TITLE: Ultrafilter on the ordinal $\omega^\omega$
QUESTION [7 upvotes]: For any ultrafilter $\mathcal{U}$ on $\omega$ and any finite $k$ we can construct tensor power $\mathcal{U}^{\otimes k}$ which is ultrafilter on $\omega^k$. Does there exist some natural extension of this construction for the ordinal $\omega^\omega$? 
Edit: my suggestion:
$$
\mathcal{U}^{\otimes\omega}=\{B\subset\omega^\omega~|~\{k<\omega~|~B\cap\omega^k\in\mathcal{U}^{\otimes k}\}\in\mathcal{U}\}
$$
But is it good idea?

REPLY [11 votes]: The relevant general construction is the sum of a family $\{\mathcal V_i:i\in I\}$ of an indexed family of ultrafilters, with respect to an ultrafilter $\mathcal U$ on the index set $I$. If $\mathcal V_i$ is an ultrafilter on $X_i$, then the sum is the ultrafilter $\mathcal W$ on the disjoint union $\bigsqcup_{i\in I}X_i$ defined by
$$
\mathcal W=\{A:\{i\in I:A\cap X_i\in\mathcal V_i\}\in\mathcal U\}.
$$
In your situation, taking $\mathcal V_i$ to be $\mathcal U^{\otimes i}$, you get a sum ultrafilter on $\bigsqcup_{i\in\omega}\omega^i$, which can be identified with the ordinal $\omega^\omega$ to produce the ultrafilter $\mathcal U^{\otimes\omega}$ in the question.