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Consider a category $\bf {Set}$ of sets and functions that admits a functor $^{*}-:\bf{Set}\to \bf {Set}$ which sends every set $S$ to an enlargement of it and every function $f:S\to T$ to its enlargement.

Is it possible that such an enlargement functor is essentially idempotent?

The problem is related to the existence of a special ultrafilter on $\mathbb {N}$. Let $g:\mathbb{N}\times \mathbb{N}\to \mathbb{N}$ be a bijection. If we write $UF(X)$ for the set of ultrafilters on the set $X$ then $g$ induces a bijection $G:UF(\mathbb{N}\times \mathbb{N})\to UF(\mathbb{N})$. Define the function $H_g:UF(\mathbb{N})\to UF(\mathbb{N})$ by sending an ultrafilter $\mathcal{F}\in UF(\mathbb{N})$ to $G(\mathcal{F}\times\mathcal{F})$. Call the pair $(g,\mathcal{F})$ 'good' if $H_g(\mathcal{F})=\mathcal{F}$ and $\mathcal{F}$ is non-principal.

Is there a good pair $(g,\mathcal{F})$?

Remark: The existence of a good pair $(g,\mathcal{F})$ is related to the enlargement problem as follows. Assume $ZF$ for the sets in $\bf {Set}$ and consider the usual ultrapower construction with respect to $\mathcal{F}$ to obtain an enlargement functor $^{*}-:\bf{Set}\to \bf {Set}$. $g$ can then rather straightforwardly be used to obtain the components of a natural isomorphism from the double enlargement to the enlargement.

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up vote 5 down vote accepted

If one unwraps the packaging, it appears that the question about good pairs is asking the following:

Question. Is there a nonprincipal ultrafilter $F$ on $\mathbb{N}$ such that $F\times F$ is isomorphic to $F$ via a bijection $g:\mathbb{N}\times\mathbb{N}\to \mathbb{N}$?

The answer to this is no. I think about it in terms of seed theory: the ultrapower by $F\times F$ amounts to doing the ultrapower by $F$ twice, and therefore has two distinct seeds generating $F$, but the ultrapower by $F$ has a unique seed generating $F$, by the unique seed lemma, and so they cannot be isomorphic.

But let me also give a direct argument. Assume that $g:\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ is an isomorphism of $F\times F$ with $F$. Let $p:\mathbb{N}\times\mathbb{N}\to \mathbb{N}$ be the projection function $p(x,y)=x$, and let $f=p\circ g^{-1}:\mathbb{N}\to\mathbb{N}$. Using the definition of the product ultrafilter, the fact that $g$ is an isomorphism, and the definition of $f$, it follows that $$A\in F\iff p^{-1}A\in F\times F\\ \iff g[p^{-1}A]\in F\\ \iff f^{-1}A\in F$$ Now, it is a completely general fact about ultrafilters, proved by Solovay, that if $A\in F\iff f^{-1}A\in F$ for all $A$, then $f$ must be the identity on a set in $F$. (I can post a proof of this if it is desired. This fact is the essence of the unique seed lemma.) Thus, $f(n)=n$ on a set $A$ in $F$. What this means is that $g^{-1}(n)$ is a pair with first coordinate $n$, for all $n\in A$. It follows that $g^{-1}A$ has at most one element on each column, and consequently is not in $F\times F$. Since $A\in F$, this contradicts the assumption that $g$ is an isomorphism of $F\times F$ with $F$. QED

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The same argument works with ultrafilters on any set, not just ultrafilters on a countable set. – Joel David Hamkins Mar 25 '12 at 9:30
I think this result should be attributed to Frolik. – Juris Steprans Mar 25 '12 at 12:51
Juris is probably right about the specific result concerning product ultrafilters. The general result that you attribute to Solovay was also found by Katetov, but I´m not sure who got it first. – Andreas Blass Mar 25 '12 at 15:10
I've forgotten that in fact Solovay told me by email last year that it appears he was not first, that Katetov had it earlier. Solovay produced an account of his argument here: – Joel David Hamkins Mar 25 '12 at 15:22
Thank you very much Joel for the simplification of my question and the very decisive answer. – Ittay Weiss Mar 29 '12 at 9:16

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