Cofinal Limit Ordinals

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Theorem

Let $x$ and $y$ be ordinals.

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

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


Then:

$\operatorname{cof} \left({ x,y }\right) \implies \left({ x \in K_{II} \iff y \in K_{II} }\right)$


Proof

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 $\operatorname{cof} \left({ x,y }\right)$, it follows that $\forall a \in x: \exists b \in y: f\left({ b }\right) \ge a$ by the definition of cofinal.

So $f\left({b}\right) \ge z$ for some $b \in y$.


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

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

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

$\Box$


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 $\operatorname{cof} \left({ x,y }\right)$, it follows that $\forall a \in x: \exists b \in y: f\left({ b }\right) \ge a$ by the definition of cofinal.


But $z \in y$, so $f\left({z}\right) \in x$ by the definition of $f$.

Therefore, $f\left({z}\right)^+ \in x$ by the fact that $x$ is a limit ordinal.

This means that $\exists b \in y: f\left({b}\right) \ge f\left({z}\right)^+$


This would mean that $f\left({b}\right) > f\left({z}\right)$ by Ordinal is Less than Successor.

But this means that $b \ge z$ since otherwise, $f\left({b}\right) < f\left({ z }\right)$ by the definition of strictly increasing.

Furthermore, $b \ne z$, since otherwise, $f\left({b}\right) = f\left({z}\right)$ by Substitutivity of Equality.

It follows that $b > z$.


This contradicts the fact that $z$ is an upper bound of $y$.

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

$\blacksquare$


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