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I've a problem with a passage of the proof of Claim 14.7 of the paper "Cofinality spectrum theorems in model theory, set theory, and general topolgy" by Malliaris and Shelah, or equivalently Proposition 4D of Fremlin "p=t, following Malliaris-Shelah and Steprans", https://www1.essex.ac.uk/maths/people/fremlin/n14528.pdf (I'll refer to this second one). I have problems with the argument by contradiction inside the proof (starting from "P?"). In particular, I don't understand why $D\in G$: I know that by definition $D\Vdash \check{D}\in F$ for every $F$ generic over $\mathfrak{M}$ with $D\in F$, but in this case to me it seems that $G$ was fixed at the beginning, and then the sequence $\langle \dot{p}_\varepsilon\rangle_{\varepsilon<\alpha}$ was fixed accordingly to the chosen $G$ and thus we cannot chose again a different $G$ so that $D\in G$.

I will explain my difficulty with an example:

Let $\mathfrak{M}$ be a (CT) model of ZFC, $(B,\leq,\land,\lor,\neg,0,1)\in\mathfrak{M}$ a complete boolean algebra and let $G\subset B$ be a fixed ultrafilter generic over $\mathfrak{M}$ and $\mathfrak{M}[G]$ the generic model extending $\mathfrak{M}$ and containing $G$ obtained using forcing. Then $\mathfrak{M}[G]$ is also a model of ZFC and $G\in\mathfrak{M}[G]$.

Given $A\subset B$ and $x\in B$, use the notation $x\leq A$ to denote $\forall a\in A[x\leq a]$. Working in $\mathfrak{M}[G]$, find a subset $A\subset G$ such that $A$ is unbounded in $G$, that is $\forall x\in B[x\leq A\to x\notin G]$ (e.g. by induction or taking $A$ to be a maximal chain in $G$). Let $\check{A}$ be the $B$-name of $A$ and $\check{G}$ be the $B$-name of $G$. Then by forcing theorem, $\mathfrak{M}[G]\vDash A\subset G$ if and only if there exists a $p\in G$ such that $p\Vdash \check{A}\subset \check{G}$.

Since it is not possible that $p\leq A$, hence $p\nleq A$ that means there exists $a\in A$ such that $(p\land a)<p$ and thus $((\neg a)\land p)>0$. But now $((\neg a)\land p)<p$ and thus $((\neg a)\land p) \Vdash \check{A}\subset \check{G}$ although $((\neg a)\land p)\notin G$.

So here for example we should have a formula that depended on a certain fixed $G$, true in $\mathfrak{M}[G]$, and is forced true also by $p\notin G$ and this does not cause contradiction.

I don't understand if there is a mistake in my example or if the situation in the proof of Fremlin (and Malliaris-Shelah) is a different context so things behave differently from the example, and why. Can someone give me a hint?


In particular, to me the context in Fremlin's (Malliaris-Shelah's) proof seems closely related to the previous example for the following reason:

Define $D_{\eta,\varepsilon}=\{n\in\omega:(p_{\eta})_n<(p_{\varepsilon})_n\}$, and $A=\{D_{\eta,\varepsilon}:\eta<\varepsilon<\alpha\}$ (they depend on the starting sequence). To ask that $\langle \dot{p}_\varepsilon\rangle_{\varepsilon<\alpha}$ is an increasing sequence in $\prod_{n\in\omega} \dot{P}_n|G$ means to ask $A\subset G$. If I understood correctly, if $E_\varepsilon$ are the sets defined in the proof by contradiction of Fremlin, we have $E_\eta\setminus E_\varepsilon$ finite for every $\varepsilon<\eta<\alpha$ if and only if $A$ is not unbounded in $G$ and $C$ (defined in Fremlin) is a bound for $A$. I cannot see why for any starting sequence $\langle \dot{p}_\varepsilon\rangle_{\varepsilon<\alpha}$ is impossible for such an $A$ to be unbounded in $G$. I know this would be trivial if $A\in\mathfrak{M}$, but this should not be always the case if I'm not wrong.


EDIT: For Malliaris-Shelah notation, the problem is the following: at the end of Claim 14.7, it is said that $Y_\beta\subseteq^\ast Y_\alpha$ for every $\alpha<\beta$. The argument is that otherwise that would be $B'\subseteq^\ast B$ contradicting $B\Vdash_\mathcal{Q}(\mathcal{N}\vDash "f_\alpha/G\trianglelefteq f_\beta/G")$. I don't get why such a $B'$ should contradict that statement: if $B'\in G$ this would cause contradiction. But I cannot figure out why this should be so.

The translation "Malliaris-Shelah$\to$Fremlin" should be the following: $B$ is $C$, $B'$ is $D$, $Y_\beta$ is $E_\beta$, $f_\beta/G$ is $\dot{p}_\beta$ and $\langle f_\alpha/G:\alpha<\theta\rangle$ is $\langle \dot{p}_\varepsilon\rangle_{\varepsilon<\alpha}$, while the definable downward closed subtree of $({\omega}^{<\omega})$ in which Malliaris and Shelah are working has been replaced by $\prod_{n\in\omega} \dot{P}_n|G$ and $s_n$ intuitively should be $\dot{P}_n$.

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  • $\begingroup$ I am familiar with the original paper by Malliaris and Shelah, but not with the second reference you quote. Are you able to phrase the problem in the notation of Malliaris and Shelah? $\endgroup$ – Douglas Ulrich Sep 1 '18 at 6:52
  • $\begingroup$ Sure. I will edit the question to add this. $\endgroup$ – Cla Sep 1 '18 at 9:13
  • $\begingroup$ If it's incomprehensible stated this way really shortly, I can add the translation into the new notation of everything, but this would make the question a lot longer. If needed I'll do it. $\endgroup$ – Cla Sep 1 '18 at 9:43
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I think the problem is slightly more basic. Fremlin has the fully correct $$ D \Vdash \check D\in\dot{\mathcal{G}} $$ Also, Fremlin did not fix one generic $G$ at the outset; he works with names and the forcing relation everywhere.

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Suppose $\alpha < \beta$ and yet $Y_\beta \not \subseteq_* Y_\alpha$. Let $B'$ be the infinite set of all $n \in B$ such that $Y_{\beta, n} := Y_\beta \cap (\{n\} \times \omega^{<\omega}) \not \subseteq Y_{\alpha, n} := Y_\alpha \cap (\{n\} \times \omega^{<\omega})$. For each $n \in B'$, choose $t_n \in Y_{\beta, n} \backslash Y_{\alpha, n}$. Then $t_n$ extends $f_\beta(n)$ but does not extend $f_\alpha(n)$, so $f_\alpha(n) \not \trianglelefteq f_\beta(n)$, so $B'$ forces that $f_\alpha/\dot{G} \not \trianglelefteq f_\beta/\dot{G}$.

This is more intuitive if we reason inside of $\mathbb{V}^\omega/\dot{G}$: let $\hat{\omega}$ denote the $\omega$ of $\mathbb{V}^\omega/\dot{G}$ (this is a $Q$-name). For each $\alpha < \theta$, write $\hat{s}_\alpha = f_\alpha/\dot{G} \in \hat{\omega}^{<\hat{\omega}}$ and write $\hat{n} = g/\dot{G} \in \hat{\omega}$. The choice of $g$ means that each $\hat{s}_\alpha \in \hat{n}^{<\hat{n}}$ (i.e., this is forced by $B$).

For each $\alpha < \theta$, consider $Y_\alpha$ as a function from $\omega$ to $[\omega]^{<\omega}$, given by $n \mapsto Y_{\alpha, n}$; let $\hat{a}_\alpha = Y_\alpha/\dot{G} \in [\hat{\omega}]^{<\hat{\omega}}$. Then $\hat{a}_\alpha = \{\hat{s} \in \hat{n}^{<\hat{n}}: \hat{s}_\alpha \subseteq \hat{s}\}$ and so now it is clear $\hat{a}_\alpha \supseteq \hat{a}_\beta$ for $\alpha < \beta$. Since this is forced by $B$, we get that $Y_\alpha \supseteq^* Y_\beta$ for $\alpha < \beta$.

Incidentally, the forcing extension is superficial. At the beginning of the proof we can suppose by passing to a Levy collapse that $\mathfrak{t} = 2^{\aleph_0}$. This allows us to build a sufficiently generic tower $(A_\gamma: \gamma < 2^{\aleph_0})$; the ultrafilter $\mathcal{U}$ generated by $(A_\gamma: \gamma < 2^{\aleph_0})$ will be as needed.

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  • $\begingroup$ (If I'm not answering your question, I'm happy to revise) $\endgroup$ – Douglas Ulrich Sep 1 '18 at 19:56
  • $\begingroup$ Thank you really much for your answer. I still don't get something: why $B'$ should force $f_\alpha/G\not \trianglelefteq f_\beta/G$? I understand that $B'$ forces $f_\alpha/F\not \trianglelefteq f_\beta/F$ for every $F$ generic ultrafilter with $B'\in F$, but $G$ is one fixed ultrafilter and we do not know weather $B'\in G$ or not right? I'm still thinking about the second part. $\endgroup$ – Cla Sep 2 '18 at 10:17
  • $\begingroup$ $\dot{G}$ is not a fixed ultrafilter, it is the name for the $Q$-generic ultrafilter. We are assuming that $B \Vdash_P f_\alpha/ \dot{G} \trianglelefteq f_\beta/\dot{G}$; this means that whenever $G$ is $\mathbb{V}$-generic with $B \in G$, then $f_\alpha/G \trianglelefteq f_\beta/G$. We end up constructing $B' \subseteq B$ such that $B'$ forces the opposite statement. Does this make sense? $\endgroup$ – Douglas Ulrich Sep 2 '18 at 17:49
  • $\begingroup$ Sorry but I don't see how that should be possible: if you say "$\langle f_\alpha/G\rangle$ is an increasing sequence for a $G$ not fixed" then $\langle f_\alpha\rangle$ must be an increasing sequence as well and everything is trivial, while instead it is possible to build in $\mathfrak{M}[G]$ sequences $\langle f_\alpha/G\rangle$ such that $\langle f_\alpha\rangle$ is not increasing. $\endgroup$ – Cla Sep 2 '18 at 19:29
  • $\begingroup$ Also, assume you can state logic sentences for $G$ without fixing it, then let $\mathcal{Q}=\mathcal{P}(\omega)/\mbox{FIN}$ consider the sentence "Let $\mathcal{A}\subset G$ and $\mathcal{I}=\{X\in \mathcal{Q} : \forall A\in \mathcal{A} [X\subseteq^\ast A]\}$ and $\mathcal{I}\cap G=\emptyset$", then there exists by forcing theorem a $B\in G$ that force that sentence. I'll proceed as in the proof assuming that the $G$ is just one of the many ultrafilters: assume there exists $A\in \mathcal{A}$ such that $B\not\subseteq^\ast A$, then $(\omega\setminus A)\cap B$ infinite. (Continue next comment) $\endgroup$ – Cla Sep 2 '18 at 19:30
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I'll close the question collecting here what helped me understanding the problem (thank you really much KP Hart and Douglas Ulrich for the help).

The thing that caused me great confusion is that I wasn't aware of the following fact: given a model $\mathfrak{M}$ and a $\lambda$-closed forcing notion $\mathbb{P}\in \mathfrak{M}$ and $G\subset \mathbb{P}$ ultrafilter generic over $\mathfrak{M}$, then $\mathfrak{M}[G]\vDash$ "$G$ (considered as a suborder of $\mathbb{P}$) is $\lambda$-closed".


I'll rewrite part of the statement of my example. I made a lot of confusion and used the wrong notation and symbols so I'll correct them here.

Let's consider $G$ an ultrafilter generic over $\mathfrak{M}$. $G$ need not to be fixed: it can be actually any of them.

Working in $\mathfrak{M}[G]$, find a subset $A\subset G$ such that $A$ is unbounded in $G$, that is $\forall x\in B[x\leq A\to x\notin G]$. Let $\dot{A}$ be the $B$-name of $A$ and $\dot{G}$ be the $B$-name of $G$ ($\check{x}$ should be used only for the canonical name of some $x\in\mathfrak{M}$, $G$ and $A$ does not belong to $\mathfrak{M}$). Then by forcing theorem, $\mathfrak{M}[G]\vDash$ "$A\subset G$ and $A$ is unbounded in $G$" if and only if there exists a $p\in G$ such that $p\Vdash$ "$A\subset G$ and $A$ is unbounded in $G$".

Here it might be useful to think $A=G$, and so we can take also $\dot{A}=\dot{G}$, since this satisfy all the requirement.

It is true that it is not possible that $\mathfrak{M}[G]\vDash p\leq A$, hence $\mathfrak{M}[G]\vDash p\nleq A$. For example in case $A=G$, $p$ might be the maximum of the boolean algebra, so this is not surprising.

That means $\mathfrak{M}[G]\vDash$ "there exists $a\in A$ such that $(p\land a)<p$", so we can fix an $a\in G$ and $p'\in G$ such that $(p\land a)<p$ and $p'\Vdash \check{a}\in \dot{A}$.
It is true that $((\neg a)\land p)>0$ and $((\neg a)\land p)<p$ and thus $((\neg a)\land p) \Vdash$ "$\dot{A}\subset \dot{G}$ and $\dot{A}$ is unbounded in $\dot{G}$", although a statement like "$((\neg a)\land p)\notin G$" makes sense only if we fixed $G$ so that $a\in G$, that is equivalent to say $a\Vdash ((\neg a)\land p)^{\check{}}\notin \dot{G}$.

So this alone does not create problem: it just means that $a$ and $p'$ incompatible and $p\not \Vdash \check{a}\in \dot{A}$, which is trivial. In fact, it's always true that given $q,r\in B$ if $q<r$ then $r\not \Vdash \check{q}\in \dot{G}$.

But in Fremlin's proof there were additional conditions that make this last stament less trivial (and hence surprising for me).
Assume $A$ wellordered and there exists $q\in B$, $\alpha$ ordinal and $b_i\in B$ for every $i<\alpha$ such that $q\Vdash$ "the order type of $\dot{A}$ is $\check{\alpha}$ and $\check{b_i}$ is the $i$-th element of $\dot{A}$ for any $i\in\check{\alpha}$".

Then we have a contradiction: in fact, since $q\in G$ and $q\Vdash$ "$\dot{A}$ unbounded in $\dot{G}$", we may find $a\in A$ such that $(q\land a)<q$ and $q'<q$ such that $q'\Vdash \check{a}\in\dot{A}$. Then there exists an $i\in\alpha$ such that $q'\Vdash \check{a}=\check{b_i}$ and so $a=b_i$, hence $q\Vdash \check{a}\in \dot{A}$. But now $((\neg a)\land q)>0$ and $((\neg a)\land q)<q$, hence $((\neg a)\land q)\Vdash \check{a}\in \dot{A}$ and at the same time $((\neg a)\land q)\Vdash \check{a}\notin \dot{A}$, contradiction.

This contradiction only proves that for every $A$ unbounded there is no such $q$. But if $\mathbb{P}$ is a $\lambda$-closed forcing notion such a $q$ can be found for every chain $A'$ of cardinality less than $\lambda$. This two things together were the cause of my confusion, while actually they just imply what stated at the beginning, i.e. that for every $\lambda$-closed forcing notion $\mathbb{P}$ and $G\subset \mathbb{P}$ generic ultrafilter over a model $\mathfrak{M}$, then $\mathfrak{M}[G]\vDash$ "$G$ is $\lambda$-closed".

My problem was that I didn't recognize what this contradiction meant at the beginning and thought there could have been generic ultrafilters with small unbounded chains, so I assumed $G$ fixed to avoid this contradiction, while actually it is not.

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