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



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 $c \in z$.

$\blacksquare$


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