergent to L, then the sequence converges to the same limit. Any ideas? Thanks.

 

[HallsofIvy]

Homework Statement

If {s_n} is a Cauchy sequence of real numbers which has a subsequence converging to L, prove that {s_n} itself converges to L.

Homework Equations

The Attempt at a Solution

I know that all Cauchy sequences are convergent, and I know that any subsequences of a convergent sequence are convergent to the same limit as the sequence, but I am not sure if I can turn the second part of the statement around to say that if a subsequence is convergent to L, then the sequence converges to the same limit. Any ideas? Thanks.

Actually, you can't just say "if a subsequence converges to L, then the sequence must converge to L". For example, the sequence \{a_n\} with a_n= 1- 1/n for n even and a_n= 1/n for n odd has a subsequence (\{a_n\} for a even) that converges to 1 but the sequence itself does not converges.

What you can say is that if the sequence converges, then because, as you say, all subsequences must converge to that limit, yes, the sequence converges to whatever limit any subsequence converges to. The difference is that you must know the sequence does converge. Which you know here because it is a Cauchy sequence.

 

[hgfalling]

Well, suppose that the Cauchy sequence converges to a number M different than L. Show that this leads to a contradiction by showing that there exists an epsilon such that some sequence value is always further away from M than epsilon.

 

[tarheelborn]

Thank you!