# Selective ultrafilter and bijective mapping

For arbitrary (selective) ultrafilter $$\mathcal{F}$$ does there exist bijection $$\phi:[\omega]^2\to\omega$$ with the property: $$\forall B\in\mathcal{F} : \phi([B]^2)\in\mathcal{F}$$ ?

No, this fails not only for selective ultrafilters but for all non-principal ultrafilters $$\mathcal F$$ on $$\omega$$.

The main ingredient in the proof is the theorem that, if an ultrafilter $$\mathcal U$$ on a set $$X$$ is sent to itself by some map $$g:X\to X$$ (meaning that $$\mathcal U=g(\mathcal U):=\{A\subseteq X:g^{-1}(A)\in\mathcal U\}$$), then the set of points in $$X$$ fixed by $$g$$ has to be in $$\mathcal U$$. This easily implies that, if $$\phi:X\to Y$$ is a bijection and $$g:X\to Y$$ is an arbitrary map and $$\mathcal U$$ is an ultrafilter on $$X$$ with $$\phi(\mathcal U)=g(\mathcal U)$$, then $$\{x\in X:\phi(x)=g(x)\}\in\mathcal U$$.

In your situation, suppose toward a contradiction that $$\phi$$ were a bijection as in the question. Let $$\mathcal G$$ be the filter on $$[\omega]^2$$ generated by the sets $$[B]^2$$ for $$B\in\mathcal F$$. ($$\mathcal G$$ may or may not be an ultrafilter, depending on whether or not $$\mathcal F$$ is selective.) Your requirement on $$\phi$$ implies that each set in $$\mathcal G$$ meets each set of the form $$\phi^{-1}(A)$$ for $$A\in\mathcal F$$. Since the intersection of two sets of the form $$[B]^2$$ for $$B\in\mathcal F$$ is again a set of that form, and since the intersection of two sets of the form $$\phi^{-1}(A)$$ for $$A\in\mathcal F$$ is again a set of that form, it follows that the sets of these two forms generate a proper filter. Let $$\mathcal U$$ be an ultrafilter on $$[\omega]^2$$ extending that filter. So we have $$[B]^2\in\mathcal U$$ and $$\phi^{-1}(A)\in\mathcal U$$ for all $$A,B\in\mathcal F$$.

It follows that $$\phi(\mathcal U)=\mathcal F$$. It also follows that $$g(\mathcal U)=\mathcal F$$, where $$g:[\omega]^2\to\omega$$ is the function sending each pair $$\{x to its smaller member $$x$$. By the theorem cited above, we have that the set $$E=\{\{x is in $$\mathcal U$$. Since $$\phi$$ is one-to-one, $$E$$ contains, for any $$x\in\omega$$, at most one pair whose smaller element is $$x$$. Thus, there is a function $$f:\omega\to\omega$$ such that $$E\subseteq\{\{x,f(x)\}:x\in\omega$$. Note that $$x$$ here is the smaller element (the one given by $$g$$) in the pair $$\{x,f(x)\}$$, and so we have $$x for all $$x$$.

Consider any two sets $$B,C\in\mathcal F$$. Since $$[B\cap C]^2$$ and $$E$$ are both in $$\mathcal U$$, they have a nonempty intersection. In particular, there is some $$x\in C$$ with $$f(x)\in B$$, i.e., with $$x\in f^{-1}(B)$$. I've just shown that $$f^{-1}(B)$$ intersects every set $$C\in\mathcal F$$; since $$\mathcal F$$ is an ultrafilter, it follows that $$f^{-1}(B)\in\mathcal F$$. And since this holds for all $$B\in\mathcal F$$, we have $$f(\mathcal F)=\mathcal F$$. By the theorem cited at the beginning, $$f(x)=x$$ for $$\mathcal F$$-almost all $$x$$. This contradicts the fact that $$x for all $$x$$.

• Thank you for your answer. Where can I read about the theorem you mention in beginning? – ar.grig Feb 27 at 20:24
• The theorem used in the solution is due to Katetov (1967), you can find the short paper where it appeared through the link: dml.cz/bitstream/handle/10338.dmlcz/105124/…. You can also find Solovay's rendition of Katetov's theorem through the link: math.berkeley.edu/~solovay/Preprints/Rudin_Keisler.pdf – Ali Enayat Feb 28 at 0:14
• @AliEnayat: Thank you. I got the links. But what about reference of Solovay's rendition which could be cited in article ? – ar.grig Mar 2 at 9:16
• @ar.grig To my knowledge, Solovay's rendition has never appeared in print; his argument is "essentially the same" as Katetov's. As pointed out in the following source, Katetov's theorem can be viewed as a special case of the Bruijn-Erdős theorem: dml.cz/bitstream/handle/10338.dmlcz/126182/… – Ali Enayat Mar 3 at 17:44