# 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:

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

## 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*: $\S 10.52 \ (2)$