# Non-tensor-representable ultrafilters on $\omega$

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.
• 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. – Andreas Blass Mar 11 at 17:56
• 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. – Dominic van der Zypen Mar 12 at 6:12