Completeness is the condition that lets infinite algebraic constructions converge inside the space. Most structural theorems below fail without it, so Banach spaces are the natural ambient category.
Complete normed spaces
The definition of a Banach space is simple, but its force is that every internally Cauchy approximation has an actual limit in the same space.
Definition. A Banach space is a normed space in which every Cauchy sequence converges in norm. Completeness is not a linear-algebraic property of the vector space alone; it belongs to the chosen norm.
Proposition. A closed linear subspace \(M\) of a Banach space \(X\) is Banach. Conversely, a Banach subspace \(M\subset X\), with the inherited norm, is closed in \(X\).
Proof. If \((m_n)\) is Cauchy in \(M\), it is Cauchy in \(X\), hence converges to some \(x\in X\). Closedness gives \(x\in M\). Conversely, if \(m_n\in M\) and \(m_n\to x\) in \(X\), then \((m_n)\) is Cauchy in \(M\). Completeness of \(M\) gives a limit \(m\in M\), and uniqueness of limits in \(X\) yields \(x=m\).
Absolute convergence criterion
Series are the main way Banach spaces are built inside proofs. Absolute convergence gives a convenient and very common completeness test.
Theorem. A normed space \(X\) is Banach if and only if every series \(\sum_{n=1}^\infty x_n\) with \(\sum_n\|x_n\|<\infty\) converges in \(X\).
Proof. If \(X\) is Banach and \(\sum\|x_n\|<\infty\), the partial sums \(s_N=\sum_{n\le N}x_n\) satisfy \(\|s_N-s_M\|\le\sum_{M<n\le N}\|x_n\|\), which tends to \(0\). Hence \(s_N\) converges.
Conversely, assume absolute convergence of series. Let \((y_n)\) be Cauchy. Choose a subsequence \(y_{n_k}\) such that \(\|y_{n_{k+1}}-y_{n_k}\|\le2^{-k}\). Then \(y_{n_1}+\sum_{k\ge1}(y_{n_{k+1}}-y_{n_k})\) converges by hypothesis, so \(y_{n_k}\) converges to some \(y\). Since the original sequence is Cauchy and has a convergent subsequence, the full sequence converges to \(y\).
Completion
Completion adds the missing limits while preserving the original metric geometry. It lets one prove results first on dense simple objects and then pass to limits.
Theorem. Every normed space \(X\) admits a completion: there are a Banach space \(\widehat X\) and an isometric linear map \(i:X\to\widehat X\) such that \(i(X)\) is dense in \(\widehat X\). The pair is unique up to unique surjective linear isometry fixing \(X\).
Proof. Let \(\mathcal C\) be the vector space of Cauchy sequences in \(X\), and let \(\mathcal N\subset\mathcal C\) be the subspace of sequences converging to \(0\). Define \(\widehat X=\mathcal C/\mathcal N\) and \(\|[(x_n)]\|=\lim_n\|x_n\|\), which exists because \((\|x_n\|)\) is Cauchy in \(\mathbb R\). This is a norm, and the map \(i(x)=[(x,x,\ldots)]\) is an isometry.
To prove completeness, take a Cauchy sequence of equivalence classes. Choose representatives successively so that a diagonal sequence approximates the classes within \(2^{-k}\). The diagonal representative is Cauchy in \(X\) and defines the limit in \(\widehat X\). Density follows because every class \([(x_n)]\) is the limit of the constant classes \(i(x_n)\). Uniqueness follows by extending the identity on \(X\) to the closure using uniform continuity of isometries.
Spaces of operators
Operator spaces are themselves normed spaces, and completeness of the target space is exactly what makes operator-norm limits meaningful.
Theorem. If \(Y\) is Banach and \(X\) is normed, then \(\mathcal L(X,Y)\) is Banach under the operator norm.
Proof. Let \((T_n)\) be Cauchy in operator norm. For each \(x\in X\), \(\|T_nx-T_mx\|\le\|T_n-T_m\|\|x\|\), so \((T_nx)\) is Cauchy in \(Y\); define \(Tx=\lim_nT_nx\). Linearity follows by passing limits through linear combinations.
Given \(\varepsilon>0\), choose \(N\) such that \(\|T_n-T_m\|<\varepsilon\) for \(m,n\ge N\). Letting \(m\to\infty\) pointwise gives \(\|(T_n-T)x\|\le\varepsilon\|x\|\), hence \(T_n\to T\) in operator norm and \(T\in\mathcal L(X,Y)\).
Neumann series
The Neumann series is the basic inverse construction. It is the functional-analytic version of summing a geometric series.
Theorem. Let \(X\) be Banach and \(A\in\mathcal L(X)\). If \(\|A\|<1\), then \(I-A\) is invertible in \(\mathcal L(X)\) and \((I-A)^{-1}=\sum_{n=0}^{\infty}A^n\). Moreover \(\|(I-A)^{-1}\|\le(1-\|A\|)^{-1}\).
Proof. The series is absolutely convergent in the Banach space \(\mathcal L(X)\) because \(\sum_n\|A^n\|\le\sum_n\|A\|^n<\infty\). Let \(S=\sum_{n\ge0}A^n\). Multiplying partial sums gives \((I-A)\sum_{n=0}^NA^n=I-A^{N+1}\) and the analogous right identity. Since \(\|A^{N+1}\|\le\|A\|^{N+1}\to0\), passing to the limit gives \((I-A)S=S(I-A)=I\). The norm estimate follows from the geometric series.
Banach quotients
Quotients keep track of information modulo a closed subspace. Closedness is what prevents nonzero cosets from having zero distance to the origin.
Proposition. If \(X\) is Banach and \(M\subset X\) is closed, then \(X/M\) is Banach.
Proof. Let \(\sum_n(x_n+M)\) be absolutely convergent in \(X/M\). Choose representatives \(y_n\in x_n+M\) such that \(\|y_n\|\le\|x_n+M\|+2^{-n}\varepsilon\). Then \(\sum_n\|y_n\|<\infty\), so \(\sum_n y_n\) converges in \(X\). Applying the quotient map gives convergence of \(\sum_n(x_n+M)\). The absolute convergence criterion proves completeness.