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**tidus89** Problem:

Let {$\displaystyle x_{i}$} be a Cauchy sequence in a metric space (M,D). Let A={$\displaystyle x_{1}, x_{2}, x_{3}$, ...}.

Suppose that {$\displaystyle x_{i}$} doesn't converge in M. Prove that A is a closed subset of (M,D).

What I have done:

So we know {$\displaystyle x_{i}$} is Cauchy, so let $\displaystyle \epsilon$>0 be given. Then there exists a positive integer N such that the D($\displaystyle x_{i}, x_{j}$)<$\displaystyle \epsilon$ for all i,j $\displaystyle \geq$ N. Also, and this may be where I'm stuck.. Since {$\displaystyle x_{i}$} doesn't converge in M, then we know that for any y$\displaystyle \in$M, $\displaystyle \epsilon$>0, and N, we can find an n>N such that D($\displaystyle x_{n}$, y)>$\displaystyle \epsilon$.

That last statement could be wrong/unnecessary, so any help you can give me on where to go would be very helpful.