Cofinal Limit Ordinals

Theorem
Let $x$ and $y$ be ordinals.

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

Let $K_{II}$ denote the class of all limit ordinals.

Then:


 * $\map {\mathrm {cof} } {x, y} \implies \paren {x \in K_{II} \iff y \in K_{II} }$

Necessary Condition
Suppose $y \in K_{II}$.

$x \ne 0$ by Cofinal to Zero iff Ordinal is Zero.

If $x = z^+$ for some $z$, then $z = \bigcup x$ by Union of Successor Ordinal.

Therefore, $z$ would be the least upper bound of $x$.

Since $\map {\mathrm {cof} } {x, y}$, it follows by the definition of cofinal that:
 * $\forall a \in x: \exists b \in y: \map f b \ge a$

So $\map f b \ge z$ for some $b \in y$.

But since $y$ is a limit ordinal, $b^+ \in y$ by Successor in Limit Ordinal.

Therefore, $\map f {b^+} \in x$ and $\map f {b^+} > \map f b \ge z$, which contradicts the fact that $z$ is an upper bound of $x$.

Therefore, $x \ne z^+$ for any $z$ and $x \in K_{II}$.

Sufficient Condition
Suppose $x \in K_{II}$.

$y \ne 0$ by Cofinal to Zero iff Ordinal is Zero.

Moreover, if $y = z^+$ for some ordinal $z$, then $z = \bigcup y$ by Union of Successor Ordinal.

Therefore, $z$ would be the least upper bound of $y$.

Since $\map {\mathrm {cof} } {x, y}$, it follows by the definition of cofinal that:
 * $\forall a \in x: \exists b \in y: \map f b \ge a$

But $z \in y$, so by the definition of $f$:
 * $\map f z \in x$

Therefore, by the fact that $x$ is a limit ordinal:
 * $\map f z^+ \in x$

This means that:
 * $\exists b \in y: \map f b \ge \map f z^+$

This would mean that by Ordinal is Less than Successor:
 * $\map f b > \map f z$

But this means that $b \ge z$ since otherwise, by the definition of strictly increasing mapping:
 * $\map f b < \map f z$

Furthermore, $b \ne z$, since otherwise, by Substitutivity of Equality:
 * $\map f b = \map f z$.

It follows that $b > z$.

This contradicts the fact that from Union of Ordinals is Least Upper Bound, $z$ is an upper bound of $y$.

Therefore, $y \ne z^+$ for any $z$ and $y \in K_{II}$.