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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?

  1. $\alpha^{\beta + \gamma} = \alpha^\beta \times \alpha^\gamma$
  2. $\alpha^{\beta \times \gamma} = (\alpha^\beta)^\gamma$

And what would be, for example, $2^\omega$, $3^\omega$, and $\omega^\omega$?

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  • $\begingroup$ There may be some way to define a commutative exponentiation but I doubt it would behave well. The problem with this one is that any well-founded order on a set extends to a well-order. Any such well-order on say $2^{\omega}$ must be larger than $2^{\aleph_0}$, which I suppose is not what you are looking for. $\endgroup$
    – nombre
    Feb 14, 2020 at 19:37
  • $\begingroup$ @nombre Sorry, what do you mean by a commutative exponentiation? $\endgroup$
    – user76284
    Feb 14, 2020 at 20:40
  • $\begingroup$ Poor choice of words on my part: I mean with $\alpha^{\beta \gamma}=\alpha^{\gamma \beta}$ (I tend to think of Hessenberg operations as "commutative versions of ordinal operations", although this is misleading). $\endgroup$
    – nombre
    Feb 14, 2020 at 20:50
  • $\begingroup$ @nombre Ah, but isn't $\beta \gamma = \gamma \beta$ always true (in Hessenberg arithmetic)? $\endgroup$
    – user76284
    Feb 14, 2020 at 21:01
  • $\begingroup$ Hehe, indeed it is, I guess I don't even know what I meant! $\endgroup$
    – nombre
    Feb 14, 2020 at 21:09

1 Answer 1

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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 $2^\omega$.

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  • $\begingroup$ Posted a follow-up here. $\endgroup$
    – user76284
    Jun 9, 2020 at 0:56

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