# Cauchy Sequence and Completeness

## Homework Statement

let (X,d) be a metric space and let A be a dense subset of X such that every Cauchy sequence in A converges in X. Prove that (X,d) is complete.

## Homework Equations

(X,d) is complete if all Cauchy sequences in X converge.
A is a dense subset of X => closure(A) = X

## The Attempt at a Solution

Since for all x in A there exists <x_n> in A s.t. <x_n> -> x, x is an element of closure(A) so x is in X, but since all Cauchy sequences in A converge in X for all x in X, doesn't this mean that all Cauchy sequences in X converge in X? Thank you for your help anyone!

## Answers and Replies

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Hurkyl
Staff Emeritus
Science Advisor
Gold Member
since all Cauchy sequences in A converge in X for all x in X, doesn't this mean that all Cauchy sequences in X converge in X?
Well, yes -- what I've quoted is a restatement of what you have been asked to prove.

But if you're asking if what you wrote is actually a proof, then no; other than those contained entirely in A, you've not said anything about Cauchy sequences in X, so you certainly cannot have proven Cauchy sequences in X converge!

Ive been trying to show that there is a subset of the original <x_n> such that all of the elements of that set (call it <x_ni>) are in closure(A) meaning that that sequence is in X. if I can say that, then cant I say that <x_ni> converges to x (because the greater set <x_n> converges to x). then i can say that since <x_ni> -> x, and those are both in X, then X is complete?

lanedance
Homework Helper
however something along the lines of any cauchy sequence in x can be estimated arbitrarily closely by a sequence in A should help