Question

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

Answer 1

Let us prove that the subspace $S$ of a complete metric $X$ is complete if and only if it is closed.

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

Let $S$ be closed and $(x_n)_{n∈\N}$ be a Cauchy sequence in $S$. Since $X$ is complete, $(x_n)_{n∈\N}$ converges to some $x ∈ X$. But as $S$ is closed, $x$ has to be in $S$.