Hessenberg power of ordinals

Solution 1:

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 $2^\omega$.