Question #217042

Prove that the subspace.of a complete metric is complete if and only if it is closed


1
Expert's answer
2021-07-26T06:38:06-0400

Let us prove that the subspace SS of a complete metric XX is complete if and only if it is closed.


Let SS be complete and xS.x\in\overline{S}. Then there exists a sequence(xn)nNS(x_n)_{n∈\N} ⊆ S converging to xx. Obviously, this sequence is a Cauchy sequence, and, since SS is complete, it converges to some ySy\in S. Since the limit of a sequence is unique in a metric space, we see that x=yS.x = y∈ S.


Let SS be closed and (xn)nN(x_n)_{n∈\N} be a Cauchy sequence in SS. Since XX is complete, (xn)nN(x_n)_{n∈\N} converges to some xXx ∈ X. But as SS is closed, xx has to be in SS.


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