Suppose \(X\) is a Banach space and \(U\) a closed subspace. Then \(U\) is complete .
Proof
Let \((u_n)_n\) be a Cauchy sequence
in \(U\). Since \(X\) is a Banach space \((u_n)_n\) converges as a sequence in \(X\), i.e. there is a \(x\in X\) such that \(u_n\to x\). But \(x\in U\), because \(U\) is closed (see (0x677c3162)
) and therefore \(U\) is complete.