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A student of mine pointed out to me that if we define ordinal exponentiation the usual way (by recursion), then we have $0^0=1$ (which is fine), and thus $$ 0^\omega=\sup\{0^n:n<\omega\}=1, $$ instead of the expected answer $0$.

What is the common resolution of this issue?

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    $\begingroup$ Jech Set Theory (3rd Millennium ed.), Hrbacek & Jech Introduction to Set Theory, Just & Weese Discovering Modern Set Theory, vol I, Roitman, Introduction to Modern Set Theory, Kunen, Set Theory (new ed.) and Komjáth & Totik Problems and Theorems in Classical Set Theory all seem to fall victim to this oversight. It seems the common resolution of this issue is either pretend it doesn't exist or assume that everyone knows what we "really mean". $\endgroup$
    – user642796
    Dec 13, 2016 at 16:38
  • $\begingroup$ Boaz, you could probably replace $\beta$ in all the recursive definitions by $\beta+1$ or $\beta>0$. Or just use the order-theoretic definitions: sum of orders, reverse lexicographic order, and that terrible dictionary-like order on finite sequences. In that second case, $0^\omega=0$ by the same argument that $0^{\aleph_0}=0$ as cardinal exponentiation. $\endgroup$
    – Asaf Karagila ♦
    Dec 13, 2016 at 16:39
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    $\begingroup$ @arjafi I was almost sure that Jech defined the operations using limits (so it is the correct definition), and I was right. But in turn, he only defines limits of nondecreasing sequences, which is incomplete. So he's a victim of an issue of a different kind. $\endgroup$
    – Pedro Sánchez Terraf
    Dec 14, 2016 at 14:08
  • $\begingroup$ (... running to correct some notes of mine!) $\endgroup$
    – Pedro Sánchez Terraf
    Dec 14, 2016 at 14:09
  • $\begingroup$ @PedroSánchezTerraf Yes, technically in Jech $0^\alpha$ isn't defined for $\alpha \geq \omega$. In detail it's a slightly different issue, but it boils down to overlooking that $\langle 0^n : n < \omega \rangle$ isn't nondecreasing, which to me is the central failing these have in common. $\endgroup$
    – user642796
    Dec 14, 2016 at 14:30

3 Answers 3

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You probably just need to define $0^\omega$ as the "limit" of the $0^n$ for $n < \omega$, instead of just the supremum. Or you can define $0^\beta$ as a special case in the definition of the exponentiation.

For any ordinal $\alpha$ other than $0$, the sequence $\alpha^n$ is increasing so the limit of the sequence is its sup. But of course for $\alpha = 0$ we get a decreasing sequence $(1, 0, 0, 0\dots)$ and its limit is its minimum, which is $0$.

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In fact, there is this trick/general idea which is often neglected, but it comes in handy in this case.

Let's look at sum first: instead of separating three cases for non-zero limit ordinals, successor ordinals and $0$, we can define $$\alpha+\beta:=\begin{cases} \alpha&\text{if }\beta=0\\ \sup\left\{(\alpha+\gamma)^+\,:\,\gamma<\beta\right\}&\text{if }\beta\ne0\end{cases}$$ in just two cases.

Often, these constructions go like this:

  1. define case $g(0)$

  2. construct $g(\beta^+)$ in terms of $g(\beta)$

  3. take the "limit as $\gamma\to\beta$" of $g(\gamma)$ for limit ordinals

  4. (optional) realise, in hindsight, that you were taking the limit of $g(\gamma^+)$ all along, and that this approach is actually comprehensive of $g(\beta^+)$ as well.

For the concrete example, you can use the trick

$$\alpha^\beta=\begin{cases}1&\text{if }\beta=0\\ \sup\left\{\alpha^\gamma\cdot\alpha\,:\,\gamma<\beta\right\}&\text{if }\beta\ne 0\end{cases}$$

Though I agree that the introduction of a notion of $\limsup\limits_{\gamma\to\beta}:=\min\limits_{\gamma<\beta}\sup\limits_{\gamma\le\delta<\beta}$ could be useful in the grand scheme of things.

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Let $x,y$ be ordinals.

Let $F(x,y)$ be the set of functions $f:x\to y$ such that $$\{z\in x: f(z)\ne 0\}\; \text { is finite.}$$ (In the absence of the axiom of choice, by a finite set I mean a bijective image of a member of $\omega$ .)

For $f,g \in F(x,y)$ let $f<^*g$ iff for the LARGEST $z$ such that $f(z)\ne g(z)$ we have $f(z)\in g(z).$

DEFINITION. $y^x$ is the ordinal whose $\epsilon$-order is isomorphic to the $<^*$-order on $F(x,y).$

From this it is immediate that $0^x=0.$

It is convenient in analysis to (sometimes) have $0^0=1,$ as in " Let $p(x)=\sum_{j=0}^n a_jx^j$ " where it is understood that $a_0x^0=a_0$ when $x=0.$ And also because $x^x$ converges to $1$ as $x\to 0^+.$

Of course you can define the ordinal $0^0$ to be $1$ if you want. This will introduce its own notational inconveniences.

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