Axiom:Axiom of Continuity
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Axiom Schema
This is more properly an collection of axioms rather than a single axiom.
Let $a,b,x,y,$ be points.
Let $\mathsf{B}$ be the relation of betweenness.
Let $\alpha, \beta$ be first-order formulas.
Let $\alpha$ contain no free occurrences of $a,b,y$.
Let $\beta$ contain no free occurrences of $a,b,x$.
This axiom asserts:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\) | \(\displaystyle \exists a: \forall x,y: \left({ \alpha \land \beta \implies \mathsf{B}axy }\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\implies\) | \(\displaystyle \exists b: \forall x,y: \left({ \alpha \land \beta \implies \mathsf{B}xby }\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
Alternative form
Let $\mathsf{B}$ be the relation of betweenness.
This axiom asserts:
- $\left({\exists a : \forall x,y : \left({x \in X \land y \in Y}\right)\implies \mathsf{B}axy}\right) \implies \left({\exists b : \forall x,y : \left({x \in X \land y \in Y}\right)\implies \mathsf{B}xby}\right)$
where $a,b,x,y$ are points and $X, Y$ are point sets.
Note that this alternate form of the Axiom of Continuity assumes the existence of sets.
As such, it cannot be expressed in the framework of first-order logic.
Intuition
Consider an infinite line.
Consider it as consisting of a left side and a right side, where the transition is any place you care to choose.
No matter what place you choose, that place of transition is a point.
Also see
Sources
- June 1999: Alfred Tarski and Steven Givant: Tarski's System of Geometry (The Bulletin of Symbolic Logic Vol. 5, no. 2: 175 – 214) : Page 185 : Axiom $11$