A specialization generic kills a tree

Fix $t\in T$ and $p\in P(T)$. If $t\in \mathrm{dom}(p)$, take $q=p$. Otherwise put $N=\{\begin{array}{ll}0,& \mathrm{ran}(p)=\mathrm{\varnothing },\\ 1+\max \mathrm{ran}(p),& \mathrm{ran}(p)\mathrm{\ne }\mathrm{\varnothing },\end{array}q=p\cup \{(t,N)\}\mathrm{.}$ The finite range has a maximum in the second case, and $N$ differs from every old label. Therefore every new comparable pair involving $t$ has unequal labels, while old pairs still satisfy [F2]. Thus $q\in P(T)$, $q\le p$, and $q\in D_t$. This calculates the density of every domain requirement, including the empty-condition case where the added label is $0$.

Since $G$ is a nonempty directed generic filter, it meets every ground dense set $D_t$. If $(t,i),(t,j)\in \bigcup G$, directedness gives a common stronger condition containing both pairs, so [F2] gives $i=j$. Hence $f=\bigcup G$ is a function. Meeting $D_t$ for each $t$ makes its domain all of $T$. If $s{<}_{T}t$, choose conditions in $G$ containing $s$ and $t$ and a common stronger condition; [F2] gives $f(s)\mathrm{\ne }f(t)$. This is the total specializing map promised by [F3].

For each $n<\omega$, define $A_n=f^{-1}(\{n\})$. If $s,t\in A_n$ are distinct, then $f(s)=f(t)=n$, so step 2.1 says they cannot be comparable; hence $A_n$ is an antichain. Totality gives $T={\bigcup }_{n<\omega }{A}_n$. The fiber $A_0$ is included but may be empty, and no argument singles out a predetermined uncountable fiber.

Suppose for contradiction that every $A_n$ were countable in $M[G]$. Then [F5] would make $T$ countable. The height map sends its nodes onto a cofinal subset of the preserved ${\omega }_1^M$: [F1] preserves ${\omega }_1$, and the old tree still has a node at every countable height by [F4]. The image of a countable set is countable, contradicting [F6]. Thus some, but not necessarily the zero, fiber $A_n$ is uncountable.

By step 2.1, the restriction of $f$ to any branch is injective into $\omega$. A cofinal branch would therefore have a countable cofinal set of node heights in the preserved ${\omega }_1$, contrary to [F6]. Hence $T$ remains Aronszajn. Step 3.1 witnesses that it is special, while step 4.1 supplies an actual uncountable antichain, so it is not Suslin. “Kills” refers only to the Suslin property: the tree set, order, height, and countable old levels remain.

The calculation was made in a supplied generic extension only to display the objects. By [F1], its dense-set, directed-union, preservation and ZFC arguments are already encoded by the internal forcing relation below the greatest empty condition. No $M$-generic over the universe is asserted to exist. The empty function, singleton extensions, label zero, possibly empty fibers, and absence of a top level at the limit height ${\omega }_1$ create no exception. The density and union calculations in steps 1.1-3.1 are choice-free; AC is used through preservation and the countable-union and boundedness conclusions in steps 4.1-5.1.