The Milner-Rado paradox from walks on ordinals

Let $\kappa$ be a cardinal. The Milner-Rado paradox states:

Every ordinal $\beta<\kappa^+$ can be written as $\bigcup_{n<\omega} X_n$, where $X_n$ is of order-type at most $\kappa^n$.

It's surprising because it goes against our intuition of the pigeonhole principle: the ordinals $\kappa^n$ don't go very far up to $\kappa^+$.


There are some nice short proofs on mathstackexchange. But the one I find easiest to remember uses the method of walks on ordinals. For each $\alpha<\kappa^+$, fix an unbounded set $C_\alpha\subseteq \alpha$ of order-type at most $\kappa$ (and for successor ordinals $\alpha$ this just means that the predecessor must be in $C_\alpha$). Now suppose $\alpha\le\beta$. The walk from $\beta$ to $\alpha$ is a finite sequence decreasing of ordinals $\alpha_0>\alpha_1>\cdots>\alpha_n$ so that $\alpha_0=\beta$, $\alpha_{i+1}=\min(C_{\alpha_i}\setminus \alpha)$, and the walk ends at $\alpha_n=\alpha$. In words, we start from $\beta$ and follow $C_\beta$ down as close to $\alpha$ we can get without overshooting it, then jump to the set indexed by that ordinal and repeat the process. We must eventually get to $\alpha$ by the well-ordering principle, and call $n$ the length of the walk.

Now suppose $\beta<\kappa^+$. Let $X_n$ be the set of $\alpha$ so that the walk from $\beta$ to $\alpha$ has length $n$. A moment's thought shows that $X_n$ has order-type at most $\kappa^n$. So we have proved the Milner-Rado paradox!

History at the BGU awards ceremony

At the BGU department awards ceremony, a certain name appeared with spooky regularity, like subtle clues in a novel.

Profs. Miriam Cohen and Daniel Sternheimer at the ceremony in 2016.

This figure was mentioned in various connections by mathematicians working in different countries---Israel, France, Japan, and Belgium---and in different areas. He was a friend, a colleague, and an advisor. And the truth turns out to be even more interesting than the imagination. A "sabra bull in a china shop", Moshe Flato was also an Air Force lieutenant who worked on Israel's nuclear program, an outstanding pianist trained by Daniel Barenboim's mother, and a first-rate mathematician and physicist.

http://www-groups.dcs.st-and.ac.uk/history/Biographies/Flato.html

Balogh's small Dowker space, part 2

We are continuing the construction of Balogh's Dowker space of size continuum. From the paper, he says:

"The heart of the proof of Theorem 1.1 is the following combinatorial lemma."

 Lemma 1.2 [Balogh] Let $\lambda=2^{\mathfrak{c}}$, and let $\langle c_\xi:\xi<\lambda\rangle$ be a one-to-one enumeration of ${}^\mathfrak{c}2$. Then there is a sequence $\langle d_\xi:\xi<\lambda\rangle$ of functions $d_\xi:\mathfrak{c}\rightarrow 2$ in such a way that for every $g:\mathfrak{c}\rightarrow [\lambda]^{<\omega}$, $f:\mathfrak{c}\rightarrow \omega$ and $h:\mathfrak{c}\rightarrow [\mathfrak{c}]^{<\omega}$, there are $\alpha<\beta$ in $\mathfrak{c}$ such that $f(\alpha)=f(\beta)$, $\beta\not\in h(\alpha)$ and for every $\xi\in g(\alpha)$, $c_\xi(\alpha)=d_\xi(\beta)$.

1 Explanation

Quite a lot to take in! But let's break down how it works, given that we know we want to construct a system of filters satisfying $(*)$ and $(**)$ from last time.

First $(**)$: For any function $f:X\rightarrow \omega$ and any assignment $\alpha\mapsto A_\alpha\in \mathcal{F}_\alpha$, there are $\alpha\neq \beta$ so that $f(\alpha)=f(\beta)$ and $\beta\in A_\alpha$.

This is roughly what is expressed by Lemma 1.2. But we don't know how to build the filters yet.

We want to build them to satisfy $(*)$: For every $A\subseteq \mathfrak{c}$ there exists $B\subseteq \mathfrak{c}$ such that $B\in \mathcal{F}_\alpha \textrm{ if }\alpha\in A$ and $\mathfrak{c}\setminus B\in \mathcal{F}_\alpha \textrm{ if }\alpha\not\in A$.

Think of the sequence $\langle c_\xi:\xi<\lambda \rangle$ as enumerating the characteristic functions of all possible $A$. For each $A$, we will find a set $B$ as in $(*)$; its characteristic function is $d_\xi$.

We will use the $d_\xi$'s to generate our filters $\mathcal{F}_\alpha$. So for each $\xi$, $\{\beta:d_\xi(\beta)=1\}\in \mathcal{F}_\alpha$ if $c_\xi(\alpha)=1$ and $\{\beta:d_\xi(\beta)=0\}\in \mathcal{F}_\alpha$ if $c_\xi(\alpha)=0$.

What else has to be in the filter? We want to be able to take finite intersections of the sets given by $d_\xi$ or complement (depending on which side makes it into the filter). This is expressed by $g$. Also, the filter will contain all cofinite subsets of $\mathfrak{c}$, a necessary condition for neighborhood filters in a space where points are closed. This is expressed by $h$.

So Lemma 1.2 succintly expresses $(**)$ for filters satisfying $(*)$!

2 Proof of Lemma 1.2

Basic motivations

This section can be skipped if you are familiar with the use of elementary submodel arguments.

If we didn't care about the part involving the $d_\xi$'s the rest is very easy to arrange. By a standard argument, there is a closed unbounded set of $\beta<\mathfrak{c}$ closed under the function $h$; that is, for every $\alpha<\beta$, $h(\alpha)\subseteq \beta$. There is also a closed unbounded set of $\beta$ so that for every $n$ with $f^{-1}[n]$ unbounded, $f^{-1}[n]\cap \beta$ is unbounded in $\beta$. We will do similar, but more sophisticated arguments using the method of elementary submodels.

Let $\theta$ be a large enough regular cardinal ($(2^{2^{\mathfrak{c}}})^+$ suffices). One of the basic ideas of the elementary submodel method is that if $N\prec H(\theta)$, $N$ countable, and $\beta<\mathfrak{c}$ is greater than $\sup(N\cap \mathfrak{c})$ (an ordinal of countable cofinality), then formulas true of $\beta$ reflect to unboundedly many $\alpha$ in $N\cap\mathfrak{c}$. This is another way of stating what we did with clubs in the last paragraph.

But this idea is a bit more powerful. Suppose that we have $\beta$ and $N$ as above. Then there are unboundedly many $\alpha\in N\cap\mathfrak{c}$ which satisfy the desired properties for $f$ and $h$. Even more crucially, we can reflect some properties of $g(\beta)$, $d_\xi(\beta)$, and $c_\xi(\beta)$ down to these $\alpha$.

But we can't reflect all of those properties down, since there are $\lambda$ many $\xi$ to consider and not all of them will be in $N$, and also $g(\beta)$ is some subset of $\lambda$ that may not be in $N$. But we will reflect enough of this information down, given by what happens inside some smaller countable elementary submodel $M\in N$.

So we know what happens inside this $M$. But there is a part outside of $M$ as well. This, we will handle by knowing that there are many reflections---so many that we will be able to guess in advance what happens on at least one of them.

That's the basic idea, but we need to make sure the construction can be carried out in $\mathfrak{c}$ many steps, while we are constructing $\lambda$-many of these $d_\xi$. The construction outlined above depends in part on the choice of submodels, but not on the full information of $M$ and $N$---we shall see that there are only $\mathfrak{c}$ many choices for the crucial information here, and the construction is in some sense canonical. That is part of the magic of this technique.

Definition of the d's

We have the motivation in mind now. We will enumerate control triples $\langle (A_\beta,B_\beta,u_\beta):\beta<\mathfrak{c}\rangle$, which are ordered triples $(A,B,u)$ that satisfy the following properties:

1.  $A\in [\mathfrak{c}]^\omega$, $B\in [{}^A2]^{\le \omega}$,
2. $u$ is a function with $\mathrm{dom}(u)\in[A]^\omega$,
3. for every $\alpha\in\mathrm{dom}(u), u(\alpha)\in [{}^A2\setminus B]^{<\omega}$,
4. (disjoint images) if $\alpha\neq \beta$ in $\mathrm{dom}(u)$, then $u(\alpha)\cap u(\beta)=\emptyset$.

Furthermore, let us ensure in the enumeration that $\beta>\sup A_\beta$.

From the sketch above, $A$ corresponds to $N\cap \mathfrak{c}$, $B$ corresponds to the information we need from $M$, and $u$ captures the information that's not in $M$. But there are only $\mathfrak{c}$ such triples!

Suppose now that $\xi<\lambda$. We will define $d_\xi$. For each $\beta<\mathfrak{c}$, there are three cases.

1. If $c_\xi\upharpoonright A_\beta\in B_\beta$, then let $d_\xi(\beta)=c_\xi(\beta)$. 
2. If $c_\xi\upharpoonright A_\beta\in u_\beta(\alpha)$ for some $\alpha\in \mathrm{dom}(u_\beta)$, then let $d_\xi(\beta)=c_\xi(\alpha)$.
3. Otherwise, set $d_\xi(\beta)=0$.

Intuitively, Case 1 corresponds to the case when $\xi$ is in the part which is controlled by $M$, and Case 2 to when $u$ captures the information outside of $M$ of a reflected $\alpha$. Note that in Case 2, $c_\xi\upharpoonright A_\beta\not\in B_\beta$, and $\alpha$ is chosen uniquely by the restrictions on $u$ in the definition of a control triple.

Final proof

We will show that the sequence of $d_\xi$ we have defined above works. Let $f,g,h$ be functions as in the statement of Lemma 1.2. We will produce the required $\alpha,\beta$.

Let $M\in N$ be elementary submodels of $H(\theta)$ containing $\langle c_\xi:\xi<\lambda\rangle,  \langle d_\xi:\xi<\lambda\rangle, f,g,h$. Let $A=\mathfrak{c}\cap N$ and $B=\{c_\xi\upharpoonright A:\xi\in \lambda\cap M\}$.

We will construct $u:A\rightarrow[{}^A2\setminus B]^{<\omega}$ satisfying the requirements of the control triple such that whenever $v\in N$ is an infinite partial function $\mathfrak{c}\rightarrow [\lambda\setminus M]^{<\omega}$ and $\alpha\neq \alpha'$ in $\mathrm{dom}(v)$ implies that $v(\alpha)\cap v(\alpha')=\emptyset$, then there is $\alpha\in\mathrm{dom}(u)\cap\mathrm{dom}(v)$ such that 

$$u(\alpha)=\{c_\xi\upharpoonright A:\xi\in v(\alpha)\}.$$

Basically, $u$ agrees with $v$ on at least one point of its domain for any $v$ of the right shape. And here, agreement actually means $u(\alpha)=\{c_\xi\upharpoonright A:\xi\in v(\alpha)\}$, since $u$ cannot actually take the ordinals less than $\lambda$ in its range, as we were careful to only use $\mathfrak{c}$-many control triples.

This $u$ is easy to construct. Just enumerate the countably many such $v\in N$ and construct $u^*:\mathfrak{c}\rightarrow [\lambda\setminus M]^{<\omega}$ so that the disjoint images property holds. Then get $u$ by taking restrictions of the $c_\xi$ to $A$. The only thing that requires some argument is to make sure that by taking the restrictions, we do not accidentally break the disjoint images property for $u$. But we will not, since the relevant $\xi$ are in $N$ and if $\xi\neq\xi'$ then this is witnessed by something in $A=N\cap\mathfrak{c}$. 

Now pick $\beta$ so that $\langle A_\beta,B_\beta,u_\beta\rangle=\langle A,B,u\rangle$. Say that $\gamma$ reflects $\beta$ if

1. $f(\gamma)=f(\beta)$,
2. $g(\gamma)\cap M=g(\beta)\cap M$,
3. for every $\xi\in g(\gamma)\cap M$, $c_\xi(\gamma)=c_\xi(\beta)$.

Find a maximal $D$ which consists of $\gamma$ reflecting $\beta$ so that $\langle g(\gamma):\gamma\in D\rangle$ forms a $\Delta$-system with root $r:=g(\beta)\cap M$. Choose $D$ in $M$, possible since the definition of reflection only used parameters in $M$. This $D$ is uncountable since otherwise it would be a subset of $M$, but $\beta$ could be then be added to it, contradicting maximality. So there is also an infinite set $H$ of $\gamma\in D\cap N$ so that $g(\gamma)\setminus r$ is disjoint from the countable set $\lambda\cap M$.

Now define $v:H\rightarrow [\lambda\setminus M]^{<\omega}$ by $v(\gamma)=g(\gamma)\setminus r$. So $v\in N$ and there is $\alpha\in \mathrm{dom}(u)\cap\mathrm{dom}(v)$ with $u(\alpha)=\{c_\xi\upharpoonright A:\xi\in v(\alpha)\}$. This is the $\alpha$ we want.

Let's check: since $\alpha$ reflects $\beta$, $f(\alpha)=f(\beta)$. Since $\alpha\in N$, $h(\alpha)\subseteq N$. But $\beta>\sup(A_\beta)$, so $\beta\not\in h(\alpha)$. Finally, if $\xi\in g(\alpha)$, there are two cases depending on if $\xi$ is in the root of the $\Delta$-system or not. In the first case, $\xi\in r$, and then we defined $d_\xi(\beta)=c_\xi(\beta)=c_\xi(\alpha)$. In the second, $c_\xi\upharpoonright A\in u_\beta(\alpha)$, so we directly defined $d_\xi(\beta)=c_\xi(\alpha)$.


(These are notes from a seminar given at Bar-Ilan University on December 4, 2017.)

Balogh's small Dowker space, part 1

A Dowker space is a normal space which is not countably paracompact. Here, a (Hausdorff) space $X$ is normal if for every pair of disjoint closed sets $H,K$, there exist disjoint open sets $U,V$ with $H\subseteq U$ and $K\subseteq V$. Normality is, according to M.E. Rudin, the boundary where point-set topology changes from analysis to set theory. 

A space $X$ is countably paracompact if for every $\subseteq$-increasing sequence of open sets $\langle G_n:n<\omega\rangle$ with $\bigcup_n G_n=X$, there exist closed sets $H_n\subseteq G_n$ so that $\bigcup_n H_n=X$. An interesting history of the concept of paracompactness can be found in the Encyclopedia of General Topology.

It turns out that Dowker spaces are the spaces whose normality is not preserved by product with compact metric spaces. See Dan Ma's topology blog for more information about Dowker spaces and all of the topological concepts considered here.

Dowker spaces are considered to be quite rare, and the first ZFC example was due to M.E. Rudin in 1971. In 1996, Zoltan Balogh published an example of a Dowker space of size continuum in ZFC. The problem of whether there exists a Dowker space of size $\aleph_1$ is still open.

Balogh's example was based off an earlier example of Watson which used strongly compact cardinals, together with an earlier argument of Rudin using the submodel method. He was able to use the techniques applied in this Dowker space construction towards other Dowker spaces and the solutions of the second and third Morita conjectures. This post gives a carefully motivated presentation of Balogh's proof.

1 Basic construction

We start the construction of $X$ with an increasing sequence of open sets $\langle G_n:n<\omega\rangle$ witnessing the failure of countable paracompactness. So the underlying set of $X$ is $\omega\times\mathfrak{c}$, and we define $G_n=(n+1)\times\mathfrak{c}$. We will also consider the levels of the space, $L_n=\{n\}\times\mathfrak{c}$.

The levels $L_n$ will be relatively discrete, which helps in the normality proof later. The set of open neighborhoods of a fixed point $(n,\alpha)$ forms a filter. If $n=0$, a neighborhood base for $(n,\alpha)$ is just the singleton. If $n>0$, the filter concentrates on $\{(n,\alpha)\}\cup G_{n-1}$. Following Watson's example, we will define a filter $\mathcal{F}_{(n,\alpha)}$ on $G_{n-1}$. Then, a set $U$ is open if and only if for every $(n,\alpha)\in U$, there is some $A\in \mathcal{F}_{(n,\alpha)}$ so that $A\subseteq U$.

Motivated by the normality argument in Watson's paper (which he credits to Rudin), we will end up working on consecutive pairs of levels. So the filter $\mathcal{F}_{(n,\alpha)}$ will be expressed as a filter $\mathcal{F}_\alpha$ on $\mathfrak{c}$ which does not depend on $n$. The filter will contain all cofinite subsets of $\mathfrak{c}$, which is necessary in a space where points are closed.

Let us find the properties of this filter system which we need to make $X$ Dowker. 

2 Normality

Let us try to get normality. The first level $G_0$ is relatively normal, since it has the discrete topology. If we consider the first two levels, however, we need to separate arbitrary disjoint sets $A_0,A_1\subseteq L_1$ by disjoint open sets. We may as well assume that $A_1$ is the complement of $A_0$ in its level. In total, we are looking for: 

$$(*) \textrm{For every }A\subseteq \mathfrak{c}, \textrm{ there exists } B \subseteq \mathfrak{c} \textrm{ such that }$$

$$B\in \mathcal{F}_\alpha \textrm{ if }\alpha\in A,$$

$$\mathfrak{c}\setminus B\in \mathcal{F}_\alpha \textrm{ if }\alpha\not\in A.$$

For the next few claims, assume $(*)$.  We will show that $X$ defined as above is normal.

First, we show this for subsets of single levels.

Claim 1. Suppose that $m\le n$ and $H\subseteq L_n$ and $K\subseteq L_m$ have disjoint closures. Then there are disjoint open sets $U,V$ separating them ($H\subseteq U$ and $K\subseteq V$). Moreover, we can arrange so that $L_n\setminus H\subseteq K$.

Proof of Claim 1: Using $(*)$ repeatedly, let $U',V'$ be disjoint open sets so that $L_m\setminus K\subseteq U'$ and $K\cap L_m\subseteq V'$. Then take $U=U'\cup(X\setminus (G_m\cup \bar{K}))$ and $V=V'$. $U$ is open since $U'$ and $X\setminus \bar{K}$ are open sets which are spliced together at a single level $m$, and $U'\cap L_m=X\setminus K=(X\setminus \bar{K})\cap L_m$. Clearly $K\subseteq V$, and $H\subseteq U$ since $H\cap \bar{K}=\emptyset$. $\square$

Now we can separate a subset of a single level from an arbitrary closed set. Suppose that $H,K$ are disjoint closed subsets of $X$. By taking only finite unions and intersections from the sets produced from Claim 1, we can show that there are disjoint open $U'_n,V'_n$ and disjoint open $U''_n,V''_n$, all subsets of $G_n$, so that 

• $H\cap L_n\subseteq U'_n$, $K\cap G_n\subseteq V'_n$, and $L_n\setminus H\subseteq V'_n$.
• $H\cap G_n\subseteq U''_n$, $K\cap L_n\subseteq V''_n$, and $L_n\setminus K\subseteq U''_n$.

Finally, take $$U_n:=\bigcup_n (U'_n\setminus \bigcup_{k\le n}\mathrm{cl}(V''_k))$$ and $$V_n:=\bigcup_n (V''_n\setminus \bigcup_{k\le n}\mathrm{cl}(U'_k)).$$  

These sets are clearly open, and disjoint by the subtraction done at each step. 

For each $n$, $H\cap L_n\subseteq U'_n\setminus \bigcup_{k\le n}\mathrm{cl}(V''_k)$ and $K\cap L_n\subseteq V''_n\setminus \bigcup_{k\le n}\mathrm{cl}(U'_k)$. This is because $V^*_n:=V'_n\cup(X\setminus (G_n\cup H))$ and $U^*_n:=U''_n\cup(X\setminus (G_n\cup K))$ are open, by the splicing argument. Furthermore, we have $K\subseteq V^*_n$ and $H\subseteq U^*_n$ and $V^*_n\cap U'_n=U^*_n\cap V''_n =\emptyset$. 

So we have reduced normality to the property $(*)$ of the filter system.

Remark: In fact, the proof shows that $X$ built from a system satisfying $(*)$ is hereditarily normal.

3 Non countable paracompactness

To get non-countable paracompactness, we need to show that every sequence of closed sets $\langle H_n:n<\omega\rangle$ with $H_n\subseteq G_n$ cannot have union $X$. The levels of each $F_n$ are small in the following sense. $H_n\cap L_n$ must be measure 0 according to $\mathcal{F}_\alpha$ (that is, the complement is in the filter) for each $\alpha\in\mathfrak{c}$. Otherwise, if $H_n\cap L_n$ is positive measure according to $\mathcal{F}_\alpha$, then $(n+1,\alpha)$ is in its closure, contradicting that $H_n\subseteq G_n$ closed. For each $n$, let us define the ideal $\mathcal{I}^1$ to be the ideal of sets which are measure 0 according to $\mathcal{F}_\alpha$ for every $\alpha\in\mathfrak{c}$.

Now if $n>0$, $H_n\cap L_{n-1}$ is measure 0 according to $\mathcal{F}_\alpha$ for all $\alpha$ except for a set in $\mathcal{I}^1$. Let $\mathcal{I}^2$ be the ideal of sets which have this property, and recursively define $\mathcal{I}^m$ for every $m<\omega$. Let $\mathcal{I}$ be the $\sigma$-ideal generated by $\bigcup_{m<\omega} \mathcal{I}^m$. For technical reasons, let $\mathcal{I}^0=\{\emptyset\}$. 

We will be done if we can show that $\mathcal{I}$ is proper, since $\{\alpha:(0,\alpha)\in H_n\cap L_0\}\in\mathcal{I}$.

Suppose otherwise. Then there is a partition of $\mathfrak{c}$ into countably many sets, each in $\mathcal{I}^m$ for some $m$. By partitioning further, we can assume for each class $C$ of the partition, $C\in \mathcal{I}_m$ is witnessed by a set $D\in \mathcal{I}_{m-1}$, so that $C\cap D=\emptyset$ and $\mathfrak{c}\setminus C\in \mathcal{F}_\alpha$ for all $\alpha\not\in D$ (any set in $\mathcal{I}_m$ has such a $D$, and if $C\cap D$ is not empty, then it is in $\mathcal{I}_{m-1}$ and we can subtract off this part; repeat the procedure finitely many times to get the further partitioning). Now to each $\alpha\in C$, we can associate a set $A_\alpha=\mathfrak{c}\setminus C\in \mathcal{F}_\alpha$.

We want to ensure that there are no such partitions. So we want:

$(**)$ For any function $f:X\rightarrow \omega$ and any assignment $\alpha\mapsto A_\alpha\in \mathcal{F}_\alpha$, there are $\alpha\neq \beta$ so that $f(\alpha)=f(\beta)$ and $\beta\in A_\alpha$.

Then, $\alpha$ and $\beta$ will be in the same class $C$, but we will have $\beta\in A_\alpha$, contradicting that $A_\alpha$ was supposed to be disjoint from $C$.

We have reduced non countable paracompactness to $(**)$. Next time we will see how to get $(*)$ and $(**)$.

Note that in this part, $\mathfrak{c}$ here can be replaced by any infinite cardinal $\kappa$.

We will continue Balogh's proof in Part 2.

(These are notes from a seminar given at Bar-Ilan University on November 27, 2017.)

New material coming

I have been in Israel at Ben-Gurion University for the past year (and plan to stay here for one more year), and I'm working on some new things. Look forward to more posts soon!

UCI Summer School, part 7: Sacks forcing (Brent Cody)

This lecture will introduce some basic properties of Sacks forcing for uncountable inaccessible cardinals, and examine an Easton support iteration of such forcing.

The Sacks forcing on $\omega$ adds a real of minimal constructibility degree, and crucially satisfies a fusion property. Although this was reviewed in the summer school, I'm going to omit the discussion for this post.

Instead we will start with Sacks forcing on uncountable cardinals, which traces back to Kanamori (1980), where using $\diamond_\kappa$ it was shown that long products and iterations of $\mathrm{Sacks}(\kappa)$ preserve $\kappa^+$.

Definition: We say $p\subseteq 2^{<\kappa}$ is a perfect $\kappa$-tree if:

1. If $s\in p$ and $t\subseteq s$ then $t\in p$.
2. If $\langle s_\alpha:\alpha<\eta\rangle$ is a sequence of nodes in $p$, then $s=\bigcup_{\alpha<\eta} s_\alpha\in p$.
3. For every $s\in p$ there is $t\supset s$ with $t\frown 0, t\frown 1\in p$.
4. Let $\mathrm{Split}(p)=\{s\in p: s\frown 0, s\frown 1\in p\}$. Then for some unique club $C(p)\subseteq \kappa$, we have $$\mathrm{Split}(p)=\{s\in p: \mathrm{length}(s)\in C(p)\}.$$

$\mathrm{Sacks}(\kappa)$ is the poset of perfect $\kappa$-trees ordered by inclusion. We think of the generic subset of $\kappa$ added by $\mathrm{Sacks}(\kappa)$ as the intersection of the trees in the generic filter.

The only surprising thing in the generalization is (4): splitting happens for every node on certain levels, which form a club in $\kappa$.

Exercise: $\mathrm{Sacks}(\kappa)$ is $<\kappa$-closed.

Assume $\kappa>\omega$ is inaccessible. Then $\mathrm{Sacks}(\kappa)$ is $\kappa^{++}$-c.c. We will really only consider this case.

Definition: $\mathrm{Split}_\alpha(p)$ is the set of all nodes $s\in p$ with $\mathrm{length}(s)=\beta_\alpha$, where $\langle \beta_\alpha:\alpha<\kappa\rangle$ is an enumeration of $C(p)$, i.e., the level of $p$ at the $\alpha$th member of $C(p)$.

For $p,q\in \mathrm{Sacks}(\kappa)$, write $p\le_\beta q$ iff $p\le q$ and $\mathrm{Split}_\alpha(p)=\mathrm{Split}_\alpha(q)$ for all $\alpha<\beta$.

A descending sequence $\langle p_\alpha:\alpha<\kappa\rangle$ in $\mathrm{Sacks}(\kappa)$ is a fusion sequence if for all $\alpha<\kappa$, $p_\alpha\le_\alpha p_\alpha$.

Lemma (fusion lemma): If $\langle p_\alpha:\alpha<\kappa\rangle$ is a fusion sequence, then $p=\bigcap_{\alpha<\kappa} p_\alpha$ is a lower bound in $\mathrm{Sacks}(\kappa)$.

Proof: exercise. Hint: show that any node $p$ in the intersection is in a cofinal branch of the intersection.

This important lemma affords us a kind of $\kappa^+$ closure, with the catch that we require more of our decreasing sequence. We can see this in action in the next lemma.

Lemma: $\mathrm{Sacks}(\kappa)$ preserves $\kappa^+$.

Proof: If $\dot{f}$ is the name of a function $\kappa\rightarrow \kappa^+$, then we will find $q\le p$ with $q\Vdash \mathrm{ran}(\dot{f})$ bounded. 

Let $p_0=p$. Given $p_\alpha$, for each $s\in \mathrm{Split}_\alpha(p_\alpha)$, let $\bar{r}^s_\alpha\le (p_\alpha)_s$ be such that $\bar{r}^s \Vdash \dot{f}(\alpha)=\eta^s_\alpha$. Here the $(p)_s$ means the subtree of $p$ of nodes compatible with $s$. 

Note $\bigcup\{\bar{r}^s_\alpha:s\in \mathrm{Split}_\alpha(p_\alpha)\}$ might not be a condition by the requirement on splitting levels. Let $C=\bigcap \{C(\bar{r}^s_\alpha:s\in \mathrm{Split}_\alpha(p_\alpha)\}$ and thin each $\bar{r}^s_\alpha$ to some $r^s_\alpha\le \bar{r}^s_\alpha$ with $C(r^s_\alpha)=C$.

At limits $\gamma<\kappa$, let $p_\gamma=\bigcap_{\alpha<\gamma} p_\alpha$ by the fusion lemma. This defines a fusion sequence where the limit forces that the range of $f$ is bounded.

Exercise: Suppose ${}^\kappa M\subseteq M$, for an inner model $M$. Suppose $\mathrm{Sacks}(\kappa)\in M\subseteq V$. If $G$ is $V$-generic for $\mathrm{Sacks}(\kappa)$ then ${}^\kappa M[G]\subseteq M[G]$ in $V[G]$.

Note: This holds for $\kappa^+$-c.c. forcing, but $\mathrm{Sacks}(\kappa)$ is not $\kappa^+$-c.c.

Now we will see what happens when we iterate these Sacks forcings with Easton support below, and at, a measurable cardinal $\kappa$. Think of this like a Sacks forcing version of the Kunen-Paris iteration, where we use the nice fusion property to replace the $\gamma^+$ closure of the factors there.

Theorem (Friedman-Thompson 2008): Assume GCH holds. Suppose $\kappa$ is measurable and let $\mathbb{P}$ be the length $\kappa+1$ Easton support iteration with $\mathbb{Q}_\gamma=\mathrm{Sacks}(\gamma)$ (computed in $V^{\mathbb{P}_\gamma}$) for $\gamma\le \kappa$ inaccessible, and $\mathbb{Q}_\gamma$ is trivial forcing otherwise. Then if $G\ast H$ is $V$-generic for $\mathbb{P}= \mathbb{P}_\kappa\ast \dot{\mathbb{Q}}_\kappa$, then every normal ultrapower lifts to $V[G\ast H]$ (and in a particularly interesting way!)

Proof: Let $j:V\rightarrow M$ be a normal ultrapower by $U\in V$. Then $j(\mathbb{P}_\kappa=\mathbb{P}_\kappa\ast \dot{\mathbb{Q}}_\kappa\ast \dot{\mathbb{P}}_{\kappa+1,j(\kappa)}$. We get the actual $\dot{\mathbb{Q}}_\kappa$ factor at the $\kappa$ step by using the $\kappa$ closure of the ultrapower.

Using this closure further, and the last exercise, ${}^\kappa M[G\ast H]\subseteq M[G\ast H]$ in $V[G\ast H]$, so $M[G\ast H] \vDash \dot{\mathbb{P}}_{\kappa+1,j(\kappa)}\textrm{ is }\le \kappa-\textrm{closed.}$ So there are $\kappa^+$ maximal antichains of $\mathbb{P}_{\kappa,j(\kappa)}$ in $M[G][H]$. We can now build as usual a generic $G_{\kappa+1,j(\kappa)}\in V[G\ast H]$ for $\mathbb{P}_{\kappa+1,j(\kappa)}$ over $M[G\ast H]$. Lift to $j:V[G]\rightarrow M[j(G)]$. 

Now we have to lift $j$ through $\mathbb{Q}_\kappa=\mathrm{Sacks}(\kappa)$. Using the Silver method, $j``H$ has size $\kappa^+$, but the target model $M[j(G)]$ does not have this much closure.

The crucial point is to just take $t:=\bigcap j``H$. We claim that $t$ is a "tuning fork": by this we mean that $t$ consists of a single branch up to the level $\kappa$, at which point it splits into two branch which are cofinal (and that's everything in $t$).

1. The function $f:\kappa\rightarrow 2$ determined by $H$ is in $t$, and this is everything in $t$ below $\kappa$.
2. Every condition in $j``H$ splits at $\kappa$ since for each $p\in H$, $p$ splits at club many levels below $\kappa$, and therefore $j(p)$ splits at level $\kappa$. Therefore, $f\frown 0,f\frown 1\in t$.
3. Since $H$ is a filter, $t$ is cofinal in $j(\kappa)$.
4. We will argue that $t$ does not split anywhere else. Given a club $C\subseteq \kappa$, $D_C:=\{p\in \mathrm{Sacks}(\kappa): C(p)\subseteq C\}$ is dense. So there must be $p_C\in H$ so that $C(p_C)\subseteq C$. Now we have:

Claim: $X=\bigcap \{j(C):C\subseteq \kappa \textrm{ club in } V[G]\}=\{\kappa\}$.

Proof of Claim: Clearly $\kappa\in X$. For the other inclusion, suppose $\alpha\in X$, $\alpha>\kappa$. Then choose $f:\kappa\rightarrow \kappa$, $f\in V[G]$ so that $j(f)(\kappa)=\alpha$. Then let $C_f=\{\nu<\kappa: f``\nu\subseteq \nu\}$ is club, but $\alpha\not\in j(C_f)$ since $\alpha$ is not a closure point of $j(f)$ ($\kappa<\alpha$ maps to $\alpha$). This proves the claim.

Let $t_0, t_1$ be the leftmost and rightmost branches through $t$, respectively. Let $K_0=\{p\in j(\mathbb{Q}_\kappa):t_0\subseteq p\}$. Clearly $j``H\subseteq K_0$.

It remains to show that $K_0$ is $M[j(G)]$-generic for $j(\mathbb{Q}_\kappa)$. Let $D$ be a dense open subset of $j(\mathbb{Q}_\kappa$ in $M[j(G)]$. Then there is a sequence $\vec{D}=\langle D_\alpha:\alpha<\kappa\rangle \in V[G]$ such that $j(\vec{D})_\kappa=D$, where each $D_\alpha$ is a dense open subset of $\mathbb{Q}_\kappa$. 

Claim: Every condition $p\in \mathrm{Sacks}(\kappa)$ can be extended to $q_\infty \le p$ so that for every $\alpha<\kappa$ there is $\beta<\kappa$ so that for any node $s\in \mathrm{Split}_\beta(q_\infty)$, the condition $(q_\infty)_s$ meets $D_\alpha$. 

Proof of Claim: exercise, a fusion argument.

Let $q_\infty\in H$ be as in the claim, using genericity of $H$. By elementarity, $j(q_\infty)$ has the property that at some splitting level of $j(q_\infty)$, say $\beta<j(\kappa)$, any node $s\in \mathrm{Split}_\beta(j(q_\infty))$ is such that $(j(q_\infty))_s$ meets $D$. Now we can just take $s$ to be $t_0\upharpoonright \delta_\beta$, where $\delta_\beta$ is the $\beta$th splitting level of $j(q_\infty)$.

Therefore $K_0$ is generic as claimed, and it is in $V[G\ast H]$, so $j$ lifts to

$$j:V[G\ast H]\rightarrow M[j(G)\ast j(H)].$$  

The rearrangement inequality is everywhere

Recently, I've been talking with John Susice about some elementary Olympiad-style problems. I told him that in high school I was taught that virtually every inequality problem that appeared in this setting follows from the Rearrangement Inequality. This states that if $x_1\le \cdots\le x_n$ and $y_1\le \cdots\le y_n$ are real numbers, then the expression
$$x_1 y_{\sigma(1)}+\cdots+x_n y_{\sigma(n)}$$
for $\sigma$ a permutation on $[n]$ is maximized when $\sigma$ is the identity permutation and minimized when $\sigma$ is the reversing permutation. To me, this neatly isolates a useful and general principle that seems kind of obvious in hindsight.

The proof of the $n=2$ case of the inequality, which is all I'll need below, is just to expand $(x_2-x_1)(y_2-y_1)\ge 0$. This case also implies the AM-GM inequality (taking $x_1=y_1=\sqrt{a}$ and  $x_2=y_2=\sqrt{b}$).

Now sometime later, John told me an interesting problem about factoring numbers. He had a solution using a trick similar to Euler's product, but for me this was a chance to test my thesis about the rearrangement inequality:

Problem: Prove that every natural number $n$ has at least as many factors congruent to 1 (mod 4) as factors congruent to 3 (mod 4).

Solution: By induction. Clearly this holds for 1 and for primes. Now suppose $n$ is composite, so write $n=pq$ where $p,q<n$. For any integer $k$, let $r_k$ denote the number of factors it has which are congruent to 1 (mod 4) and $s_k$ the number of factors congruent to 3 (mod 4).

The product of two 1 (mod 4) numbers is still 1 (mod 4), and the product of two 3 (mod 4) numbers is 3 (mod 4). On the other hand, the product of a 1 (mod 4) and a 3 (mod 4) is a 3 (mod 4), and even factors can never be 1 or 3 (mod 4). This proves that $r_n=r_pr_q+s_ps_q$, which must be greater than or equal to $s_n=r_ps_q+r_qs_p$ by the rearrangement inequality!