# Ultrafilters vs Well-orderings

This question was actually asked by John Stillwell in a comment to an answer to this question. I thought I would advertise it as a separate question since no one has yet answered and I am also curious about it.

Question: Is the existence of a non-principal ultra-filter on $\omega$ a weaker assumption than the existence of a well-ordering of $\mathbb{R}$?

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It is consistent that there exists a non-principal ultrafilter over $\omega$ while $\mathbb{R}$ is not well-ordered. To see this, suppose that the partition relation $\omega \to (\omega)^{\omega}$ holds in $L(\mathbb{R})$. Then forcing with $\mathbb{P}= [\omega]^{\omega}$ adjoins a selective ultrafilter $\mathcal{U}$ over $\omega$ and $\mathcal{P}(\omega)$ cannot be well-ordered in $L(\mathbb{R})[\mathcal{U}]$. (See Eisworth's paper: Selective ultrafilters and $\omega \to (\omega)^{\omega}$.) Thus $L(\mathbb{R})[\mathcal{U}]$ is a model of $ZF$ which contains the nonprincipal ultrafilter $\mathcal{U}$ and yet $\mathbb{R}$ cannot be well-ordered in $L(\mathbb{R})[\mathcal{U}]$.

An update: it is perhaps also interesting to note that in $L(\mathbb{R})[\mathcal{U}]$, the ultraproduct $\prod_{\mathcal{U}} \bar{\mathbb{F}}_{p}$ of the algebraic closures of the fields of prime order $p$ is not isomorphic to $\mathbb{C}$.

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How about relative consistency rather than direct implication? –  Harry Gindi Apr 11 '10 at 21:55
Thanks a lot for the answer, Simon. –  John Stillwell Apr 12 '10 at 10:52
It was a very interesting question. While it is well-known that the existence of non-principal ultrafilters is in general strictly weaker than the axiom of choice, this special case is probably the one which would interest most mathematicians. Plus I now understand a conversation that I recently had with a much more knowledgeable set theorist ... –  Simon Thomas Apr 12 '10 at 12:45