Real analysis begins before limits: it begins with the algebraic and order structures that make limiting arguments meaningful. The natural numbers provide induction, the integers provide additive inverses, and the rationals provide the first ordered field, but the rationals still have gaps.
Induction and recursion on natural numbers
Induction is the proof principle; recursion is the construction principle. Together they make definitions by finite iteration legitimate.
Definition. The natural numbers are a set \(\mathbb N\) with a distinguished element \(0\) and successor map \(S\) satisfying the Peano axioms: \(0\) is not a successor, \(S\) is injective, and every set containing \(0\) and closed under \(S\) is all of \(\mathbb N\).
Theorem. If \(A\subseteq\mathbb N\), \(0\in A\), and \(n\in A\Rightarrow S(n)\in A\), then \(A=\mathbb N\). Equivalently, if a statement \(P(n)\) holds at \(0\) and \(P(n)\Rightarrow P(n+1)\), then \(P(n)\) holds for all \(n\).
Proof. Apply the Peano induction axiom to the set \(A=\{n:P(n)\text{ is true}\}\). The hypotheses say exactly that \(A\) contains \(0\) and is closed under successor, hence \(A=\mathbb N\).
Integers and cancellation
Passing from \(\mathbb N\) to \(\mathbb Z\) is the algebraic operation of adding formal additive inverses.
Proposition. The integers may be constructed as equivalence classes of pairs \((a,b)\in\mathbb N^2\), with \((a,b)\sim(c,d)\) if \(a+d=b+c\). The class of \((a,b)\) represents \(a-b\), and addition and multiplication are well-defined.
Proof. The relation is reflexive, symmetric, and transitive by associativity and commutativity in \(\mathbb N\). Define \((a,b)+(c,d)=(a+c,b+d)\) and \((a,b)(c,d)=(ac+bd,ad+bc)\). If representatives are changed by the relation, the defining equalities preserve these formulas after expanding both sides, so the operations descend to equivalence classes.
Rationals as the smallest ordered field
The rationals add multiplicative inverses for nonzero integers while preserving the order inherited from \(\mathbb Z\).
Definition. The rational numbers are equivalence classes of pairs \((m,n)\in\mathbb Z\times(\mathbb Z\setminus\{0\})\), with \((m,n)\sim(p,q)\) if \(mq=np\). The class of \((m,n)\) is written \(m/n\).
Proposition. \(\mathbb Q\) is an ordered field, and for any ordered field \(F\), there is a unique order-preserving field embedding \(\mathbb Q\hookrightarrow F\).
Proof. The field operations are inherited from cross-multiplication. In an ordered field \(F\), the image of \(n\in\mathbb N\) is \(1+\cdots+1\), and the images of integers and quotients are forced by additive and multiplicative inverses. Positivity is preserved because squares and sums of positive elements are positive.
Density and countability of rational points
The rationals are algebraically rich and order-dense, yet still too small for analysis.
Theorem. \(\mathbb Q\) is countable. Once it is embedded in an Archimedean ordered field \(F\), it is order-dense: if \(a,b\in F\) and \(a\lt b\), then some \(q\in\mathbb Q\) satisfies \(a\lt q\lt b\).
Proof. For every \(N\in\mathbb N\), there are only finitely many integer pairs \((m,n)\) with \(n\ne0\) and \(|m|+|n|=N\). Listing these finite sets successively maps \(\mathbb N\) onto \(\mathbb Q\), so \(\mathbb Q\) is countable. For density, choose \(n\in\mathbb N\) with \(n(b-a)>1\) by the Archimedean property. The set \(\{k\in\mathbb Z:k>na\}\) has a least element \(m\); then \(m-1\le na<m\). Hence \(m\le na+1<nb\), and \(q=m/n\) lies strictly between \(a\) and \(b\).
The first gap
The rationals already support limits of many sequences, but not all Cauchy sequences converge rationally.
Example. The set \(\{q\in\mathbb Q:q^2\lt 2\}\) is nonempty and bounded above in \(\mathbb Q\), but has no rational least upper bound. This is the gap repaired by Dedekind cuts or Cauchy completion.
Why no rational supremum exists. Suppose a positive rational \(s\) were the supremum. It cannot satisfy \(s^2=2\), since the usual parity argument proves that \(\sqrt2\) is irrational. If \(s^2<2\), then for sufficiently small rational \(\varepsilon>0\), \((s+\varepsilon)^2<2\), contradicting that \(s\) is an upper bound. If \(s^2>2\), then for sufficiently small rational \(\varepsilon>0\), \((s-\varepsilon)^2>2\); every \(q\) with \(q^2<2\) must then satisfy \(q<s-\varepsilon\), contradicting that \(s\) is the least upper bound.