What is $0^0$, in ordinal exponentiation?

Question

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?

Answer 1

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 $0$.

Answer 2

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.

Answer 3

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.