Cofinal Ordinal Relation is Transitive

Theorem

Let $x$, $y$, and $z$ be ordinals.

Let $\operatorname {cof}$ denote the cofinal relation.

Then:

$\map {\operatorname {cof} } {x, y} \land \map {\operatorname {cof} } {y, z} \implies \map {\operatorname {cof} } {x, z}$

Proof

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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:

$\forall a \in x: \exists b \in y: \map f b \ge a$

Let $g : z \to y$ be the strictly increasing mapping that satisfies:

$\forall a \in y: \exists b \in z: \map g b \ge a$

By Composite of Strictly Increasing Mappings is Strictly Increasing:

$f \circ g$ is a strictly increasing mapping.

Moreover, take any $a \in x$.

There is some $b \in y$ such that:

$\map f b \ge a$

But $b \in y$, so there is some $c \in z$ such that:

$\map g c \ge b$

\(\ds \map f {\map g c}\)

\(\ge\)

\(\ds \map f b\)

$f$ is strictly increasing

\(\ds \)

\(\ge\)

\(\ds a\)

Inequality above

Therefore:

$\map {f \circ g} c \ge a$

for some $c \in z$.

$\blacksquare$

Sources

• 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 10.52 \ (2)$