Limit Ordinals Closed under Ordinal Exponentiation

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Theorem

Let $x$ and $y$ be ordinals.

Let $y$ be a limit ordinal.

Let $x^y$ denote ordinal exponentiation.


Then:

If $x > 1$, then $x^y$ is a limit ordinal.
If $x \ne \varnothing$, then $y^x$ is a limit ordinal.


Proof

Suppose $x > 1$.

Suppose also that $x^y$ is the successor of some ordinal $w$.

By definition of ordinal multiplication:

$\displaystyle x^y = \bigcup_{z \mathop \in y} x^z$

Then:

\(\displaystyle w\) \(\in\) \(\displaystyle x^y\) Ordinal is Less Than Successor
\(\displaystyle \exists z \in y: \ \ \) \(\displaystyle w\) \(\in\) \(\displaystyle x^z\) by definition of ordinal exponentiation
\(\displaystyle w^+\) \(\subseteq\) \(\displaystyle x^z\) Successor of Element of Ordinal is Subset
\(\displaystyle \) \(\in\) \(\displaystyle x^{z^+}\) Membership is Left Compatible with Ordinal Exponentiation


But $z^+ \in y$ by Successor in Limit Ordinal.

So $w^+ \in x^y$ and $w^+ \in w^+$.

This creates a membership loop and thus is a contradiction by No Membership Loops.

$\Box$


For the second part, since $x$ is not the empty set it follows that $x = z^+$ for some $z$ or that $x$ is a limit ordinal.

If $x$ is a limit ordinal, then $y^x$ is a limit ordinal by the first part.


If $x$ is the successor of $z$, then:

$y^x = y^w × y$ by the definition of ordinal exponentiation.

Then, $y^x$ is a limit ordinal by Limit Ordinals Preserved Under Ordinal Multiplication.

$\blacksquare$


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