"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:
- $A\in [\mathfrak{c}]^\omega$, $B\in [{}^A2]^{\le \omega}$,
- $u$ is a function with $\mathrm{dom}(u)\in[A]^\omega$,
- for every $\alpha\in\mathrm{dom}(u), u(\alpha)\in [{}^A2\setminus B]^{<\omega}$,
- (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.
- If $c_\xi\upharpoonright A_\beta\in B_\beta$, then let $d_\xi(\beta)=c_\xi(\beta)$.
- If $c_\xi\upharpoonright A_\beta\in u_\beta(\alpha)$ for some $\alpha\in \mathrm{dom}(u_\beta)$, then let $d_\xi(\beta)=c_\xi(\alpha)$.
- 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
- $f(\gamma)=f(\beta)$,
- $g(\gamma)\cap M=g(\beta)\cap M$,
- 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.)
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