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Given an infinite family $\{\mathcal{F}_{\lambda}$, $\lambda <\kappa\}$, $\kappa \geq \omega_0$, of (ultra)filters of a set $X$, how it is defined the infinite tensor/Fubini product $$\bigotimes_{\lambda <\kappa}\mathcal{F}_\lambda$$ as a family of subsets of $X^{\kappa}$?

Recall that given two ultrafilers $\mathcal{F}$ and $\mathcal{G}$ of a set $X$, the tensor/Fubini product $\mathcal{F}\otimes \mathcal{G}$ of them is defined as the collection of sets $$\{A\subset X\times X:\{\xi \in X:\{\nu \in X:(\xi,\nu)\in A\}\in \mathcal{G}\}\in \mathcal{F}\}.$$

Is there any example in the bibliography of their use? More precisely, is there any basic bibliography in this topic?

For instance, for $\kappa=\omega_0$, if I take an ultrafilter $\mathcal{F}$ in a discrete space $X$ and I do the iterated tensor product $$\mathcal{F}\bigotimes \mathcal{F}=\mathcal{F}^2 \subset \mathcal{P}(X\times X),$$ $$\mathcal{F}^n\bigotimes \mathcal{F} =\mathcal{F}^{n+1}\subset \mathcal{P}(X^{n+1}),\, n<\omega_0,$$ how I sould define $\mathcal{F}^{\omega_0}$ in $X^{\omega_0}$?

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  • $\begingroup$ It would be better to use \bigotimes instead of \otimes — the latter is used for binary operations. $\endgroup$
    – Z. M
    Mar 9, 2023 at 19:05

1 Answer 1

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The product of ultafilters $F_\lambda$ for $\lambda<\kappa$ is defined on $\kappa\times X$, not $X^\kappa$, and it is defined relative to a fixed ultrafilter $\mu$ on the index set $\kappa$. Namely, for $Y\subseteq\kappa\times X$, we denote $Y_\lambda=\{x\in X\mid (\lambda,x)\in Y\}$ for the various sections of $Y$ and then say $$Y\in\int \mathcal{F}_\lambda d\mu\qquad\text{if and only if}\qquad \{\lambda\in\kappa\mid Y_\lambda\in \mathcal{F}_\lambda\}\in\mu.$$ This directly generalizes the product measure case you defined, in the event that all $F_\lambda$ are the same, taking $X=\kappa$ and looking at subsets of $X\times X$.

This kind of product ultrafilter occurs routinely in the set-theoretic large cardinal literature, since many large cardinals involve this kind of measure, particularly in the case where these ultrafilters are countably complete.

In your comment below you seek to generalize to infinite exponents. One common way to do this in the large cardinal context is with extenders, which are in effect the natural limit of the finite products. If we have measures $\mathcal{F}_i$ defined for $i\in I$, then we can define the natural measure on $X^I$ concentrating on sets with finite support as follows. For each finite $s\subseteq I$, we have a natural measure $\mathcal{F}_s=\Pi_{i\in s}\mathcal{F}_i$ defined by the Fubini product. And now we say that a set $Y\subseteq X^I$ has support $s$ if there is a set $Y_0\subseteq X^s$ such that $Y$ consists of all $I$-tuples extending a sequence in $Y_0$. And we define that $Y$ has measure one for the extender if $Y_0\in\mathcal{F}_s$. That is, a set in the full product is large, if it contains a set with finite support that is large for the finite Fubini product on that finite set of indices.

The main point is that one can define ultrapowers by an extender using the finite-support functions on $I$, and then the extender ultrapower is the direct limit of the system of finite Fubini products. This conception of limits of ultrapowers is central in the large cardinal inner model theory, particularly in the case of strong cardinals, whose embeddings are realized by such extender measures.

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  • $\begingroup$ A naif question: so, what is there in $X^\kappa$? For instance, for $\kappa=\omega_0$ if I take an ultrafilter $\mathcal{F}$ in a discrete space $X$ and I do the iterated tensor product $\mathcal{F}\otimes \mathcal{F}=\mathcal{F}^2\in X\times X$, $\mathcal{F}^n \otimes \mathcal{F}=\mathcal{F}^{n+1}\in X^{n+1}$ how I should define $\mathcal{F}^{\omega_0}\in X^{\omega_0}$ $\endgroup$
    – BTN
    Mar 9, 2023 at 18:01
  • $\begingroup$ I updated my answer with an answer to this question. What you seek is the notion of a finite-support extender. $\endgroup$ Mar 9, 2023 at 23:05
  • $\begingroup$ Thank you very much Joel. $\endgroup$
    – BTN
    Mar 10, 2023 at 11:02

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