Tarski's axiomatization of the reals

Axioms of order (primitives: R, <):

Axiom 1

If x < y, then not y < x. That is, "<" is an asymmetric relation. This implies that "<" is not a reflexive relationship, i.e. for all x, x < x is false.

Axiom 2

If x < z, there exists a y such that x < y and y < z. In other words, "<" is dense in R.

Axiom 3

"<" is Dedekind-complete. More formally, for all X, Y ⊆ R, if for all x ∈ X and y ∈ Y, x < y, then there exists a z such that for all x ∈ X and y ∈ Y, if z ≠ x and z ≠ y, then x < z and z < y.

To clarify the above statement somewhat, let X ⊆ R and Y ⊆ R. We now define two common English verbs in a particular way that suits our purpose:

X precedes Y if and only if for every x ∈ X and every y ∈ Y, x < y.

The real number z separates X and Y if and only if for every x ∈ X with x ≠ z and every y ∈ Y with y ≠ z, x < z and z < y.

Axiom 3 can then be stated as:

"If a set of reals precedes another set of reals, then there exists at least one real number separating the two sets."

The three axioms imply that R is a linear continuum.

Axioms of addition (primitives: R, <, +):

Axiom 4

x + (y + z) = (x + z) + y.

Axiom 5

For all x, y, there exists a z such that x + z = y.

Axiom 6

If x + y < z + w, then x < z or y < w.

Axioms for one (primitives: R, <, +, 1):

Axiom 7

1 ∈ R.

Axiom 8

1 < 1 + 1.

This axiomatization does not give rise to a first-order theory, because the formal statement of axiom 3 includes two universal quantifiers over all possible subsets of R. Tarski proved that these 8 axioms and 4 primitive notions are independent.

These axioms imply that R is a totally ordered[2] abelian group under addition with distinguished element 1. R is also Dedekind-complete, divisible, and Archimedean. The submonoid of R generated by {1} is isomorphic to the free monoid on one generator, which is the natural numbers ${\displaystyle \mathbb {N} }$. From there, one could recursively define multiplication as a partial binary operation restricted on the domain to ${\displaystyle \mathbb {N} }$, then construct ${\displaystyle \mathbb {Z} }$ and ${\displaystyle \mathbb {Q} }$ from ${\displaystyle \mathbb {N} }$ and show that ${\displaystyle \mathbb {Z} }$ is a commutative ring and ${\displaystyle \mathbb {Q} }$ is a Archimedean ordered field, and that the subgroup generated by {1} in R is isomorphic to ${\displaystyle \mathbb {Z} }$ and the divisible subgroup generated by {1} is isomorphic to ${\displaystyle \mathbb {Q} }$. Since R is a strictly ordered divisible group, R is a torsion-free group and thus a ${\displaystyle \mathbb {Q} }$-vector space, and left and right scalar multiplication by elements of ${\displaystyle \mathbb {Q} }$ is possible in R.

Because ${\displaystyle \mathbb {Q} }$ is an Archimedean ordered field, and the Dedekind completion of any Archimedean ordered field is terminal in the concrete category of Dedekind complete Archimedean ordered fields,[3] one could construct the Dedekind completion of ${\displaystyle \mathbb {Q} }$, and then could define the field operations on the Dedekind completion of ${\displaystyle \mathbb {Q} }$ in the usual way, and the fact that the Dedekind completion of ${\displaystyle \mathbb {Q} }$ is terminal shows that R is a field.[3]

Tarski stated, without proof, that these axioms gave a total ordering. The missing component was supplied in 2008 by Stefanie Ucsnay.[2]