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Recall that Mathias forcing is the poset $\mathbb{M}$ whose elements are pairs $(s,A)$, where $s$ is a finite subset of $\omega$, $A$ an infinite subset of $\omega$, and $\max(s)<\min(A)$. Extension is defined by $(s,A)\leq(t,B)$ if and only if $s\supseteq t$, $A\subseteq B$ and $s\setminus t\subset B$. If $U$ is an ultrafilter, then one can define Mathias forcing relative to $U$, $\mathbb{M}_U$, likewise by ensuring that the side condition $A\in U$.

By a Mathias real over $M$, I mean an infinite $x\subset\omega$ such that $x=\bigcup\{s:(s,A)\in G\}$, where $G$ is $M$-generic for $\mathbb{M}$.

I have a question about the proof of the following theorem, 26.35 in Jech's Set Theory:

Theorem (Mathias): Let $M$ be a transitive model of ZFC. An infinite set $x\subset\omega$ is a Mathias real over $M$ if and only if for every maximal almost disjoint family $\mathcal{A}\in M$ of subsets of $\omega$, there exists an $X\in\mathcal{A}$ such that $x\setminus X$ is finite.

The forward direction is an easy density argument. For the converse, fix $D\in M$ a dense open subset of $\mathbb{M}$. Given $s$, we say that an infinite set $X\subset\omega$ captures $(s,D)$ if $\max(s)<\min(X)$ and for every infinite $Y\subset X$, there is an initial segment $t$ of $Y$ such that $(s\cup t,X\setminus|t|)\in D$, where $|t|=\sup\{n+1:n\in t\}$.

A key lemma to proving the above theorem is:

Lemma: For every infinite set $A\subset\omega$ and finite $s\subset\omega$, there exists an infinite set $X\subset A\setminus|s|$ such that $X$ captures $(s,D)$.

The proof proceeds by constructing a sequence $Y_0\supset Y_1\supset\cdots$ of infinite sets, and $m_0<m_1<\cdots$, with $m_n=\min(Y_n)$ as follows:

Let $Y_0=A\setminus|s|$. For the inductive case, the proof claims that given $Y_n$, we can find $Y_{n+1}\subset Y_n\setminus\{m_n\}$ such that for any $t\subset\{m_0,\ldots,m_n\}$, if there exists $Y\subset Y_n$ such that $(s\cup t,Y)\in D$, then $(s\cup t,Y_{n+1})\in D$.

This I do not see. In the case of the analogous claim for $\mathbb{M}_U$, where $U$ is a Ramsey ultrafilter (as in section 2 of Mathias' Happy Families paper), one can do this, because the witnesses $Y$ for $(s\cup t,Y)\in D$, will have $U$-large intersection, and $D$ is open. However, for $\mathbb{M}$, isn't it possible that there are $t_0,t_1\subset\{m_0,\ldots,m_n\}$ with $(s\cup t_0,Y),(s\cup t_1,Y')\in D$ for some $Y,Y'\subset Y_n$, but so that the only such witnesses are (almost) disjoint?

Edit: For context, here is how the rest of the proof of the lemma proceeds. Once we have such a sequence of $Y_n$ and $m_n$, let $Y=\{m_n:n\in\omega\}$. Let $O=\bigcup\{[t,S]^\omega:(t,S)\in D\}$, where $[t,S]^\omega=\{X\in[\omega]^\omega:t\subset X \land X\setminus|t|\subset S\}$. This is a dense open set in the Ellentuck topology (hence, its complement is Ramsey null), thus there is an infinite $X\subset Y$ such that $[s,X]^\omega\subset O$. We claim that $X$ captures $(s,D)$. Let $Z\subset X$ be infinite, then $s\cup Z\in O$, so there is an initial segment $t$ of $Z$ and infinite $S\subset\omega$ such that $(s\cup t,S)\in D$, and $s\cup Z\in[s\cup t,S]^\omega$. Since $D$ is open in $\mathbb{M}$, it follows that $(s\cup t,Z\setminus|t|)\in D$, and if $\max(t)=m_n$, then $Z\setminus|t|$ is an infinite subset of $Y_n$, and so, by the claimed choice of $Y_{n+1}$, we have that $(s\cup t,Y_{n+1})\in D$. $X\setminus|t|\subset Y_{n+1}$, so it follows that $(s\cup t,X\setminus|t|)\in D$, as desired.

Edit: The question seems to have been answered by Francois Dorais in the comments.

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    $\begingroup$ There might be a typo in the subscripts, the expected ending for the next to last paragraph is: "if there exists $Y \subset Y_{n+1}$ such that $(s \cup t,Y) \in D$, then $(s\cup t,Y_{n+1}) \in D$." $\endgroup$ – François G. Dorais Jun 5 '14 at 3:22
  • $\begingroup$ @FrançoisG.Dorais Is it clear that we can do that in this case? Also, I've added the rest of the proof in an edit, showing where this part of the construction is being used. $\endgroup$ – Iian Smythe Jun 5 '14 at 3:43
  • $\begingroup$ Do you mean "... $s\setminus t\subseteq B$" (not "$...\in B$")? $\endgroup$ – John Bentin Jun 5 '14 at 9:53
  • $\begingroup$ The corresponding change in the addendum is: "[...] then $Z\setminus|t|$ is an infinite subset of $Y_{n+1}$, and so, by the claimed choice of $Y_{n+1}$, we have that $(s\cup t,Y_{n+1})\in D$." This seems to work fine. $\endgroup$ – François G. Dorais Jun 5 '14 at 13:15
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    $\begingroup$ List the $t$'s as $t_1,t_2,\dots,t_{2^{n+1}}$, then deal with each one in order building a sequence $Y_n = Z_0 \supseteq Z_1 \supseteq \cdots \supseteq Z_{2^{n+1}} = Y_{n+1}$. At each step, if there is a way to extend $(s \cup t_i,Z_{i-1})$ in to $D$ then do that for $Z_i$, otherwise let $Z_i = Z_{i-1}$. $\endgroup$ – François G. Dorais Jun 5 '14 at 15:23

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