Ordered-field algebra lets one manipulate inequalities; completeness lets one take limits of bounded approximations. Real analysis is the study of what becomes possible when both structures coexist.
Order-compatible field operations
The order must interact with addition and multiplication strongly enough to justify algebraic inequality arguments.
Definition. An ordered field is a field \(F\) with a total order \(\le\) such that \(x\le y\Rightarrow x+z\le y+z\), and \(0\le x,0\le y\Rightarrow 0\le xy\).
Proposition. In an ordered field, \(x^2\ge0\), \(1>0\), multiplication by a positive element preserves inequalities, and taking reciprocals reverses inequalities among positive elements.
Proof. Since either \(x\ge0\) or \(-x\ge0\), the product rule gives \(x^2\ge0\). If \(1\lt 0\), then \((-1)(-1)>0\) gives \(1>0\), contradiction, so \(1>0\). The remaining claims follow by adding \(-ay\) or comparing \(1/x\) and \(1/y\) after multiplying by the positive product \(xy\).
Least upper bounds
Suprema are the formal language for saying that a bounded increasing approximation has an endpoint.
Definition. If \(A\subset F\), an upper bound is \(u\in F\) with \(a\le u\) for all \(a\in A\). A supremum \(\sup A\) is an upper bound \(s\) such that every upper bound \(u\) satisfies \(s\le u\).
Lemma. If \(s=\sup A\), then for every \(\varepsilon>0\) there is \(a\in A\) with \(s-\varepsilon\lt a\le s\).
Proof. If no such \(a\) existed, then \(s-\varepsilon\) would still be an upper bound for \(A\), contradicting the minimality of \(s\) among upper bounds.
Archimedean order
Completeness rules out real infinitesimals and infinitely large real numbers.
Theorem. Every complete ordered field is Archimedean: for every \(x\in F\), there exists \(n\in\mathbb N\) such that \(x\lt n\).
Proof. If \(\mathbb N\) were bounded above, let \(s=\sup\mathbb N\). By the approximation property for suprema, there is \(n\in\mathbb N\) with \(s-1\lt n\). Then \(s\lt n+1\), contradicting that \(s\) is an upper bound for \(\mathbb N\).
Density of rational numbers
Archimedean order makes rational grids arbitrarily fine.
Proposition. If \(x\lt y\) in a complete ordered field containing \(\mathbb Q\), then there is \(q\in\mathbb Q\) with \(x\lt q\lt y\).
Proof. Choose \(n\in\mathbb N\) so large that \(1/n\lt y-x\). By Archimedeanness choose \(m\in\mathbb Z\) with \(m-1\le nx\lt m\). Then \(x\lt m/n\le x+1/n\lt y\).
Equivalent faces of completeness
Different areas of analysis prefer different formulations of the same no-gap principle.
Theorem. On an Archimedean ordered field, the least-upper-bound property, monotone convergence for bounded monotone sequences, Cauchy completeness, nested closed interval completeness, and Bolzano-Weierstrass are equivalent.
Proof. Suppose first that least upper bounds exist. If \(x_n\) is increasing and bounded, put \(s=\sup_nx_n\). For every \(\varepsilon>0\), some \(x_N>s-\varepsilon\), and monotonicity then traps the tail in \((s-\varepsilon,s]\).
For nested intervals \(I_n=[a_n,b_n]\), the left endpoints increase and are bounded above. Their limit \(x=\sup_na_n\) satisfies \(a_m\le x\le b_m\) for every \(m\), so \(x\in\cap_mI_m\). If the lengths tend to zero, this point is unique.
Given a Cauchy sequence, choose indices \(N_k\) so that \(|x_m-x_n|<2^{-k}\) for \(m,n\ge N_k\). The closed intervals \[ I_k=\overline{\{x_n:n\ge N_k\}} \] may be replaced by nested closed intervals of length at most \(2^{1-k}\) containing the corresponding tails. Their unique common point is the Cauchy limit.
For Bolzano-Weierstrass, bisect a bounded interval containing the sequence and retain a half containing infinitely many terms. The resulting nested intervals have lengths tending to zero; choosing successively indexed terms from them produces a convergent subsequence.
Finally assume Bolzano-Weierstrass and let \(A\ne\varnothing\) be bounded above. Archimedeanness gives, for each \(n\), a grid point \(k_n/2^n\) that is an upper bound for \(A\) while \((k_n-1)/2^n\) is not. These upper bounds form a bounded sequence and have a convergent subsequence with limit \(s\). For every \(a\in A\), \(a\le k_n/2^n\), hence \(a\le s\); and every \(u<s\) is eventually below \((k_n-1)/2^n\), so it cannot be an upper bound. Thus \(s=\sup A\).