Asaf Karagila
I don't have much choice...

Earlier this morning I received an email question from Yair Hayut. Is it consistent without the axiom of choice, of course, that there are free ultrafilters on the natural numbers but none on the real numbers?

Well, of course that the answer is negative. If $$\cal U$$ is a free ultrafilter on $$\omega$$ then $$\{X\subseteq\mathcal P(\omega)\mid X\cap\omega\in\cal U\}$$ is a free ultrafilter on $$\mathcal P(\omega)$$. But that doesn't mean that the question should be trivialized. What Yair asked was actually slightly subtler than that: is it consistent that there are free ultrafilters on $$\omega$$, but no uniform ultrafilters on the real numbers?

Let me remind you, in case that it has slipped your mind in this sleep deprived hour, that a uniform filter is one that all its sets have the same cardinality. For example, in $$\ZFC$$ we can extend the filter which is dual to $$[\Bbb R]^{\lt\frak c}$$ and the ultrafilter will be uniform (and therefore free). So, can we have a model where there are free ultrafilters on $$\omega$$ but no uniform ultrafilters on $$\Bbb R$$?

Unfortunately, or fortunately, this is the side of "choice references" I am less knowledgeable about. I know less references, and I am slightly less familiar with the proofs. Of course that my educated guess is that the answer is positive. It is consistent. But off the top of my head, I can't think of any reference. So instead of relying on searching for proof, I decided it would be good practice to generalize the solution to an exercise from Jech The Axiom of Choice (Ch. 5, Problem 24, p.82) and prove this thing myself.

Off the bat, however, I could tell you that we know of many models where there are no free ultrafilters on $$\omega$$. We know about models where there are no free ultrafilters on any set. We know about models where every set has a free ultrafilter, but not all filters can be extended to an ultrafilter. But I couldn't quite recall a proof for this problem, although I'm quite certain it hides within these known models.

In the following proof I will use notation and terminology about symmetric extensions, these are more or less standard (and where they are not, I hope you can infer from context). But feel free to ask for further clarification if needed and I'll be happy to extend. You can find a reasonable overview of this topic in either my M.Sc. thesis, or my paper "Embedding Orders Into The Cardinals With $$\DC_\kappa$$", both appear on the Papers page of this site.

#### Theorem. If $\ZFC$ is consistent then $\ZF+2^{\aleph_0}=\aleph_1+\text{No uniform ultrafilters on }\omega_1$ is consistent.

##### Proof.
Suppose [without loss of generality] that $$M$$ is a model of $$\ZFC+2^{\aleph_0}=\aleph_1$$. We shall construct a symmetric extension $$N$$ of $$M$$ such that with the same initial ordinals, $$\mathcal P(\omega)^M=\mathcal P(\omega)^N$$ (and therefore both sets have size $$\aleph_1$$) and in $$N$$ there are no uniform ultrafilters on $$\omega_1$$ and therefore on $$\Bbb R$$.

Let $$\PP$$ be the forcing adding $$\omega_1\times\omega_1$$ subsets to $$\omega_1$$ with countable conditions. Namely $$p\in\PP$$ is a function $$p\colon\omega_1\times\omega_1\to2$$ and $$\dom p$$ is countable, of course $$q\leq p$$ if $$p\subseteq q$$ (so $$q$$ is stronger).

It is a standard fact, by today anyway, that $$\PP$$ does not change cofinalities and therefore does not collapse cardinals. It is also countably closed so it does not add reals. Therefore any symmetric extension defined by $$\PP$$ automatically satisfies the first two conditions.

We shall write $$\dot x_\alpha=\{\tup{p,\check\beta}\mid p(\alpha,\beta)=1\}$$ as the name for the $$\alpha$$-th subset added.

Let $$\cG$$ be the group of automorphisms defined as follows, $$\pi\in\cG$$ if and only if there exists $$A\subseteq\omega_1\times\omega_1$$ such that $$\pi p(\alpha,\beta)=|\chi_A(\alpha,\beta)-p(\alpha,\beta)|$$. In other words, if $$\tup{\alpha,\beta}\in A$$ then $$\pi p(\alpha,\beta)$$ negates its value, and otherwise it keeps the value in place. If $$A\subseteq\omega_1\times\omega_1$$ we will write $$\pi_A$$ as the automorphism defined here with $$A$$ as the witness.

For $$A\subseteq\omega_1$$ we write $$\fix(A)=\{\pi_X\mid X\cap (A\times\omega_1)=\varnothing\}$$, this is a subgroup of $$\cG$$, and let $$\cF$$ be the normal filter of subgroups generated by the collection $$\{\fix(A)\mid A\in[\omega_1]^{\lt\omega_1}\}$$ (one can easily see that the group of generators is closed under conjugation and therefore the filter it generates is normal).

We denote by $$\HS$$ the class of hereditarily $$\cF$$-symmetric $$\PP$$-names. Let $$G\subseteq\PP$$ be an $$M$$-generic filter, and let $$N=\HS^G$$. Then $$M\subseteq N\subseteq M[G]$$. Observe that for each $$\alpha$$ the name $$\dot x_\alpha$$ is in $$\HS$$, since $$\fix(\{\alpha\})=\sym(\dot x_\alpha)$$. However, unlike the "usual models", the name $$\{\dot x_\alpha\mid\alpha\lt\omega_1\}^\bullet$$ is not in $$\HS$$, and indeed the standard argument in fact shows that the interpretation of this name is not in $$N$$ either.

Finally we wish to show that there are no uniform ultrafilters on $$\omega_1$$ in $$N$$. Suppose that $$\dot D\in\HS$$ and $$p\forces\dot D\text{ is an ultrafilter on }\omega_1$$. Let $$A$$ be some countable subset of $$\omega_1$$ such that $$\fix(A)$$ is a subgroup of $$\sym(\dot D)$$. Let $$\alpha\notin A$$, then if $$p\forces\dot x_\alpha\in\dot D$$ then there is some $$\beta$$ such that $$(\alpha,\gamma)\notin\dom p$$ for all $$\gamma\geq\beta$$, and let $$X=\{\tup{\alpha,\gamma}\mid\beta\leq\gamma\}$$ then $$\pi_X p=p$$ and we have that $$1\forces\pi_X\dot x_\alpha\cap\dot x_\alpha\subseteq\check\beta$$ and is therefore countable. But now we have this: \begin{align} \pi_X p&\forces\pi_X\dot x_\alpha\in\pi_X\dot D&\implies\\ p&\forces\pi_X\dot x_\alpha\in\dot D&\implies\\ p&\forces\pi_X\dot x_\alpha\cap\dot x_\alpha\in\dot D&\implies\\ p&\forces\dot D\text{ is not uniform}. \end{align}

Similarly, if $$p\forces\dot x_\alpha\notin D$$ then it must force that $$\pi_X\dot x_\alpha\notin\dot D$$ either, and therefore their union which is co-countable cannot be forced to be in $$\dot D$$. In either case we find that a countable set must be in $$\dot D$$.

Of course, if $$p$$ does not decide for any $$\alpha\notin A$$ whether $$\dot x_\alpha\in\dot D$$ or not, then by the above it cannot have any extension which forces that $$\dot D$$ is uniform, therefore it does not force that either.

And lest not forget, since $$\power(\omega)$$ can be well-ordered we can extended every filter on $$\omega$$ to an ultrafilter, in particular the cofinite filter. $$\qquad\square$$

Remarks. The sharp eyed reader will notice that the forcing is countably closed, as well the set of generators for $$\cF$$ is countably closed, so $$\cF$$ itself is countably closed. And under these two conditions we also have that $$\DC$$ holds in $$N$$, so we have a bit more than just $$\ZF$$. This can also be easily generalized to any case where $$2^{\aleph_0}$$ is a regular cardinal. It could be a nice exercise to figure out what happens if $$2^{\aleph_0}$$ is singular, since in that case just adding Cohen subsets won't work, and things get far more complicated. However since we just wanted a consistency result, it's fine like that.

### There are 4 comments on this post.

By
(Jul 25 2014, 21:21)

Nice "negative image" support symmetric model!

Just a question; should $$\pi p(\alpha,\beta)$$ be the absolute value of $$\chi_A(\alpha,\beta)-p(\alpha,\beta)$$ or is the minus a "setminus"? So, when you say $$1\forces \pi_X \dot{x}_\alpha\cap\dot{x}_\alpha\subset\check{\beta}$$, is this because $$1\forces \pi_X \dot{x}_\alpha\cap\dot{x}_\alpha=\dot{x}_\alpha\cap\beta$$?

Also, I really like the observation that countably closed forcings with countably closed normal filters create symmetric models of DC! For which other AC consequences have you found such restrictions on the symmetric model ingredients?

By
(Jul 25 2014, 23:54 In reply to Ioanna M. Dimitriou)

Ioanna, note that $$1-1=0$$ and $$1-0=1$$, so there's no need for absolute value here since $$p(\alpha,\beta)$$ takes either $$1$$ or $$0$$ as a value. Regarding the intersection, you're right. This is the reason and it is in fact clearer to see it this way.

As for the second question. I know that in Blass' paper about SVC he points out that Pincus observed that if certain conditions hold for a set for which SVC holds, then the Boolean Prime Ideal theorem holds.

Beyond that I haven't really put too much thought into this question. I know that the closure of the forcing is essential for the "obvious proof", but under certain conditions can be omitted. I recall when I last thought about this, by looking at the relevant proofs in Jech "The Axiom of Choice", it seemed that $$\AC_\kappa$$ can be obtained by assuming the filter has a certain... breadth, which is some undefined measurement of something similar to closure. I can't really explain it, though. But it should be there. I'm not quite sure as for trichotomy for $$\kappa$$, but I think that a careful analysis can yield similar results.

All the above not withstanding, I don't believe there are "nice" optimal conditions, though, for either of these principles. (By the way, you can find proof of this preservation of $$\DC$$ in both my thesis and the paper I mentioned in the post; or you can do it yourself since it's quite straightforward.)

By
(Jul 28 2014, 11:22)

About the definition of the automorphisms, I was thinking about the case when $$(\alpha,\beta)\not\in A$$ and $$p(\alpha,\beta)=1$$.

About the restrictions on the symmetric model ingredients, yes, it is quite straighforward to see how it works for DC. I haven't thought about this before, so thanks for the remark and the pointers, I like this sort of results. I guess even non-optimal conditions would be nice!

By
(Jul 28 2014, 11:34 In reply to Ioanna M. Dimitriou)

Ah, you're right of course. I've corrected the definition.

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