Condition for Cofinal Nonlimit Ordinals

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

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

Let $\le$ denote the subset relation.

Furthermore, let $x$ and $y$ satisfy the condition:


 * $0 < x \le y$

Then:


 * $\operatorname{cof} \left({ y,x }\right)$

Proof
Both $x$ and $y$ are non-empty, so by the definition of a limit ordinal:


 * $x = z^+$ for some $z$.


 * $y = w^+$ for some $w$.


 * $\bigcup z^+ \le \bigcup w^+$ follows by Set Union Preserves Subsets/General Case.


 * $z \le w$ follows by Union of Successor Ordinal.

Define the function $f : x \to y$ as follows:


 * $f \left({a}\right) = \begin{cases}

a &: a \ne z \\ w &: a = z \end{cases}$

$a < b \le w$, so $f \left({a}\right) = a$.

Take any $a,b \in x$ such that $a < b$.


 * $f\left({a}\right) < f\left({b}\right)$ shall be proven by cases:

Case 1: $b \ne z$
If $b \ne z$:


 * $f\left({a}\right) < f\left({b}\right)$ is simply a restatement of $a < b$.

Case 2: $b = z$
If $b = z$, then $f\left({b}\right) = w$ by the definition of $f$.

Since $a < z \le w$, $f\left({a}\right) < f\left({b}\right)$.

It follows that $f$ is strictly increasing.

Moreover, since $\bigcup y = w$ is the least upper bound, $f\left({z}\right) \ge a$ for all $a \in y$.