According to these notes on ordinal arithmetic:
-
The Hessenberg sum $\alpha + \beta$ is the supremum of ordinals that are isomorphic to some well-order on $\alpha \sqcup \beta = (\{0\} \times \alpha) \cup (\{1\} \times \beta)$ extending the union partial-order:
$$x \leq y \Longleftrightarrow (x_1 = y_1) \land (x_2 \leq y_2)$$ -
The Hessenberg product $\alpha \times \beta$ is the supremum of ordinals that are isomorphic to some well-order on $\alpha \times \beta$ extending the product partial-order:
$$x \leq y \Longleftrightarrow (x_1 \leq y_1) \land (x_2 \leq y_2)$$
Does it make sense to define the Hessenberg power $\alpha^\beta$ as the supremum of ordinals that are isomorphic to some well-order on $\beta \rightarrow \alpha$ extending the following partial-order?
$$x \leq y \Longleftrightarrow \forall z \in \beta: x(z) \leq y(z)$$
If so, does it satisfy the following properties?
- $\alpha^{\beta + \gamma} = \alpha^\beta \times \alpha^\gamma$
- $\alpha^{\beta \times \gamma} = (\alpha^\beta)^\gamma$
And what would be, for example, $2^\omega$, $3^\omega$, and $\omega^\omega$?
Best Answer
Your construction cannot work.
Consider the case $2^\omega$. The functions in that case would be the characteristic functions of subsets of $\omega$, and the partial order you give then simply refers to the subset order of the corresponding sets.
Now consider the set $M=\{\omega\setminus n\mid n\in\omega\}$. The elements of this set have the form $S_n=\omega\setminus n=\{n,n+1,n+2,n+3,\ldots\}$. Now it is easily seen that the inclusion partial order already gives a total order on $S_n$, namely $S_m\subseteq S_n\iff m\ge n$.
Note the reversal of the order direction here, which means that because $\omega$ has no maximal element, $V$ has no minimal element. And since $V$ is already totally ordered, no extension of the partial order on $2^\omega$ will cause its order to change.
Therefore no extension of that partial order on $2^\omega$ to a well-order can exist, which means your exponentiation function is not defined for $2^\omega$.