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If ${\scr U}$ and ${\scr V}$ are ultrafilters on non-empty sets $A$ and $B$ respectively, then the tensor product ${\scr U}\otimes{\scr V}$ is the following ultrafilter on $A\times B$: $$\big\{X\subseteq A\times B: \{a\in A:\{b\in B: (a,b)\in X\}\in {\scr V}\}\in {\scr U}\big\}.$$ Is there a non-principal ultrafilter ${\scr U}$ on $\omega$ and a function $f:\omega \to (\omega\times\omega)$ such that $f({\scr U}) = {\scr U}\otimes {\scr U}$?

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  • $\begingroup$ No. . . . . . . $\endgroup$
    – Joseph Van Name
    Commented Mar 25, 2019 at 11:55
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    $\begingroup$ Actually, now that I think about it, shouldn't the answer be "yes" because you can just use a projection mapping $\omega \times \omega \rightarrow \omega$ to recover $\mathcal U$ from $\mathcal U \otimes \mathcal U$? $\endgroup$
    – Will Brian
    Commented Mar 25, 2019 at 13:15
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    $\begingroup$ @WillBrian Projections give you $f(\mathcal U\otimes\mathcal U)=\mathcal U$, but the OP is asking for the other direction, $f(\mathcal U)=\mathcal U\otimes\mathcal U$, and that is indeed impossible. $\endgroup$
    – Andreas Blass
    Commented Mar 25, 2019 at 14:18
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    $\begingroup$ @AndreasBlass: Thanks -- I was reading that in the wrong direction. $\endgroup$
    – Will Brian
    Commented Mar 25, 2019 at 14:27
  • $\begingroup$ The answer on your question is negative as Andreas Blass mentioned. But Katetov proved the existence of filter with property $f(\mathcal{F})=\mathcal{F}\otimes\mathcal{F}$. See details here. $\endgroup$
    – ar.grig
    Commented Mar 25, 2019 at 18:21

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As Will Brian noted in a comment, either projection $p:\omega^2\to\omega$ satisfies $p(\mathcal U\otimes\mathcal U)=\mathcal U$. If you also had a function $f$ as in the question, with $f(\mathcal U)=\mathcal U\otimes\mathcal U$, then the composite function $f\circ p$ would send $\mathcal U\otimes\mathcal U$ to itself. By a result that I quoted at Selective ultrafilter and bijective mapping, and for which Ali Enayat supplied a reference, it would follow that $f\circ p$ becomes the identity function when restricted to a suitable set in $\mathcal U\otimes\mathcal U$. That is absurd, because $p$ is not one-to-one on any set in $\mathcal U\otimes\mathcal U$.

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