Cofinal Ordinal Relation is Transitive
Jump to navigation
Jump to search
Theorem
Let $x$, $y$, and $z$ be ordinals.
Let $\operatorname{cof}$ denote the cofinal relation.
Then:
- $\operatorname{cof} \left({ x,y }\right) \land \operatorname{cof} \left({ y,z }\right) \implies \operatorname{cof} \left({ x,z }\right)$
Proof
The conditions for $z$ being cofinal with $x$ shall be verified individually.
Let $\le$ denote the subset relation.
Since $x \le y$ and $y \le z$, it follows that $x \le z$ by Subset Relation is Transitive
Let $f : y \to x$ be the strictly increasing mapping that satisfies:
- $\forall a \in x: \exists b \in y: f \left({b}\right) \ge a$
Let $g : z \to y$ be the strictly increasing mapping that satisfies:
- $\forall a \in y: \exists b \in z: g \left({b}\right) \ge a$
Then $f \circ g : z \to x$ is a strictly increasing mapping.
Moreover, take any $a \in x$.
There is some $b \in y$ such that $f \left({b}\right) \ge a$.
But $b \in y$, so there is some $c \in z$ such that:
- $g \left({c}\right) \ge b$.
\(\displaystyle f\left({g \left({c}\right)}\right)\) | \(\ge\) | \(\displaystyle f \left({b}\right)\) | $f$ is strictly increasing | ||||||||||
\(\displaystyle \) | \(\ge\) | \(\displaystyle a\) | Inequality above |
Therefore, $f \circ g \left({c}\right) \ge a$ for some $c \in z$.
$\blacksquare$
Sources
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 10.52 \ (2)$