A subspace $Y$ of a Banach space $X$ is complete iff the set $Y$ is
closed in $X$.
Proof:⇒) If $x∈cl(Y)$ there exists a sequence $x_
n$ in Y which converges to x and as a consquence is a Cauchy sequence (in X and Y). But Y is complete, so xn converges in Y to a point $x′$. According to the uniqueness of the limit, $x=x′∈Y$. That is, $cl(Y)⊂Y$ and hence, $Y$ is closed.
⇐) Let $x_n$ be a Cauchy sequence in $Y$. The limit $x$ of $x_n$ exists in $X$ because $X$ is complete. As $Y$ is complete, contains all its limit points, so $x∈Y$ and so, $Y$ is complete.
It may be obvious but i'm not getting Why $x_n$ is a cauchy sequence??
Best Answer
This is basic metric space theory. Suppose $x_n\to x$. Given $\varepsilon>0$ there is some $N\in\mathbb N$ such that $\|x_n-x\|<\varepsilon/2$ for $n\geq N$. Then for $n,m\geq N$ we have $$\|x_n-x_m\|\leq\|x_n-x\|+\|x-x_m\|<\varepsilon.$$ Therefore the sequence $(x_n)$ is Cauchy.