If ${\cal U}$ and ${\cal V}$ are ultrafilters on non-empty sets $A$ and $B$ respectively, then the tensor product ${\cal U}\otimes{\cal V}$ is the following ultrafilter on $A\times B$: $$\big\{X\subseteq A\times B: \{a\in A:\{b\in B: (a,b)\in X\}\in {\cal V}\}\in {\cal U}\big\}.$$ It is a standard exercise to verify that this is an ultrafilter on $A\times B$.

We say two non-principal ultrafilters ${\cal U, V}$ are Rudin-Keisler equivalent if there is a bijection $\varphi:\omega\to\omega$ such that ${\cal U} =\varphi(\cal V)$, where $\varphi({\cal V}) = \{\varphi(R): R\in {\cal V}\}$.

We say an ultrafilter ${\cal Z}$ on $\omega$ is Tensor-representable if there are non-Keisler-Rudin-equivalent ultrafilters ${\cal U}, {\cal V}$ and a bijection $\psi:\omega^2\to \omega$ such that ${\cal Z} =\psi({\cal U}\otimes {\cal V})$.

What is an example of an ultrafilter ${\cal Z}$ on $\omega$ that is not Tensor-representable?


Recall that $\mathcal Z$ is a weak $P$-point if it is not in the closure of any countable subset of $\omega^* \setminus \{\mathcal Z\}$. A weak $P$-point is never the tensor product of two non-principal ultrafilters.

To see this, suppose $\mathcal U$ and $\mathcal V$ are two non-principal ultrafilters on $\omega$, and let $\mathcal Z = \mathcal U \times \mathcal V$. I will show that $\mathcal Z$ is not a weak $P$-point in the space of non-principal ultrafilters on $\omega \times \omega$. (This implies that $\psi(\mathcal Z)$ is not a weak $P$-point in $\omega^*$ for any bijection $\psi: \omega \times \omega \rightarrow \omega$.)

For each $i \in \omega$, let $$v_i = \{\{i\} \times B \,:\, B \in \mathcal V\},$$ and observe that $v_i$ is a non-principal ultrafilter on $\omega \times \omega$. The set $\{v_i \,:\, i \in \omega\}$ is a countable, relatively discrete subset of $(\omega \times \omega)^*$. (It is relatively discrete because for each $i \in \omega$, $U_i = (\{i\} \times \omega)^*$ is a (cl)open subset of $(\omega \times \omega)^*$ such that $U_i \cap \{v_i \,:\, i \in \omega\} = \{v_i\}$.) Clearly $\mathcal Z$ is not equal to any of the $v_i$. Yet we have $\mathcal Z \in \overline{\{v_i \,:\, i \in \omega\}}$. Indeed, we see from the definition of $\mathcal Z$ that every neighborhood of $\mathcal Z$ in $(\omega \times \omega)^*$ contains $\mathcal U$-many of the $v_i$. (In what might be more familiar notation, this means $\mathcal Z \,=\, \mathcal U\hspace{-.5mm}-\lim_{i \in \omega} v_i$.) Thus $\mathcal Z$ is not a weak $P$-point.

Finally, let me mention that Kunen famously proved that $\omega^*$ contains weak $P$-points. Thus the above observation shows that some ultrafilters are not representable as tensor products.

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    $\begingroup$ Let me add that "weak P-point" implies "not a tensor product" with some room to spare. A tensor product is, as this answer shows, in the closure of a countable discrete set of pairwise isomorphic nonprincipal ultrafilters. I believe it was known before Kunen's theorem (and probably due to Frolik) that there are points in $\omega^*$ not in the closure of any countable discrete subset of $\omega^*$ (provably in ZFC); these are the Rudin-Frolik-minimal ultrafilters. $\endgroup$ – Andreas Blass Mar 11 at 17:56
  • $\begingroup$ Thanks Will for your really nice answer, I was amazed that the topology on $\omega^*$ would come into play! Also thank you Andreas for your helpful addition. $\endgroup$ – Dominic van der Zypen Mar 12 at 6:12

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