A countably closed forcing adds a normal Suslin tree

Let $(p_{\xi }{)}_{\xi <\eta }$ be descending, where $\eta <{\omega }_1$. The empty sequence has lower bound $1_P$, and a successor-length sequence has its last member as a lower bound. Suppose $\eta$ is nonzero limit and put $\delta ={\sup }_{\xi <\eta }({\alpha }_{p_{\xi }}+1)$ and $U={\bigcup }_{\xi <\eta }{T}_{p_{\xi }}$. The ordinal $\delta$ is below ${\omega }_1$ by F16. Using a surjection from $\omega$ onto the countable ordinal $\eta$, F6 makes $U$ countable. Literal end extension makes its levels coherent. If $\delta$ is a successor, its value is attained by some ${\alpha }_{p_{\xi }}+1$; all later conditions then have the same height and hence are equal to $p_{\xi }$, so $p_{\xi }$ is a lower bound. If $\delta$ is limit, $U$ has height $\delta$: every node has extensions on all higher levels because some later condition reaches each such level, and every nontop successor level already occurs in a condition, so normality and full $\omega$-splitting persist. Its root singleton is a maximal antichain. Apply F2 to that singleton and identify each new branch-top with the union of its sequence branch; the resulting fixed-coded tree $q$ has top level $\delta$, is a condition, and end extends every $p_{\xi }$.

Step 1.1 supplies a lower bound for every descending sequence of every length $\eta <{\omega }_1$, including lengths zero, one, successor, and nonzero limit. By F3, $P$ is ${\mathrm{\aleph }}_1$-closed, that is, countably closed.

F5 verifies the regularity hypothesis needed to apply F4 at $\kappa ={\mathrm{\aleph }}_1$, so step 2.1 implies that $P$ preserves ${\omega }_1$ and adds no countable sequence of ground-model elements. For every $\beta <{\omega }_1$, let $D_{\beta }=\{p:{\alpha }_p\ge \beta \}$. A one-level extension is obtained by adjoining $t^{⌢}⟨n⟩$ for every old top node $t$ and every $n<\omega$; it remains countable by F6. Iterating this operation, and using step 2.1 for lower bounds at countable limit stages, F7 and A1 produce below any condition a member of $D_{\beta }$. Thus every $D_{\beta }$ is dense.

If two conditions have a common extension, their heights are comparable and literal restriction from that common extension shows that the taller end extends the shorter. Hence a generic filter's conditions form an end-extension chain and $T_G$ is coherent. By density of every $D_{\beta }$, it has a level at every ground ordinal $\beta <{\omega }_1$; F4 and F11 say that this is still exactly the extension's ${\omega }_1$. For a fixed $\beta$, once $p\in G$ has ${\alpha }_p\ge \beta$, every condition in $G$ is compatible with $p$ and all taller ones have exactly $(T_p{)}_{\beta }$, so $(T_G{)}_{\beta }=(T_p{)}_{\beta }$ is countable. The same directed common-extension argument supplies every higher-level extension of each node, while the literal successor levels retain full $\omega$-splitting and function extensionality retains limit uniqueness. Thus $T_G$ is a normal $\omega$-splitting ${\omega }_1$-tree with countable levels.

Fix $p_0$ and a name $\dot{A}$ with $p_0⊩$ “$\dot{A}$ is a maximal antichain of $\dot{T}$.” If $r\le p_0$ and $s\in T_r$, then $r$ forces that some member of $\dot{A}$ is comparable with $s$. By F8 and F9, strengthen to choose a name for such a member. It is forced to be a node of $\dot{T}$, hence a natural-valued sequence whose ordinal domain is below ${\omega }_1$; F9 and F11 first decide that ground ordinal domain, and F4 then lets a further extension decide the whole sequence as a ground node $t$. Finally F10 and the displayed canonical-union name say that conditions whose tree contains $t$ are dense below a condition forcing $t\in \dot{T}$: a membership coefficient is a condition containing $t$, and a common extension contains it by end extension. We may therefore find $r^{\mathrm{\prime }}\le r$ with $t\in T_{r^{\mathrm{\prime }}}$ and $r^{\mathrm{\prime }}⊩$ “$t\in \dot{A}$ and $t$ is comparable with $s$.” All strengthenings preserve earlier decisions by F9.

Starting below an arbitrary $r_0\le p_0$, use A1 to enumerate the countable tree of the current condition. Apply step 5.1 successively to every node in that enumeration, take a lower bound of the resulting descending omega-sequence by step 2.1, and then strengthen into the dense set whose top is strictly higher. Repeat this outer construction for $n<\omega$. Let $U={\bigcup }_n{T}_{p_n}$ and let $A^{\ast }$ be the set of all ground nodes decided into $\dot{A}$ during the construction. The strictly increasing top heights make $U$ a countable normal $\omega$-splitting tree of nonzero countable limit height. Every node of $U$ occurred at some stage and is comparable with a member of $A^{\ast }$. Distinct members of $A^{\ast }$ are incomparable: a later common condition forces both into the antichain $\dot{A}$, and persistence forbids it from forcing two distinct comparable members. Hence $A^{\ast }$ is a countable maximal antichain of $U$. Apply F2 to seal $A^{\ast }$ with a new top level, yielding a condition $q\le r_0$ and below every decision condition.

Persistence gives $q⊩A^{\ast }\subseteq \dot{A}$. Every node of $q$ is comparable with $A^{\ast }$, and each new top node extends a member of $A^{\ast }$ by F2. Consequently every node added by a future end extension extends one of those top nodes and remains comparable with $A^{\ast }$; so $q$ forces that $A^{\ast }$ is maximal in $\dot{T}$. Since $q$ also forces that $\dot{A}$ is an antichain containing the maximal antichain $A^{\ast }$, it forces $\dot{A}=A^{\ast }$ and therefore countable. Because $r_0\le p_0$ was arbitrary, such q's are dense below $p_0$, and F9 gives $p_0⊩$ “$\dot{A}$ is countable.”

By F12 the extension satisfies ZFC, so F13 extends every antichain of $T_G$ to a maximal one; step 7.1 makes that maximal antichain countable. Thus $T_G$ has no uncountable antichain. If it had a cofinal branch, F15 applied inside the ZFC extension to the splitting ${\omega }_1$-tree from step 4.1 would produce an uncountable antichain, a contradiction. F14 now identifies $T_G$ as a normal splitting Suslin tree.

Steps 2.1, 3.1, and 8.1 prove the closure, preservation, and forced-tree claims. F8 converts the dense local conclusions to the displayed internal forcing assertion; it does not supply or assert a generic over the universe. AC is used exactly through F5 and F6, the recursive choices in steps 3.1 and 6.1, Zorn in step 8.1, and F15.