# Ultrafilter on the ordinal $\omega^\omega$

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?

• Note that it is consistent with ZF that the reals are well ordered (and CH holds) but there is no uniform ultrafilter on the continuum, so it is likely that there is no natural construction. – Yair Hayut Jan 27 '20 at 11:24
• @Yair: I think that a "natural extension" would be on the ordinal $\omega^\omega$, which may exist as some sort of ultralimit of $\mathcal U^{\otimes k}$ with $\cal U$ being the ultrafilter used for the limit. – Asaf Karagila Jan 27 '20 at 13:39
• Also, to clarify Yair's comment, the fact that the continuum is well-ordered means that there are free ultrafilters on $\omega$, so there is essence to this question. I'd go on to add that the above fact means also that there is no definable way of having such extension (so in particular no real natural extension). – Asaf Karagila Jan 27 '20 at 13:40
• @YCor: Yes, of course, the model Yair mentioned satisfies just that. Since there are free ultrafilters on $\omega$ (as the continuum is well-orderable), these extend to free ultrafilters on any set with a countable subset, e.g. the continuum. But of course, if there are no uniform ultrafilters, so every free ultrafilter contains a small subset. Now if you add CH (which holds in the model mentioned by Yair), then easily enough such small set must be countable. – Asaf Karagila Jan 27 '20 at 18:36
• The edit suggests that the $\omega^\omega$ in the question is, in fact, ordinal exponentiation, not cardinal exponentiation. – Emil Jeřábek Jan 28 '20 at 7:33

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.
• Thank you for your answer. What about construction for $(\omega^\lambda,~\mathcal{U}^{\otimes\lambda})$, where $\lambda<\omega_1$ is an arbitrary limit ordinal ? – ar.grig Jan 28 '20 at 20:55
• Such a $\mathcal U^{\otimes\lambda}$ would depend on how you rearrange the relevant index set $\lambda$ in an $\omega$-sequence, in order to use the ultrafilter $\mathcal U$ as an ultrafilter on $\lambda$. – Andreas Blass Jan 29 '20 at 0:53
• By the way, if we replace disjoint union with union in your definition of the sum, we also get ultrafilter. Particularly, in case $X_i=X$ we get ultrafilter on $X$. – ar.grig Jan 30 '20 at 18:07
• @ar.grig Yes. But in the special case where all $X_i=X$ I would all this ultrafilter the limit (not the sum) of the $\mathcal V_i$ along $\mathcal U$. (It is, in fact, the limit in the usual topology of $\beta X$.) – Andreas Blass Jan 30 '20 at 20:00