Sweet density transfers along complete suborders

Fix the canonical surjection $n(i)=m$ where $i+1=2^m(2r+1)$, For fixed $m$, the indices $i=2^m(2r+1)-1$ as $r$ varies are unbounded, so $m$ occurs arbitrarily late; in particular $n(0)=0$.

The Boolean conditions $p$ and $e(q)$ are compatible in $B$: the quotient hypothesis applied to $p_0=p$ says exactly that they are compatible.

For each $i<\omega$, use [F4] to choose $q_i\in D$ with $q_i<msub><mi>E</mi><mi>i</mi></msub>q$ so that, whenever there exists $q^{\mathrm{\prime }}<msub><mi>E</mi><mi>i</mi></msub>q$ for which no $p^{\mathrm{\prime }}\in A_{n(i)}$ satisfies $p^{\mathrm{\prime }}\le p$ and $p^{\mathrm{\prime }}⊩q^{\mathrm{\prime }}\in Q\mathrm{/}P$, the chosen $q_i$ has that property. The admissible set is nonempty: choose a bad witness if one exists, and otherwise use $q$ itself.

There is $r\in D$ with $r\le q$ and $e(r)\le p$. Indeed step 1.2 gives a nonzero $b\le p\wedge e(q)$ in $B$. Density of $e[Q]$ supplies $s\in Q$ with $e(s)\le b$. Then $e(s)$ and $e(q)$ are compatible, so compatibility reflection in [F2] supplies a common strengthening $t\le s,q$ in $Q$. Finally density of $D$ supplies $r\in D$ with $r\le t$. Order preservation gives $e(r)\le e(s)\le p$, while $r\le q$.

There is $k<\omega$ such that every $q^{\mathrm{\prime }}<msub><mi>E</mi><mi>k</mi></msub>q$ has $e(q^{\mathrm{\prime }})$ compatible with $p$. Apply the comparable form of the transfer clause to $r\le q$ at $n=0$. It supplies $k$ such that every $q^{\mathrm{\prime }}<msub><mi>E</mi><mi>k</mi></msub>q$ has some $r^{\mathrm{\prime }}<msub><mi>E</mi><mn>0</mn></msub>r$ with $r^{\mathrm{\prime }}\le r,q^{\mathrm{\prime }}$. Hence $e(r^{\mathrm{\prime }})\le e(r)\le p$ and $e(r^{\mathrm{\prime }})\le e(q^{\mathrm{\prime }})$, so $e(r^{\mathrm{\prime }})$ witnesses the required Boolean compatibility.

The sequence $(q_i)$ extends to the diagonal witness: since $q_i<msub><mi>E</mi><mi>i</mi></msub>q$ and $E_i$ refines $E_k$ for all $i\ge k$, the sequential clause applied to the sequence with last term $q$ gives $q^{\ast }\in D$ with $q^{\ast }<msub><mi>E</mi><mi>k</mi></msub>q$ and $q^{\ast }\le q_i$ for every $i\ge k$.

Since $q^{\ast }<msub><mi>E</mi><mi>k</mi></msub>q$, step 3.1 makes $e(q^{\ast })$ compatible with $p$. We claim that some $p^{\mathrm{\prime }\mathrm{\prime }}\le p$ in $P$ forces $q^{\ast }\in Q\mathrm{/}P$. Otherwise the quotient convention makes $C=\{s\in P:s\le p\text{ and }s{\perp }_{B}e(q^{\ast })\}$ dense below $p$ in $P$. Apply [F4] to the poset of antichains contained in $C$, ordered by inclusion: the empty antichain is present and unions bound chains because any two elements of a chain union occur together in one antichain. Obtain a maximal such $A$. It is predense below $p$: otherwise a condition below $p$ incompatible with all of $A$ has a strengthening in $C$ that could be added. Apply the same argument to antichains of $P$ containing $A$ to obtain a maximal antichain $M$ of $P$. Every member of $M\setminus A$ is incompatible with $p$: if such an $m$ were compatible with $p$, a common strengthening in $P$ would be compatible with some member of the predense antichain $A$ below $p$, contradicting that $M$ is an antichain. By completeness of the suborder [F1], $M$ is maximal in $B$. But a common Boolean strengthening of $p$ and $e(q^{\ast })$ is incompatible with every member of $A$ (by the definition of $C$) and every member of $M\setminus A$ (because they are incompatible with $p$), contradicting maximality in $B$. This proves the claim. Now choose $p^{\mathrm{\prime }}\in A_j$ below $p^{\mathrm{\prime }\mathrm{\prime }}$ from the dense union of the $A_j$. Then $p^{\mathrm{\prime }}\le p$ and $p^{\mathrm{\prime }}⊩q^{\ast }\in Q\mathrm{/}P$; by upward closure in [F3], it also forces every $q^{\mathrm{\prime }}\in Q$ with $q^{\ast }\le q^{\mathrm{\prime }}$ into the quotient.

Choose $i\ge k$ with $n(i)=j$, possible by step 1.1. Then $p^{\mathrm{\prime }}\in A_{n(i)}$ satisfies $p^{\mathrm{\prime }}\le p$ and, since $q^{\ast }\le q_i$, forces $q_i\in Q\mathrm{/}P$; so $q_i$ is not bad at level $i$. By step 2.1 the existence of a bad witness at level $i$ would have forced $q_i$ to be bad, hence no $q^{\mathrm{\prime }}<msub><mi>E</mi><mi>i</mi></msub>q$ is bad for $A_{n(i)}$: for every $q^{\mathrm{\prime }}<msub><mi>E</mi><mi>i</mi></msub>q$ there is $p^{\mathrm{\prime }}\in A_j$ with $p^{\mathrm{\prime }}\le p$ and $p^{\mathrm{\prime }}⊩q^{\mathrm{\prime }}\in Q\mathrm{/}P$. Thus the pair $(A_j,i)$ has the uniform property required in part (1) of the Statement.

Let $A^{\ast }=\bigcup \{A_j:\text{for some }k,A_j\text{ has the uniform property for }k\}$. Given $p_0\le p$, the hypothesis $p_0⊩q\in Q\mathrm{/}P$ holds by monotonicity [F3], so the argument of steps 1.2 through 5.1 with $p_0$ in place of $p$ produces $j,k$ such that every $q^{\mathrm{\prime }}<msub><mi>E</mi><mi>k</mi></msub>q$, in particular $q^{\mathrm{\prime }}=q$, has a condition $p^{\mathrm{\prime }}\in A_j$ below $p_0$ forcing $q^{\mathrm{\prime }}\in Q\mathrm{/}P$; The uniform property below $p_0$ implies the one below $p$ since every witness below $p_0$ is below $p$, so this $A_j$ is included in $A^{\ast }$. Hence $A^{\ast }$ meets every strengthening of $p$ and is dense below $p$.

The steps above establish both conclusions of the Statement.