Cofinal Ordinal Relation is Transitive

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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 $c \in z$.

$\blacksquare$


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