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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??

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    $\begingroup$ $\|x_n-x_m\| \le \|x_n-x\| + \|x_m -x\|$, hence if $x_n \to x$, we see that $x_n$ is Cauchy. $\endgroup$ – copper.hat Mar 21 '18 at 15:39
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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.

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