analysis (differential/integral calculus, functional analysis, topology)
metric space, normed vector space
open ball, open subset, neighbourhood
convergence, limit of a sequence
compactness, sequential compactness
continuous metric space valued function on compact metric space is uniformly continuous
…
…
A mathematical structure used to define the real numbers in Alfred Tarski‘s axioms for the real numbers.
A Tarski group is a pointed commutative invertible semigroup $(G, +, -, 1)$ with a dense linear order $\lt$ such that $1 \lt 1 + 1$, $a \lt b + c + a$ implies $a \lt b + a$ or $a \lt c + a$, and $a \lt b$ implies $c + a \lt c + b$.
As a result, every Tarski group is an abelian group with identity element $0 \coloneqq 1 - 1$, and a nontrivial ordered group.
Tarski’s axioms for the real numbers are as follows:
Axiom 5 says that $\mathbb{R}$ is a commutative semigroup?.
Axiom 6 with axiom 5 together say that $\mathbb{R}$ is a commutative invertible semigroup.
Axiom 8 says that $\mathbb{R}$ is a pointed set, which with axioms 5 and 6 imply that $\mathbb{R}$ is an abelian group.
Axiom 1 says that $\mathbb{R}$ has a connected relation $\lt$.
Axiom 2 says that $\lt$ is an asymmetric relation and thus an irreflexive relation.
The fragment of Axiom 4 that only refers to singleton subsets says that $\lt$ is a comparison, making $\lt$ into a linear order. (the full axiom 4 is the Dedekind completeness condition).
Axiom 3 says that $\lt$ is a dense linear order.
Axiom 7 says that for all $a, b, c, d \in G$, $a + b \lt c + d$ implies that $a \lt c$ or $b \lt d$, which imply that $\mathbb{R}$ is a linearly ordered group.
Axiom 9 says that $1 \lt 1 + 1$, which indicates that $\mathbb{R}$ is not trivial.
Let us denote the Dedekind-complete Tarski group as $\mathbb{R}$. There is an archimedean field structure on $\mathbb{R}$.
$\mathbb{R}$ is a Archimedean ordered abelian group.
Since $\mathbb{R}$ is Dedekind-complete and a strictly ordered abelian group, $\mathbb{R}$ 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.
$\mathbb{R}$ has a complete metric
Since $\mathbb{R}$ is strictly ordered, it is a totally ordered abelian group. As a result, there exist maximum and minimum binary functions $\max:\mathbb{R} \times \mathbb{R} \to \mathbb{R}$ and $\min:\mathbb{R} \times \mathbb{R} \to \mathbb{R}$, with the absolute value function defined as $\vert x \vert = \max(x, -x)$.
Since $\mathbb{R}$ is Dedekind-complete, Archimedean, and a totally ordered abelian group, $\mathbb{R}$ is a metric space with respect to the absolute value $\vert x \vert$ and thus a Hausdorff space, and every Cauchy net in $\mathbb{R}$ converges to a unique element of $\mathbb{R}$, and thus the absolute value $\vert x \vert$ is a complete metric on $\mathbb{R}$.
$\mathbb{Q}$ embeds in $\mathbb{R}$.
Since $\mathbb{R}$ is an abelian group, it is a $\mathbb{Z}$-module, and since $\mathbb{R}$ is totally ordered, it is a torsion-free module and thus a torsion-free abelian group, which means that the integers $\mathbb{Z}$ embed in $\mathbb{R}$, with injective group homomorphism $f:\mathbb{Z} \to \mathbb{R}$ where $f(0) = 0$ and $f(1) = 1$. As a result, for every integer $a \in \mathbb{Z}$ and $b \in \mathbb{Z}$ the affine functions $x \mapsto a x + b$ are well defined in $\mathbb{R}$.
Since $\mathbb{R}$ is Dedekind-complete, Archimedean, and a totally ordered abelian group, any closed interval $[a, b]$ on $\mathbb{R}$ is compact and conencted. Since $\mathbb{R}$ is also a complete metric space, the intermediate value theorem is satisfied for every function from a closed interval $[a, b]$ to $\mathbb{R}$. Because $x \mapsto a x + b$ are monotonic for $a \gt 0$, and for $a \lt 0$ the function is just the negation of a monotonic function, $x \mapsto a x + b$ have a root? for $\vert a \vert \gt 0$. Thus $\mathbb{R}$ is a divisible group and a $\mathbb{Q}$-vector space, with an injective group homomorphism $f:\mathbb{Q} \to \mathbb{R}$ where $f(0) = 0$ and $f(1) = 1$, and $\mathbb{Q}$ embeds in $\mathbb{R}$.
$\mathbb{R}$ is a commutative ring.
Since every Cauchy net in $\mathbb{R}$ converges to a unique element of $\mathbb{R}$, for every directed set $A$ and Cauchy net $(a_i)_{i \in A}$ in the rational numbers, there exists a Cauchy net of linear functions $(f_i)_{i \in A}$ defined as $f_i(x) = a_i x$. The limit of the Cauchy net $\lim_{i \in A} (f_i)_i$ exists and is a unique function $g(x) = \lim_{i \in A} (a_i)_i x$. Since every real number is the limit of a Cauchy net of rational numbers, there is an $\mathbb{R}$-action $\mu:\mathbb{R} \to (\mathbb{R} \to \mathbb{R})$ which takes a real number $r$ to the linear function $x \mapsto r x$, with $\alpha(1) = x \mapsto x$ being the identity function. The uncurrying of $\alpha$ leads to a bilinear function $(-)(-):\mathbb{R} \times \mathbb{R} \to \mathbb{R}$ 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, $\mathbb{R}$ with multiplication and the multiplicative identity element $1$ is also commutative monoid, which means that $\mathbb{R}$ is a commutative ring.
$\mathbb{R}$ is a field
Since $\mathbb{R}$ is a commutative ring, power series are well defined, and because all Cauchy nets converge in $\mathbb{R}$, all Cauchy sequences and all Cauchy power series converge in $\mathbb{R}$. In particular, every geometric series is a Cauchy power series and the limit of the geometric series
and
converges in the open interval $(-1, 1)$. Thus let us define functions $f:(-1, 1) \to \mathbb{R}$ and $g:(-1, 1) \to \mathbb{R}$ as
Let us define the function
for $a \lt 0$ and
for $a \gt 0$. These are functions which converge on the open interval $(1/a, 0)$for $h(x, a)$ and $(0,1/a)$ for $k(x, a)$, and satisfy the identity $h(x, a) x = 1$ for all $a \lt 0$ and $x \in (1/a, 0)$, and $k(x, a) x = 1$ for all $a \gt 0$ and $x \in (0,1/a)$, by definition of the geometric series.
The reciprocal is piecewise defined as
As limits preserve multiplication, $\frac{1}{x} x = 1$ for all $x \in \mathbb{R}$. Thus, $\mathbb{R}$ is a field.
Alfred Tarski, Introduction to Logic and to the Methodology of Deductive Sciences (4th edition). Oxford University Press. (1994) $[$doi:10.2307/2180610, ISBN 978-0-19-504472-0$]$
Ucsnay, Stefanie (Jan 2008), A Note on Tarski’s Note. The American Mathematical Monthly, Vol 115 No. 1, pg 66–68. JSTOR 27642393
Last revised on May 19, 2022 at 10:06:03. See the history of this page for a list of all contributions to it.