# Thread: Cauchy Sequence in a Metric Space

1. ## Cauchy Sequence in a Metric Space

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.

2. Originally Posted by 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.
Maybe the best way to do this is an argument by contradiction. Suppose that y is in the closure of A but not in A. Then there must be a sequence in A (in other words, a subsequence of $\displaystyle \{x_i\}$) converging to y. But if a subsequence of a Cauchy sequence converges, then the whole sequence converges. There's your contradiction.

3. So are we supposing that A is open, and then the contradiction comes from the fact that the sequence would then have to converge in M?

4. Originally Posted by tidus89
So are we supposing that A is open, and then the contradiction comes from the fact that the sequence would then have to converge in M?
No, we are not supposing that A is open, we are supposing that A is not closed. That means that there is a point in the closure of A that is not in A.

Yes, the contradiction comes from the fact that we then prove that the sequence converges, contrary to the statement of the question.