Completeness is not only about limits of sequences; it also prevents a space from being built out of too many thin pieces. Baire category is one of the deepest consequences of completeness.
Completion of a metric space
Every metric space can be embedded densely into a complete one.
Theorem. Every metric space \(X\) has a completion \(\widehat X\): an isometric embedding \(i:X\to\widehat X\) with dense image, where \(\widehat X\) is complete. It is unique up to unique isometry fixing \(X\).
Proof. Let \(\mathcal C\) be the Cauchy sequences in \(X\), and declare \((x_n)\sim(y_n)\) when \(d(x_n,y_n)\to0\). The triangle inequality proves transitivity. For two classes define \[ \widehat d([(x_n)],[(y_n)])=\lim_{n\to\infty}d(x_n,y_n). \] The limit exists because \[ |d(x_n,y_n)-d(x_m,y_m)| \le d(x_n,x_m)+d(y_n,y_m), \] and changing representatives does not change it. The metric axioms follow from those of \(d\).
The map \(i(x)=[(x,x,\ldots)]\) is isometric. Its image is dense because \([(x_n)]\) is approximated by \(i(x_N)\): for large \(N\), \[ \widehat d(i(x_N),[(x_n)])=\lim_{n\to\infty}d(x_N,x_n) \] is arbitrarily small. To prove completeness, let \(\xi_k\) be Cauchy in \(\widehat X\). Pass to a subsequence, still denoted \(\xi_k\), with \(\widehat d(\xi_{k+1},\xi_k)<2^{-k}\). By density choose \(z_k\in X\) such that \(\widehat d(i(z_k),\xi_k)<2^{-k}\). Then \((z_k)\) is Cauchy, so it defines \(\xi=[(z_k)]\), and the same estimates show \(\xi_k\to\xi\). The original Cauchy sequence has the same limit.
If \(j:X\to Y\) is another dense isometric embedding into a complete space, define \(U([(x_n)])=\lim j(x_n)\). This is well-defined, isometric, surjective by density, and satisfies \(U\circ i=j\). Density also makes such an isometry unique.
Contraction principle
Complete spaces make iterative fixed-point schemes converge.
Theorem. If \(X\) is complete and \(T:X\to X\) satisfies \(d(Tx,Ty)\le c d(x,y)\) for some \(0\lt c\lt 1\), then \(T\) has a unique fixed point and \(T^nx_0\) converges to it for every \(x_0\in X\).
Proof. Put \(x_{n+1}=Tx_n\). Iteration gives \[ d(x_{n+1},x_n)\le c^n d(x_1,x_0), \] and therefore, for \(m>n\), \[ d(x_m,x_n) \le\sum_{k=n}^{m-1}d(x_{k+1},x_k) \le\frac{c^n}{1-c}d(x_1,x_0). \] Thus \((x_n)\) is Cauchy and has a limit \(x^\ast\). A contraction is Lipschitz, hence continuous, so \(Tx^\ast=\lim Tx_n=\lim x_{n+1}=x^\ast\). If \(x^\ast,y^\ast\) are fixed, \(d(x^\ast,y^\ast)\le c\,d(x^\ast,y^\ast)\), forcing equality to zero. Letting \(m\to\infty\) in the tail estimate also gives the a priori bound \[ d(x_n,x^\ast)\le\frac{c^n}{1-c}d(x_1,x_0), \] while \(d(x_n,x^\ast)\le c^n d(x_0,x^\ast)\) is the corresponding a posteriori geometric rate.
Example (Picard iteration for an integral equation). On \(C([0,T])\) with the sup norm, let \[ (Tu)(t)=u_0+\int_0^t F(s,u(s))\,\mathrm{d}s, \] where \(|F(t,x)-F(t,y)|\le L|x-y|\). Then \[ \|Tu-Tv\|_\infty\le LT\|u-v\|_\infty. \] For \(LT<1\), the contraction theorem gives a unique solution of \(u=Tu\), and the successive approximations \(u_{n+1}=Tu_n\) converge geometrically. Longer time intervals are handled by iteration over short subintervals.
Nowhere dense and meagre sets
Category measures topological smallness rather than measure-theoretic size.
Definition. A set \(A\) is nowhere dense if \(\operatorname{int}\overline A=\varnothing\). A set is meagre if it is a countable union of nowhere dense sets. A residual set contains the complement of a meagre set.
Remark. A dense set can be meagre, as \(\mathbb Q\) is in \(\mathbb R\). Category and measure are different notions of largeness.
Baire category theorem
A complete metric space cannot be exhausted by countably many closed sets with empty interior.
Theorem. In a complete metric space, the intersection of countably many dense open sets is dense. Equivalently, a nonempty complete metric space is not meagre in itself.
Proof. Let \(G_n\) be dense open sets and let \(U\) be nonempty and open. Since \(U\cap G_1\) is nonempty and open, choose a closed ball \(\overline B(x_1,r_1)\subset U\cap G_1\) with \(r_1<1\). Inductively, the open set \(B(x_n,r_n)\cap G_{n+1}\) is nonempty, so choose \[ \overline B(x_{n+1},r_{n+1}) \subset B(x_n,r_n)\cap G_{n+1}, \qquad r_{n+1}<2^{-n}. \] The closed balls are nested. If \(m>n\), then \(x_m\in B(x_n,r_n)\), so \(d(x_m,x_n)<r_n\to0\); hence \((x_n)\) is Cauchy and converges to some \(x\). For each fixed \(n\), all \(x_m\) with \(m\ge n\) belong to \(\overline B(x_n,r_n)\), and this ball is closed, so \(x\in\overline B(x_n,r_n)\subset G_n\). The first ball also lies in \(U\). Thus \(x\in U\cap\bigcap_nG_n\), proving density.
From pointwise boundedness to a uniform bound
The basic category mechanism already contains the core of the uniform boundedness principle.
Theorem (uniform boundedness principle). Let \(X\) be a Banach space, \(Y\) a normed space, and \(\mathcal T\subset\mathcal L(X,Y)\). If \[ \sup_{T\in\mathcal T}\|Tx\|<\infty \qquad\text{for every }x\in X, \] then \(\sup_{T\in\mathcal T}\|T\|<\infty\).
Proof. Define \[ E_m=\left\{x\in X:\sup_{T\in\mathcal T}\|Tx\|\le m\right\}. \] Each \(E_m\) is closed, and pointwise boundedness says \(X=\bigcup_{m\ge1}E_m\). By Baire, some \(E_m\) contains a ball \(B(x_0,r)\). If \(\|h\|<r\), then both \(x_0\) and \(x_0+h\) lie in the corresponding closed ball after decreasing \(r\) if necessary, and \[ \|Th\|\le\|T(x_0+h)\|+\|Tx_0\|\le2m \] for every \(T\in\mathcal T\). By homogeneity, for \(x\ne0\), apply this to \(h=(r/2)x/\|x\|\) to obtain \[ \|Tx\|\le\frac{4m}{r}\|x\|. \] The bound is independent of \(T\), as required.
Remark (category and measure do not dominate one another). The rationals are dense and meagre in \(\mathbb R\), while a fat Cantor set can be nowhere dense and have positive Lebesgue measure. Conversely, a dense \(G_\delta\) set can have measure zero. “Typical” in the Baire sense and “almost every” in the measure sense answer different questions.