# Tarski's axiomatization of the reals

In 1936, Alfred Tarski set out an axiomatization of the real numbers and their arithmetic, consisting of only the 8 axioms shown below and a mere four primitive notions:[1] the set of reals denoted R, a binary total order over R, denoted by infix <, a binary operation of addition over R, denoted by infix +, and the constant 1.

The literature occasionally mentions this axiomatization but never goes into detail, notwithstanding its economy and elegant metamathematical properties. This axiomatization appears little known, possibly because of its second-order nature. Tarski's axiomatization can be seen as a version of the more usual definition of real numbers as the unique Dedekind-complete ordered field; it is however made much more concise by using unorthodox variants of standard algebraic axioms and other subtle tricks (see e.g. axioms 4 and 5, which combine the usual four axioms of abelian groups).

The term "Tarski's axiomatization of real numbers" also refers to the theory of real closed fields, which Tarski showed completely axiomatizes the first-order theory of the structure 〈R, +, ·, <〉.

## The axioms

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 XY ⊆ 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.

## How these axioms imply a field

Theorem — ${\displaystyle \mathbb {R} }$ is a Archimedean ordered abelian group.

Proof

Tarski's axioms imply that ${\displaystyle \mathbb {R} }$ is a totally ordered[2] abelian group under addition with distinguished element ${\displaystyle 1}$. Since ${\displaystyle \mathbb {R} }$ is Dedekind-complete and a totally ordered abelian group, ${\displaystyle \mathbb {R} }$ is Archimedean, because the Dedekind-completion of any totally ordered abelian group with infinite elements or infinitesimals is not an abelian group, and the Dedekind-completion of any Archimedean ordered abelian group is still Archimedean.

Theorem — ${\displaystyle \mathbb {R} }$ has a complete metric

Proof

Tarski's axioms imply that ${\displaystyle \mathbb {R} }$ is a totally ordered abelian group under addition with distinguished element ${\displaystyle 1}$. As a result, there exist maximum and minimum binary functions ${\displaystyle \max :\mathbb {R} \times \mathbb {R} \to \mathbb {R} }$ and ${\displaystyle \min :\mathbb {R} \times \mathbb {R} \to \mathbb {R} }$, with the absolute value function defined as ${\displaystyle \vert x\vert =\max(x,-x)}$.

Since ${\displaystyle \mathbb {R} }$ is Dedekind-complete, Archimedean, and a totally ordered abelian group, ${\displaystyle \mathbb {R} }$ is a metric space with respect to the absolute value ${\displaystyle \vert x\vert }$ and thus a Hausdorff space, and every Cauchy net in ${\displaystyle \mathbb {R} }$ converges to a unique element of ${\displaystyle \mathbb {R} }$, and thus the absolute value ${\displaystyle \vert x\vert }$ is a complete metric on ${\displaystyle \mathbb {R} }$.

Theorem — ${\displaystyle \mathbb {Q} }$ embeds in ${\displaystyle \mathbb {R} }$.

Proof

Since ${\displaystyle \mathbb {R} }$ is an abelian group, it is a ${\displaystyle \mathbb {Z} }$-module, and since ${\displaystyle \mathbb {R} }$ is totally ordered, it is a torsion-free module and thus a torsion-free abelian group, which means that the integers ${\displaystyle \mathbb {Z} }$ embed in ${\displaystyle \mathbb {R} }$, with injective group homomorphism ${\displaystyle f:\mathbb {Z} \to \mathbb {R} }$ where ${\displaystyle f(0)=0}$ and ${\displaystyle f(1)=1}$. As a result, for every integer ${\displaystyle a\in \mathbb {Z} }$ and ${\displaystyle b\in \mathbb {Z} }$ the affine functions ${\displaystyle x\mapsto ax+b}$ are well defined in ${\displaystyle \mathbb {R} }$.

Since ${\displaystyle \mathbb {R} }$ is Dedekind-complete, Archimedean, and a totally ordered abelian group, any closed interval ${\displaystyle [a,b]}$ on ${\displaystyle \mathbb {R} }$ is compact and connected. Since ${\displaystyle \mathbb {R} }$ is also a complete metric space, the intermediate value theorem is satisfied for every function from a closed interval ${\displaystyle [a,b]}$ to ${\displaystyle \mathbb {R} }$. Because ${\displaystyle x\mapsto ax+b}$ are monotonic for ${\displaystyle a>0}$, and for ${\displaystyle a<0}$ the function is just the negation of a monotonic function, ${\displaystyle x\mapsto ax+b}$ have a root for ${\displaystyle \vert a\vert >0}$. Thus ${\displaystyle \mathbb {R} }$ is a divisible group and a ${\displaystyle \mathbb {Q} }$-vector space, with an injective group homomorphism ${\displaystyle f:\mathbb {Q} \to \mathbb {R} }$ where ${\displaystyle f(0)=0}$ and ${\displaystyle f(1)=1}$.

Theorem — ${\displaystyle \mathbb {R} }$ is a commutative ring.

Proof

Since every Cauchy net in ${\displaystyle \mathbb {R} }$ converges to a unique element of ${\displaystyle \mathbb {R} }$, for every directed set ${\displaystyle A}$ and Cauchy net ${\displaystyle (a_{i})_{i\in A}}$ in the rational numbers, there exists a Cauchy net of linear functions ${\displaystyle (f_{i})_{i\in A}}$ defined as ${\displaystyle f_{i}(x)=a_{i}x}$. The limit of the Cauchy net ${\displaystyle \lim _{i\in A}(f_{i})_{i}}$ exists and is a unique function ${\displaystyle g(x)=\lim _{i\in A}(a_{i})_{i}x}$. Since every real number is the limit of a Cauchy net of rational numbers, there is an ${\displaystyle \mathbb {R} }$-action ${\displaystyle \mu :\mathbb {R} \to (\mathbb {R} \to \mathbb {R} )}$ which takes a real number ${\displaystyle r}$ to the linear function ${\displaystyle x\mapsto rx}$, with ${\displaystyle \alpha (1)=x\mapsto x}$ being the idenitity function. The uncurrying of ${\displaystyle \alpha }$ leads to a bilinear function ${\displaystyle (-)(-):\mathbb {R} \times \mathbb {R} \to \mathbb {R} }$ called multiplication of the real numbers, defined on the entire domain of the binary function. Since linear functions in the function space with function composition and the identity function is a commutative monoid, ${\displaystyle \mathbb {R} }$ multiplication and the multiplicative identity element ${\displaystyle 1}$ is also commutative monoid, which means that ${\displaystyle \mathbb {R} }$ is a commutative ring.

Theorem — ${\displaystyle \mathbb {R} }$ is a field

Proof

Since ${\displaystyle \mathbb {R} }$ is a commutative ring, power series are well defined, and because all Cauchy nets converge in ${\displaystyle \mathbb {R} }$, all Cauchy sequences and all Cauchy power series converge in ${\displaystyle \mathbb {R} }$. In particular, every geometric series is a Cauchy power series and the limit of the geometric series ${\displaystyle \sum _{n=0}^{\infty }x^{n}}$ and ${\displaystyle \sum _{n=0}^{\infty }(-1)^{n}x^{n}}$ converges in the open interval ${\displaystyle (-1,1)}$. Thus let us define functions ${\displaystyle f:(-1,1)\to \mathbb {R} }$ and ${\displaystyle g:(-1,1)\to \mathbb {R} }$ as

${\displaystyle f(x):=\sum _{n=0}^{\infty }x^{n}}$

${\displaystyle g(x):=\sum _{n=0}^{\infty }(-1)^{n}x^{n}}$

Let us define the function

${\displaystyle h(x,a):=(-a)\sum _{n=0}^{\infty }a^{n}(x+f(a+1))^{n}}$

for ${\displaystyle a<0}$ and

${\displaystyle k(x,a):=a\sum _{n=0}^{\infty }(-a)^{n}(x+g(a-1))^{n}}$

for ${\displaystyle a>0}$. These are functions which converge on the open interval ${\displaystyle (1/a,0)}$ for ${\displaystyle h(x,a)}$ and ${\displaystyle (0,1/a)}$ for ${\displaystyle k(x,a)}$, and satisfy the identity ${\displaystyle h(x,a)x=1}$ for all ${\displaystyle a<0}$ and ${\displaystyle x\in (1/a,0)}$, and ${\displaystyle k(x,a)x=1}$ for all ${\displaystyle a>0}$ and ${\displaystyle x\in (0,1/a)}$, by definition of the geometric series.

The reciprocal function is piecewise defined as ${\displaystyle {\frac {1}{x}}={\begin{cases}\lim _{a\to 0^{-}}h(x,a)&{\text{if }}x<0\\\lim _{a\to 0^{+}}k(x,a)&{\text{if }}x>0\\\end{cases}}}$

As limits preserve multiplication, ${\displaystyle {\frac {1}{x}}x=1}$. Thus, ${\displaystyle \mathbb {R} }$ is a field.

Otto Hölder showed that every Archimedean group is isomorphic (as an ordered group) to a subgroup of the Dedekind-complete Archimedean group with distinguished element ${\displaystyle 1>0}$, ${\displaystyle \mathbb {R} }$.[3][4][5][6] Because ${\displaystyle \mathbb {Q} }$ is an Archimedean ordered field, let us define ${\displaystyle \mathbb {R} ^{'}}$ as the Dedekind completion of ${\displaystyle \mathbb {Q} }$. The Dedekind completion of any Archimedean ordered field is terminal in the concrete category of Dedekind complete Archimedean ordered fields,[7] Because ${\displaystyle \mathbb {R} ^{'}}$ is a Dedekind-complete Archimedean ordered field, every Archimedean group embeds into ${\displaystyle \mathbb {R} ^{'}}$ as well. As a result, the two sets ${\displaystyle \mathbb {R} ^{'}}$ and ${\displaystyle \mathbb {R} }$ are isomorphic to each other, which means that ${\displaystyle \mathbb {R} }$ is a field.

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