Let (X,d) be a metric space and A a dense subspace such that everyCauchy sequenc

Question

Let (X,d) be a metric space and A a dense subspace such that everyCauchy sequence in A converges in X. Prove that X iscomplete.


My comments: Me and the term 'dense' don't understand eachother too well at the moment (ironic, isn't it?) I understand(I think) that the metric space X is complete if every Cauchysequence in X converges to a point in X. Does the fact that Ais a dense subspace mean that there are no other Cauchy sequencesin X that are outside of A? I'm just not sure where to go onthis one. Thanks.

Explanation / Answer

A subspace A is dense in a metric space X if X is the closure A,meaning if X is the smallest closed subset containing A. To prove a space is complete, we need to prove that every Cauchysequence converges to a point in that space. Let {xi} be a Cauchy sequence in X. If {xi} is in A, then itconverges to a point in X and we are done. Suppose {xi} is not contained entirely in A. Then there is some subsequence of elements of X that is containedentirely in A. Then this sequence converges to a point y inA, and thus the whole sequence converges to y. If {xi} is in X-A (the complement of A in X), then it must be aconstant sequence, i.e. {xi}={xn,xn,xn,xn...} for some xn inX. Then it converges to xn. Therefore every Cauchy sequence in X converges to a point in X,therefore X is complete.

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