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How to prove that a closed subspace of a Banach space is a Banach space?

A subspace is closed if it contains all of its limit points. But in the proof of the above question how can use this idea to get a Cauchy sequence and show that it is convergent in the subspace?

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  • $\begingroup$ You don't need to get a Cauchy sequence. You just assume a sequence is Cauchy and prove it converges and the limit point is in the subspace. $\endgroup$ – Hui Yu Nov 19 '12 at 16:15
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Note: you don't need to "get a Cauchy sequence" -- you assume you have one and then you need to show that its limit is in the subspace.

Let $C$ be a closed subspace of a Banach space $B$, let $x_n$ be a Cauchy sequence in $C$ (with respect to the norm on $B$). Then the limit $x$ of $x_n$ exists in $B$ since $B$ is complete. Also, if it is not already in $C$, it must be a limit point of $C$. But $C$ is closed and hence contains all its limit points. Hence $C$ is complete, too.

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In general, a closed subset of a complete metric space is also a complete metric space. In this case, the metric is given by the prescribed norm on the given Banach space. Hence, a closed subspace of a Banach space is a normed vector space that is complete with respect to the metric induced by the norm. By definition, this makes it a Banach space.

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Let Y be closed subspace of banach space X. Now consider a sequence(cauchy) in Y. As X is banach the series converges in X. But Y contains all of it limit points. By uniqueness of limits we'll get for any cauchy sequence in Y, it is convergent in Y and so it is Banach space.

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