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Proving a metric space is complete?

So, I am given a metric space. I have to prove it is complete. I know complete means that every cauchy sequence is convergent.

So you let {x_n} be a sequence of elements in the space and prove it converges.

My issue is, to prove convergence you state:

for every epsilon > 0, there exists N such that for every n >= N, d(x_n, x) < epsilon.

But how do I prove the existence of such an x? (a limit point?)

I think that is my only issue because once I construct the limit point, I should be able to prove the given cauchy sequence is convergent.

(also, since one is trying to prove d(x_n, x) < epsilon, I thought you might be able to do:

d(x_n, x) <= d(x_n, x_m) + d(x_m, p)

and then since cauchy you know:

d(x_n, x) < (epsilon/2) + d(x_m, p)

but then you are just back where you started since now you have to prove d(x_m, p) < epsilon/2 which is the same as proving d(x_n, x) < epsilon.

So, once one has a limit point, which is part of my question since I don't know how to construct one, does being cauchy play a role in proving convergence? (the latter question stemming from the above).

1 Answer

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  • Anonymous
    6 years ago
    Favorite Answer

    You can't prove it since it's not true. The rational numbers with the usual metric is a metric space, but is not complete. Also see https://en.wikipedia.org/wiki/Metric_space#Complet... for a different metric.

    You nailed the issue: how do you know the point x exists. With the rationals, you don't have enough points, as you do with the reals.

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