Is $0^\omega=1$?

Question

According to a definition of ordinal exponentiation defined in Kunen's Set Theory: An Introduction to Independence Proofs (pp. 26), we define

$$\begin{align} \alpha^0&=1\\ \alpha^{(\beta+1)}&=\alpha^\beta\cdot\alpha\\ \alpha^\beta&=\bigcup\{\alpha^\gamma:\gamma<\beta\}&\text{if }\beta\text{ is a limit}. \end{align}$$

So, it is correct to conclude that $0^\omega=1$, right? (In fact, we have $0^\beta=1$ for all limit $\beta$).

This is quite surprising for me since I never think that other exponentiations of $0$, except $0^0$, can be $1$.

Furthermore, there is another definition of ordinal exponentiation defined in Kunen's (pp. 43). It can be defined as follows.

Let

$$F(\alpha,\beta)=\{f\in{^\beta\alpha}:|\{\xi:f(\xi)\ne0\}|<\omega\}.$$

For any $f,g\in F(\alpha,\beta)$ such that $f\ne g$, say $f\lhd g$ iff $f(\xi)<g(\xi)$, where $\xi$ is the largest ordinal such that $f(\xi)\ne g(\xi)$. Then $\alpha^\beta=\operatorname{type}(\langle F(\alpha,\beta),\lhd\rangle)$.

The book claims that this definition is equivalent to the above definition.

But, as we can see, for $0^\omega$, we consider $F(0,\omega)$, which is $\emptyset$. So $\operatorname{type}(\langle F(0,\omega),\lhd\rangle)=0$.

Which one is a correct one, or if you think none of them is wrong, which one is much more make sense to you?

Answer 1

Where did you get that definition? That shouldn't be right. It's fine if you replace the 3rd line (limit $\beta$ ordinal) by $$\alpha^\beta=\lim_{\gamma<\beta} \alpha^\gamma$$ Then the sequence you get is $$1,0,0,0,\ldots$$ and the limit is $0$ as it should be.

Answer 2

Ordinal exponentiation is actually defined as a limit when $\beta$ is a limit ordinal: $\alpha^\beta = \lim_{\gamma < \beta} \alpha^\gamma$. It just so happens that:

• When a sequence $(\delta_\gamma)_\gamma$ of ordinals is nondecreasing, $\lim_\gamma \delta_\gamma = \bigcup_\gamma \delta_\gamma = \sup_\gamma \delta_\gamma$;
• When $\alpha \neq 0$, the sequence $(\alpha^\gamma)_\gamma$ is nondecreasing.

So that formula you gave is valid in every case except when $\alpha = 0$. But of course when $\alpha = 0$ that sequence decreases once, because $0^0 = 1$ and $0^1 = 0$.

Answer 3

Enderton's Elements of Set Theory uses a definition equivalent to yours, except that he special-cases $0$ to avoid this issue:

Consider ... a fixed nonzero ordinal $\alpha$. We propose to define $\alpha^\beta$ in such a way ... $\alpha^\lambda=\sup\{\alpha^\beta\,|\,\beta\in\lambda\}$ for a limit ordinal $\lambda$ ...

There is a special problem in defining $0^\beta$. If we were to follow blindly [the above definitions], we would have $0^\omega=1$. This is undesirable, which is our reason for having specified in the foregoing that $\alpha$ is a fixed nonzero ordinal. We can simply define $0^\beta$ directly: $0^0=1$ and $0^\beta=0$ for $\beta\neq0$.

Answer 4

I just look in Jech's Set Theory (pp. 23), as those above comments, it defines

$$\alpha^\beta=\lim\limits_{\xi\to\beta} \alpha^\xi \text{ for all limit }\beta>0,$$

Where the definition of limit is in pp. 22. For a limit $\alpha$, the limit of a nondecreasing sequences of ordinals $\langle\gamma_\xi:\xi<\alpha\rangle$ is defined (same as @Najib Idrissi stated) by

$$\lim\limits_{\xi\to\alpha}\gamma_\xi=\sup\{\gamma_\xi:\xi<\alpha\}.$$

Still I don't get the answer whether $0^\omega$ is $0$ or $1$ from this definition since the sequence $\langle 0^n:n<\omega\rangle$ decreases once as $user2345214 said.

I agree with @Chris Culter that $0^\beta$ must be defined separately from those definitions, and so, I feel that $0^\omega$ should be $0$.

Answer 5

Recall that

\begin{align} \lim_{\beta \rightarrow \alpha} f(\beta) = \bigcup_{\beta < \alpha} \bigcap_{\beta \leq \gamma < \alpha} f(\gamma) \end{align}

Therefore

\begin{align} 0^\omega &= 0^{\lim_{\beta \rightarrow \omega} \beta} \\ &= \lim_{\beta \rightarrow \omega} 0^\beta \\ &= \bigcup_{\beta < \omega} \bigcap_{\beta \leq \gamma < \omega} 0^\gamma \\ &= \bigcap_{\beta < \omega} 0 \\ &= 0 \end{align}

This is something the other answers have noted. However, we can also answer this question more directly with the order-theoretic definition of exponentiation for ordinals:

$\alpha^\beta$ is the order-type of the set of all functions $\beta \rightarrow \alpha$ of finite support, under lexicographic order. There is no function $\omega \rightarrow 0$ (or more generally, $\beta \rightarrow 0$ for $\beta > 0$). Therefore, $0^\omega = 0$.