# Cofinal Ordinal Relation is Transitive

## 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 $c \in z$.

$\blacksquare$