No uniform ultrafilters

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.


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:
\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}.

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. $\qquad\square$

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.

Why Carl Sagan was better than Neil deGrasse Tyson, and from the most of us too

I’ve recently watched the finale of Cosmos, the new version, presented by Neil deGrasse Tyson. It was a very nice series which seem to push forward the fact that science is based on not knowing, rather than knowing, and the will to know. No, not will, the need to know. We need to know, and this is why we go on searching the answers to questions that haunt us.

Neil deGrasse Tyson pushed a lot on the point that we really push the planet to its limits, and we might be close to the point of no return from which there is only a terrible Venus-like fate to this planet. And that is an important issue, no doubt.

But in the finale, a recording of Carl Sagan was given, narrating over “The Pale Blue Dot” after which deGrasse Tyson gave his final speech. DeGrasse Tyson gave a speech about the importance of science, about how it can better our lives and that science means not knowing all the answers but looking for them endlessly and beyond. Yes, this is an important note. But his speech was a pale blue dot compared to that of Sagan.

It touches me, especially in these troubled times in Israel. Where sirens are ablaze and an ongoing war goes on and on and on. Sagan, in his awe inspiring speech, reminds us how insignificant we are. How little all this matters, and how foolish we are to keep engaging these battles. Here, and everywhere else. But I can’t win Sagan, so instead, I will give you, Carl Sagan and the “Pale Blue Dot”.

You can also read his speech on the Wikipedia entry of the Pale Blue Dot.

I will point out in advance, that any attempt to bring political issues in the comments will be promptly deleted.