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A subspace $Y$ of a Banach space $X$ is complete iff the set $Y$ is closed in $X$.

Proof:⇒) If $x\in\mathrm{cl}(Y)$ 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, $x=x′\in Y$. That is, $cl(Y)\subset Y$ 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 $x\in Y$ and so, $Y$ is complete.

It may be obvious but i'm not getting Why $x_n$ is a cauchy sequence??

Sahan Manodya
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Styles
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1 Answers1

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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.

Aweygan
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