Suslin Trees, Lines, Algebras, and Independence

Put $T^{\ast }=\{t\in T:T\uparrow t\text{ is uncountable}\}$. Some root belongs to $T^{\ast }$, since the countable level of roots cannot have only countable cones while $T$ has height ${\omega }_1$. If $t\in T^{\ast }$ and $\mathrm{ht}(t)<\beta <{\omega }_1$, then some extension of $t$ on level $\beta$ has uncountable cone: otherwise the part of $T\uparrow t$ below $\beta$, together with the countably many countable cones based on that level, would be countable by F4. Thus $T^{\ast }$ still has height ${\omega }_1$, countable levels, and extension to every higher level. Being a subtree, it has no forbidden branch or antichain.

First repair possible non-Hausdorff limit splitting; doing this before passing to branching points is essential. For each downward-closed chain $C\subseteq T^{\ast }$ of limit order type ${\alpha }_C$, where at least two nodes of $T^{\ast }$ at height ${\alpha }_C$ lie above every member of $C$, adjoin one history node ${\mathrm{\ell }}_C$. Keep the old order and declare $x<{\mathrm{\ell }}_C$ exactly when $x\in C$, declare ${\mathrm{\ell }}_C<x$ exactly when every member of $C$ is below $x$, and put ${\mathrm{\ell }}_C<{\mathrm{\ell }}_D$ exactly when $C⊊D$. The eight old/history cases verify transitivity. An infinite descending sequence of history nodes would, by choosing a point in each successive set difference, give a descending sequence in $T^{\ast }$; mixed descending sequences reduce to the same contradiction. Predecessors of any node are linearly ordered by inclusion of their histories, so the enlarged order $H$ is a tree.

The enlargement remains Suslin. An uncountable chain containing uncountably many history nodes gives a strictly increasing ${\omega }_1$-sequence of histories; choosing one point from each successive difference gives an uncountable chain in $T^{\ast }$. For an uncountable antichain of history nodes, choose for each $C$ an old upper bound $u_C$ at height ${\alpha }_C$ lying above every member of $C$, as guaranteed by the definition in step 2.1. Comparability of two such bounds would make their predecessor histories comparable, so the $u_C$ form an uncountable antichain in $T^{\ast }$. If uncountably many members are old nodes, the contradiction is immediate. AC makes these choices and permits thinning the old/history partition. Thus every forbidden set in $H$ lifts to $T^{\ast }$ and then to $T$. Every old node retains its uncountable cone, and each ${\mathrm{\ell }}_C$ lies below two incompatible old upper bounds with uncountable cones; hence every node of $H$ has an uncountable cone. Extension to higher levels is inherited from $T^{\ast }$.

The history insertion makes $H$ Hausdorff at nonzero limit levels. Suppose a limit-length chain $⟨w_{\xi }:\xi <\rho ⟩$ has two distinct upper nodes at its level. If old nodes occur cofinally in the chain, those old nodes and all their old predecessors form a downward-closed $T^{\ast }$-chain $C$ of limit length with the two upper nodes above it; step 2.1 inserted ${\mathrm{\ell }}_C$ strictly between the chain and those upper nodes. If the chain is eventually made of history nodes ${\mathrm{\ell }}_{C_{\xi }}$, then the increasing union $C=\bigcup C_{\xi }$ has the same properties, and again ${\mathrm{\ell }}_C$ is a missing further predecessor. Either case contradicts the assumed level of the two upper nodes. This is Monk's two-case verification of Hausdorff uniqueness; in particular, competing old limit nodes move to a successor level immediately above their common history node.

Now branching points are cofinal above every $t\in H$. Suppose instead that the branching points above $t$ were bounded below some countable level. At a higher level, the cone above each node is a chain: two incomparable extensions would have a first divergence, and the Hausdorff property from step 3.2 rules out a first divergence at a limit level, so their last common predecessor would be a branching point. Each such chain is countable by Suslinity, and the level is countable because it is an antichain in $H$; F4 would make the uncountable cone above $t$ countable, a contradiction. Let $B$ be the induced tree of branching points of $H$. Above two incompatible immediate successors of $x\in B$, choose branching points of least possible height. They are distinct immediate successors of $x$ in $B$. Thus every node of $B$ branches, and cofinality of branching points plus the extension property of $H$ gives extensions to every higher $B$-level. As a subtree of $H$, $B$ remains Suslin.

The Hausdorff property passes to $B$. Indeed, if two nodes at a nonzero limit $B$-level had the same strict $B$-predecessors but different $H$-predecessor histories, their first divergence in $H$ would yield a branching point strictly above all their common $B$-predecessors and below one of the two nodes. That branching point belongs to $B$, contradicting equality of their $B$-predecessor sets.

Retain the nodes of $B$ on its limit levels, ordered as before, and reindex those levels increasingly by ${\omega }_1$. Between a retained level and the next retained level lie $\omega$ successive branching levels of $B$. Iterating the two-successor choice through the first $n$ of them gives at least $2^n$ incompatible extensions, and the extension property carries all of them to the next retained level. Hence every node has countably infinitely many immediate successors in the retained tree. Its levels are countable because each is an antichain in the Suslin tree $B$, and it keeps the extension property and the limit-history uniqueness from step 5.1. Choose a root of the retained tree and take its cone; every node of $B$ has an uncountable cone by steps 3.1 and 4.1, so this cone is cofinal and the resulting tree has a unique root.

Call the resulting cone $S$. It is normal and infinitely splitting by steps 5.1 and 6.1. Restriction to levels and a cone cannot create a branch or antichain, so $S$ is Suslin by steps 3.1 and 4.1; and the lifting transformations in step 3.1 apply to any hypothetical uncountable forbidden set in $S$. This proves both the refinement and the stated lifting clause. The uses of AC were the simultaneous countability enumerations, history witnesses, branching-point selections, and final root cone; no weaker-choice claim is made.

For each node $t$, its immediate-successor set is countably infinite by F1 and the countability of the next level. Using F4 and A1, choose a bijection from that set to $\mathbb{Q}$ and transport the usual dense no-endpoint order from F5. Use F2 to regard a branch as a coherent sequence of successor choices. The poset of branches through a fixed node, ordered by inclusion, is nonempty and the union of every chain is an upper bound, so F7 supplies a maximal branch through every node. Such a branch has countable limit length: it cannot have length ${\omega }_1$ because $S$ is Suslin, and it cannot have a last node because normality extends that node higher. Two distinct maximal branches cannot be proper initial segments of one another, so they have a least differing coordinate $d(B,C)$.

Define $B{<}_{\mathrm{l}\mathrm{e}\mathrm{x}}C$ when, at $d(B,C)$, the successor chosen by $B$ precedes the successor chosen by $C$. The usual first-difference argument is valid because F2 identifies all earlier coordinates and F3 makes their common predecessor unique. If $A<B<C$, the least of the two relevant first-difference levels determines the same orientation for $A$ and $C$; hence the relation is transitive. Exactly one orientation holds for distinct branches, so this is a linear order.

If $B<C$, choose at their first difference a successor strictly between their two successors in the dense local order and extend it to a maximal branch $E$. Then $B<E<C$. Given a branch $B$, take one of its successor choices and choose local successors immediately below and above it in the no-endpoint local order; maximal branches through them lie respectively below and above $B$. Thus the lexicographic branch order is dense and has no endpoints.

Suppose $(I_{\xi }{)}_{\xi <{\omega }_1}$ were pairwise disjoint nonempty open branch intervals, writing $I_{\xi }=(B_{\xi },C_{\xi })$. Choose $E_{\xi }\in I_{\xi }$. Since its length is limit, choose $d(B_{\xi },E_{\xi }),d(E_{\xi },C_{\xi })<{\alpha }_{\xi }<\mathrm{len}(E_{\xi })$ and let $x_{\xi }$ be the height-${\alpha }_{\xi }$ node of $E_{\xi }$. If $x_{\xi }{\le }_S{x}_{\eta }$, then the two middle branches agree through ${\alpha }_{\xi }$; comparing them at the two earlier first-difference coordinates puts $E_{\eta }$ strictly between $B_{\xi }$ and $C_{\xi }$. This contradicts disjointness of $I_{\xi }$ and $I_{\eta }$. Hence the $x_{\xi }$ form an uncountable tree antichain, contrary to F1. The branch order is ccc.

Fix a nonempty open interval $(B,C)$ and a countable set $D$ of branches in it. By F6 choose a countable ordinal $\delta$ strictly above $d(B,C)$ and above the lengths of every member of $D$. At the first difference of $B,C$, choose an intermediate successor, extend its cone to a node $x$ at height $\delta$, and then to a maximal branch; every branch through $x$ lies in $(B,C)$. The cone above $x$ is not a chain, for normality would otherwise give a cofinal branch. Choose incomparable $y,z>x$, and incomparable $u,v>y$, and extend $u,v,z$ to branches $P,Q,R$. After interchanging $P,Q$ and, if needed, reversing the picture, either $(P,R)$ or $(R,Q)$ is a nonempty interval contained in $(B,C)$ whose two endpoints share $x$. Any branch lying strictly between those endpoints must agree with one endpoint through height $\delta$, so has length greater than $\delta$. It therefore is not in $D$. Thus $D$ is not dense in $(B,C)$.

Steps 2.1-5.1 prove linearity, density, absence of endpoints, ccc, and failure of separability in every nonempty interval. AC is used through F7 and to choose the family of local rational orders, maximal branches, interval witnesses, and the countable ordinal bound; no claim is made in ZF alone.

Let $\mathcal{C}(L)$ consist of the subsets $X\subseteq L$ such that (i) $b<a\in X$ implies $b\in X$, and (ii) whenever $X$ has a least upper bound $a$ in $L$, one has $a\in X$. Order $\mathcal{C}(L)$ by inclusion. These are the downward-closed cuts with every already-existing $L$-supremum closed in.

The inclusion order on $\mathcal{C}(L)$ is linear. Indeed, if $X,Y\in \mathcal{C}(L)$ and $a\in X\setminus Y$, then every $b\in Y$ satisfies $b<a$ (otherwise downward closure would put $a$ in $Y$), and hence $b\in X$; thus $Y⊊X$.

Every family $\mathcal{X}\subseteq \mathcal{C}(L)$ has a supremum. Put $U=\bigcup \mathcal{X}$. If $U$ has no least upper bound in $L$, then $U\in \mathcal{C}(L)$ and is the union-supremum. If $a={\sup }_{L}U$ exists, downward closure gives $U\cup \{a\}=(-\mathrm{\infty },a]$, which lies in $\mathcal{C}(L)$ and is the least cut above every member of $\mathcal{X}$. This includes $\mathcal{X}=\mathrm{\varnothing }$. Infima then exist as suprema of sets of lower bounds, so $\mathcal{C}(L)$ is complete.

Send $a\in L$ to $j(a)=(-\mathrm{\infty },a]$. Each $j(a)$ is a cut, and $a<b$ holds exactly when $j(a)⊊j(b)$, so $j$ is an order embedding. Replacing $L$ by its named copy $j[L]$ if necessary gives literal inclusion as required by C1.

Every cut $X$ is $\sup \{j(a):a\in X\}$, including the empty cut. If $a={\sup }_{L}A$ for $A\subseteq L$, then $j(a)$ is an upper bound of $j[A]$; any cut $Y$ above every $j(x)$ for $x\in A$ contains every $b<a$, and it contains $a$ either because $a\in A$ or because clause (ii) closes $Y$ under the supremum $a$. Hence $j(a)=\sup j[A]$. Thus the constructed order satisfies C2-C4.

Let $P$ be any completion satisfying C1-C4 and define ${\tau }_P(x)=\{a\in L:a{\le }_{P}x\}$. This is a cut: downward closure is immediate, while if $b={\sup }_L{\tau }_P(x)$, C4 makes $b$ its supremum in $P$, which also equals $x$ by C3, so $b=x\in {\tau }_P(x)$. If $x<y$, C3 gives an $a\in L$ with $x<a\le y$, so ${\tau }_P(x)⊊{\tau }_P(y)$. Conversely the traces reflect order. For every cut $X$, if $x={\sup }_{P}X$, then ${\tau }_P(x)=X$: an $a<x$ in the trace cannot upper-bound $X$, and if $a=x\in L$, then $a={\sup }_{L}X$ and cut closure puts $a$ in $X$. Thus ${\tau }_P$ is an onto isomorphism from $P$ to $\mathcal{C}(L)$ and fixes $L$. Applying this to two completions gives the unique isomorphism over $L$, since C3 forces any such isomorphism to send each ${\sup }_{P}A$ to ${\sup }_{N}A$.

Suppose now that $L$ is dense and $M$ is a completion. If $x<y$ in $M$, C3 supplies $b\in L$ with $x<b\le y$. If $x\in L$, density in $L$ gives $x<a<b$ for some $a\in L$. If $x\mathrm{\notin }L$ and there were no $a\in L$ with $x<a<b$, then $b$ would be the least upper bound in $L$ of the $L$-points below $x$; C4 would make that supremum equal both $b$ and $x$, a contradiction. Hence in all cases some $a\in L$ satisfies $x<a<y$, so $L$ is order-dense in $M$.

If $(I_{\xi }{)}_{\xi <{\omega }_1}$ were pairwise disjoint nonempty open intervals of $M$, order-density and A1 would choose $a_{\xi }<b_{\xi }$ in $L\cap I_{\xi }$. The nonempty $L$-intervals $(a_{\xi },b_{\xi })$ would remain pairwise disjoint, contradicting the corresponding hypothesis on $L$. Thus the interval ccc passes to $M$.

Suppose a nonempty open interval $I$ of $M$ had a countable dense set $D$. Choose $a<b$ in $L\cap I$. The set $D\cap (a,b)$ is nonempty and countable; list it as $(d_n{)}_{n<\omega }$, repeating entries in the finite case. For every pair $d_i<d_j$, use order-density and A1 to choose $e_{ij}\in L$ with $d_i<e_{ij}<d_j$. The set $E=\{e_{ij}:d_i<d_j\}$ is countable by diagonal enumeration of the pairs of natural indices. Given $u<v$ in $L\cap (a,b)$, density of $D$ first gives $d_i\in (u,v)$ and then $d_j\in (d_i,v)$, so $u<e_{ij}<v$. Therefore $E$ is dense in the nonempty $L$-interval $(a,b)$, contradicting the hypothesis on $L$. No nonempty open interval of $M$ is separable.

Finally assume that $L$ has no endpoints, and delete from $M$ its first and last elements when they exist. Neither deleted point belongs to $L$. The remainder $M^{\circ }$ still contains the order-dense copy of $L$, is dense and has no endpoints, and retains the conclusions of steps 6.1-6.2. If a nonempty $A\subseteq M^{\circ }$ is bounded above there, then ${\sup }_{M}A$ lies above a member of $A$ and below an upper bound in $M^{\circ }$, so it is neither deleted endpoint and belongs to $M^{\circ }$; hence $M^{\circ }$ is boundedly complete. This proves every assertion and records the precise use of AC.

Apply F2 to $T$ and obtain an infinitely splitting normal Suslin refinement $S$.

By F3, the maximal branches of $S$, ordered at their first differing successor, form a nonempty dense linear order $L$ without endpoints. The order has no uncountable pairwise disjoint family of nonempty open intervals, and every nonempty open interval of $L$ is nonseparable.

Take the exact completion of $L$ and delete its possible first and last points. By F4 the resulting order $M$ is nonempty, dense, has no endpoints, is boundedly complete, satisfies the interval ccc, and has no separable nonempty open interval. In particular $M$ itself has no countable order-dense subset: if such a set existed, it would be dense in every nonempty open subinterval, contradicting the preceding property. Thus every clause of F1 holds, so $M$ is a Suslin line in the published convention. All three constructions are in ZFC and their uses of choice are exactly those recorded by the supplying lemmas; the implication is not asserted in ZF.

Reflexivity and symmetry of $\sim$ are immediate. For transitivity, the closed interval between $x$ and $z$ is contained in the union of the closed intervals between $x,y$ and $y,z$, regardless of their order; dense sets for those two intervals, together with their finitely many endpoints, restrict to a countable dense set in the first interval. F3 therefore proves transitivity. If $x<z<y$ and $x\sim y$, a dense set for $[x,y]$, restricted to $[x,z]$ and augmented by $x,z$, makes $[x,z]$ separable; hence $x\sim z$. Thus every equivalence class is convex.

Fix an equivalence class $K$. If $K$ has at most two points it is separable. Otherwise let ${\mathbb{P}}_K$ be the inclusion poset of pairwise disjoint nonempty intervals $(a,b)$ with $a,b\in K$. It is nonempty, and the union of a chain in ${\mathbb{P}}_K$ is again such a disjoint family, so F4 gives a maximal family $\mathcal{M}$. The intervals in $\mathcal{M}$ are pairwise disjoint open intervals of $S$, hence F1 makes $\mathcal{M}$ countable. For each $(a_n,b_n)\in \mathcal{M}$, the definition of $\sim$ makes $[a_n,b_n]$ separable; choose a countable dense $D_n$ there. Then $D={\bigcup }_n{D}_n$, augmented by the first and last points of $K$ when they exist, is countable by F3. If $c<d$ in $K$ and $(c,d)$ is nonempty, maximality makes $(c,d)$ meet some $(a_n,b_n)$, and $D_n$ meets that open intersection. Endpoint rays meet $D$ by the same argument unless they end at an included endpoint. Hence $D$ is dense in $K$, so every class is separable.

Let $Q=S\mathrm{/}\sim$ and order its classes by $I<J$ when one, equivalently every, member of $I$ is below one, equivalently every, member of $J$. Convexity makes this well defined and gives a linear order. If $I<J$ had no class strictly between them, then for $a\in I$ and $b\in J$ the interval $[a,b]$ would lie in $I\cup J$; step 2.1 and F3 would make it separable, forcing $a\sim b$, a contradiction. Thus $Q$ is dense.

Fix $I<J$ in $Q$ and suppose the nonempty quotient interval $(I,J)$ had a countable dense set $\mathcal{A}$. Let $\mathcal{B}$ be the classes $K$ strictly between $I,J$ having more than two points. For each $K\in \mathcal{B}$, convexity supplies a nonempty open interval of $S$ contained in $K$; these intervals are pairwise disjoint, so F1 and A1 make $\mathcal{B}$ countable. Put $\mathcal{C}=\mathcal{A}\cup \mathcal{B}\cup \{I,J\}$. By step 2.1 choose a countable dense set $D_K\subseteq K$ for every $K\in \mathcal{C}$, and let $E={\bigcup }_{K\in \mathcal{C}}{D}_K$, countable by F3. Choose $a\in I$ and $b\in J$. If $a\le c<d\le b$ and $(c,d)$ is nonempty, then either $c,d$ lie in the same endpoint class, in the same member of $\mathcal{B}$, or in distinct classes; in the last case quotient density and density of $\mathcal{A}$ put a class of $\mathcal{A}$ strictly between them. In every case $E\cap (c,d)$ is nonempty. Thus $E\cap (a,b)$ together with $a,b$ is countable and dense in $[a,b]$, so $a\sim b$, contradicting $I<J$. No nonempty quotient interval is separable.

Suppose $Q$ had an uncountable pairwise disjoint family of nonempty open intervals $(I_{\xi },J_{\xi })$. Choose one representative $s_K\in K$ for every endpoint class $K$ that occurs. By density of $Q$, $(s_{I_{\xi }},s_{J_{\xi }})$ is a nonempty open interval of $S$. The resulting original intervals are pairwise disjoint because the quotient intervals are, contradicting the ccc of $S$. Hence $Q$ is ccc.

The quotient $Q$ is not a singleton, since otherwise step 2.1 would make all of $S$ separable, contrary to F1. Nor can $Q$ have exactly two classes: then $S=I\cup J$ is the union of the two separable classes, and countable dense subsets $D_I\subseteq I$ and $D_J\subseteq J$ meet every nonempty open interval of $S$, because such an interval is infinite, lies in $I\cup J$, and therefore has one part that is infinite, hence contains a nonempty interval of that dense class. That would make $S$ separable, again contradicting F1. So $Q$ has at least three classes, and since $Q$ is dense, deleting its possible first and last classes leaves a nonempty dense no-endpoint order $Q^{\circ }$; steps 4.1-4.2 persist under this deletion. Apply F2 to $Q^{\circ }$, take its exact completion, and delete the possible completion endpoints. The resulting order is nonempty, dense, has no endpoints, is boundedly complete, is ccc, and has no separable nonempty open interval. This is the asserted nowhere-separable complete line. The only choice costs are F4 and the simultaneous selections explicitly charged to A1; no ZF claim is made.

Recursively choose $a_{\alpha }<b_{\alpha }$ in $L$ for every $\alpha <{\omega }_1$. At stage $\alpha$, the earlier endpoints form a countable set by F4. They cannot be dense in $L$, so some nonempty open interval $(c,d)$ misses all of them; density of $L$ supplies $c<a_{\alpha }<b_{\alpha }<d$. AC gives a choice operation on the nonempty sets of possible quadruples, and F5 then performs the recursion.

For $\xi <\alpha$, the connected interval $(c,d)$ selected at stage $\alpha$ avoids $a_{\xi },b_{\xi }$. Consequently either $[a_{\alpha },b_{\alpha }]\subset (a_{\xi },b_{\xi })$, or the two open intervals $(a_{\alpha },b_{\alpha })$ and $(a_{\xi },b_{\xi })$ are disjoint. Define $\xi \prec \alpha$ exactly in the first case. The relation is irreflexive and transitive. If $\xi ,\eta \prec \alpha$, their intervals both contain $[a_{\alpha },b_{\alpha }]$, so the disjoint alternative is impossible; whichever index is earlier is therefore $\prec$-below the other. Thus the predecessors of $\alpha$ are linearly ordered by the ordinal order and, being a subset of $\alpha$, are well ordered. Hence $({\omega }_1,\prec )$ is a tree.

There is no uncountable chain. Otherwise enumerate an uncountable chain increasingly as $({\alpha }_{\xi }{)}_{\xi <{\omega }_1}$. Successive nesting gives $a_{{\alpha }_{\xi }}<a_{{\alpha }_{\xi +1}}<b_{{\alpha }_{\xi +1}}<b_{{\alpha }_{\xi }}$, so the nonempty open intervals $(a_{{\alpha }_{\xi }},a_{{\alpha }_{\xi +1}})$ are pairwise disjoint. This contradicts ccc of $L$.

There is no uncountable antichain. For incomparable $\alpha ,\beta$, the dichotomy in step 2.1 makes $(a_{\alpha },b_{\alpha })$ and $(a_{\beta },b_{\beta })$ disjoint. An uncountable tree antichain would therefore give an uncountable family of pairwise disjoint nonempty open intervals in $L$, again contradicting ccc.

Each tree level is an antichain, hence countable by step 3.2. Every node has countable height because all its predecessors have smaller ordinal indices. If the tree height were a countable ordinal $\delta$, its carrier would be the union of the countably many countable levels indexed below $\delta$, hence countable by F4, contradicting that the carrier is ${\omega }_1$. Its height is therefore exactly ${\omega }_1$, not merely the length of the construction. A countable branch cannot be cofinal, because F4 makes the supremum of its countably many countable node-heights countable; an uncountable branch is excluded by step 3.1. Thus there is no cofinal branch. Steps 3.1-3.2 and F3 now show that the tree is Suslin. AC was used only for the recursion's simultaneous interval selections and the countable-union consequences recorded above.

Apply F1 to $S$ and obtain a dense no-endpoint ccc line $L$ in which every nonempty open interval is nonseparable.

Apply F2 to $L$. Its recursively nested closed intervals form a tree of height exactly ${\omega }_1$ with countable levels, no cofinal branch, and no uncountable antichain; thus it is a Suslin tree. The source lemma derives the height and every forbidden-set conclusion, so this composition does not assume that construction stage equals tree level. Its inherited choice principle is AC, and no implication over ZF is asserted.

Conditions $s,t\in P$ are compatible exactly when they are comparable in $T$: a common stronger condition is a common tree extension, which makes $s,t$ comparable by F5, while the deeper member of two comparable nodes is already a common forcing extension. Therefore forcing antichains are exactly tree antichains, and F4 makes $P$ ccc. The unique root supplied by normality makes $P$ nonempty.

Let $D\subseteq P$ be dense open. In the inclusion poset of antichains contained in $D$, the empty antichain is present and the union of every chain is an antichain contained in $D$; F7 gives a maximal member $A_D$. It is maximal as a forcing antichain: for any $p$, density gives $d{\le }_{P}p$ in $D$, and if $d$ were incompatible with every member of $A_D$ it could be adjoined. By step 1.1, $A_D$ is countable. F6 therefore gives ${\alpha }_D<{\omega }_1$ above the heights of all its nodes. If $t$ has height greater than ${\alpha }_D$, maximality makes it compatible with some $a\in A_D$; comparability and the height inequality give $a{<}_{T}t$, hence $t{\le }_{P}a$, and openness puts $t$ in $D$. Thus every node above level ${\alpha }_D$ lies in $D$.

Let $(D_n{)}_{n<\omega }$ be dense open and fix $p\in P$. Apply step 2.1 to each $D_n$, using A1 for the simultaneous maximal-antichain and bound choices. By F6 the countable set $\{{\alpha }_{D_n}:n<\omega \}$ has a bound $\beta <{\omega }_1$. Normality gives a tree extension $t{>}_{T}p$ of height greater than $\beta$. Then $t\in D_n$ for every $n$ by step 2.1, and $t{\le }_{P}p$. Hence ${\bigcap }_n{D}_n$ is dense; it is open because every $D_n$ is open. This is ${\mathrm{\aleph }}_1$-distributivity by F1.

By F9, ${\mathrm{\aleph }}_1$ is regular in the stated ZFC setting. Apply F8 with $\kappa ={\mathrm{\aleph }}_1$ to the separative quotient of $P$; the quotient has the same dense-open distributivity and is forcing equivalent to $P$. Step 3.1 therefore implies that no sequence of ground-model elements of length below ${\mathrm{\aleph }}_1$ is added. By F10 every ordinal of a forcing extension is already a ground-model ordinal, and $\omega <{\mathrm{\aleph }}_1$, so forcing with $P$ adds no new $\omega$-sequence of ordinals. This last assertion comes from F8 and F10, not from the definition of distributivity.

By F3-F4, $B$ is a complete Boolean algebra and each $0<b\in B$ has some tree condition $t$ with $0<e(t)\le b$. Since $P$ is nonempty, its underlying space is nonempty, so the regular-open bounds $0=\mathrm{\varnothing }$ and $1=P$ are distinct. Thus $B$ is nontrivial.

Let $0<b\in B$ and choose $t$ with $e(t)\le b$. Splitting supplies two distinct immediate tree successors $s,u$ of $t$. They are incompatible, so F3 gives nonzero disjoint elements $e(s),e(u)\le e(t)\le b$. Therefore $0<e(s)<b$, proving atomlessness.

Let $X\subseteq B^{+}$ be pairwise disjoint. By A1 and density of $e$, choose $t_b\in P$ with $e(t_b)\le b$ for every $b\in X$. If $b\mathrm{\ne }c$, compatibility of $t_b,t_c$ would make $e(t_b)\wedge e(t_c)$ nonzero by F3, while it lies below $b\wedge c=0$. Thus the $t_b$ form a forcing antichain; they are distinct and F2 makes $X$ countable. Hence $B$ is ccc.

Fix a double sequence $(b_{n,m}{)}_{n,m<\omega }$ in $B$ and put $c={\bigwedge }_n{\bigvee }_m{b}_{n,m}$ and $r={\bigvee }_{f\in {\omega }^{\omega }}{\bigwedge }_n{b}_{n,f(n)}$. Always $r\le c$. In any complete Boolean algebra, $a\wedge \bigvee X={\bigvee }_{x\in X}(a\wedge x)$: the right side is below $a\wedge \bigvee X$, and if $y$ bounds all $a\wedge x$, then $\mathrm{\neg }a\vee y$ bounds $X$, giving $a\wedge \bigvee X\le y$. Suppose $0<d=c\wedge \mathrm{\neg }r$ and choose $p_0$ with $e(p_0)\le d$. For each $n$, let $D_n$ contain the conditions $q$ such that either $q$ is incompatible with $p_0$, or $e(q)\le b_{n,m}$ for some $m$. This set is open. It is dense: if $q$ is compatible with $p_0$, take $q^{\mathrm{\prime }}\le q,p_0$; since $e(q^{\mathrm{\prime }})\le c\le {\bigvee }_m{b}_{n,m}$, the just-proved distributive identity makes some $e(q^{\mathrm{\prime }})\wedge b_{n,m}$ nonzero, and density of $e$ plus compatibility reflection gives an actual $s\le q^{\mathrm{\prime }}$ with $e(s)\le b_{n,m}$. By F2 choose $q\le p_0$ in every $D_n$. Then incompatibility with $p_0$ is impossible. Let $f(n)$ be the least $m$ with $e(q)\le b_{n,m}$. Completeness gives $e(q)\le {\bigwedge }_n{b}_{n,f(n)}\le r$, while $e(q)\le e(p_0)\le \mathrm{\neg }r$, a contradiction. Therefore $d=0$, so $c\le r$ and the exact diagonal law holds.

Steps 1.1-2.3 give nontriviality, completeness, atomlessness, ccc, and precisely the identity in F1. Therefore $B$ is a Suslin algebra. AC is used only at step 2.2 for a set-indexed simultaneous selection and through the already declared distributivity supplier; the regular-open construction itself is choice free.

By AC well-order $B$. For each $a>0$, atomlessness makes $S_a=\{c:0<c<a\}$ nonempty; let $s(a)$ be its least member and put $a^0=s(a)$ and $a^1=a\wedge \mathrm{\neg }s(a)$. Then $a^0,a^1$ are nonzero, disjoint, and have join $a$: if $a^1=0$, then $a\le s(a)$, contradicting $s(a)<a$. Thus this one fixed selector gives a genuine binary split of every positive element, including $1$; zero is never a node.

Prescribe $A_0=\{1\}$; prescribe $A_{\alpha +1}=\{a^0,a^1:a\in A_{\alpha }\}$; and, for every nonzero limit $\lambda <{\omega }_1$, prescribe $A_{\lambda }$ to be exactly the positive meets ${\bigwedge }_{\xi <\lambda }{a}_{\xi }$ of coherent sequences with $a_{\xi }\in A_{\xi }$ and $a_{\eta }\le a_{\xi }$ whenever $\xi <\eta <\lambda$. These clauses are determined by the earlier levels and the fixed selector from step 1.1, so F4 gives a unique sequence $(A_{\alpha }{)}_{\alpha <{\omega }_1}$.

Inductively, each $A_{\alpha }$ is a countable maximal Boolean antichain and every later level refines every earlier one. This is clear for $A_0$; the split identities of step 1.1 prove it at successors and prove strict refinement. Let $0<\lambda <{\omega }_1$ be limit and assume the assertion below $\lambda$. The tagged union of the earlier levels is countable by F5, since $\lambda$ and all its levels are countable. Choose a nondecreasing cofinal sequence $({\xi }_n{)}_{n<\omega }$ in $\lambda$ and enumerate each nonempty countable antichain $A_{{\xi }_n}$ as $(a_{n,m}{)}_{m<\omega }$, repeating entries when necessary. Each row has join $1$, so F1 gives $1={\bigwedge }_n{\bigvee }_m{a}_{n,m}={\bigvee }_{f\in {\omega }^{\omega }}{\bigwedge }_n{a}_{n,f(n)}$. A positive diagonal meet can use only compatible entries; refinement and the antichain property then make these entries a decreasing cofinal selection, which extends uniquely to a coherent choice through every earlier level. Its meet over all $\xi <\lambda$ equals its meet on the cofinal sequence, so every positive diagonal meet belongs to $A_{\lambda }$. Hence $\bigvee A_{\lambda }=1$. Distinct coherent branches first differ in some earlier antichain and therefore have disjoint meets, so $A_{\lambda }$ is an antichain; join $1$ makes it maximal, and ccc makes it countable. Its definition gives refinement. This proves the induction, and also proves that every limit level is precisely the displayed continuous branch-meet level rather than a subsequent maximal extension.

Let $T=\{(\alpha ,a):\alpha <{\omega }_1\text{ and }a\in A_{\alpha }\}$ and define $(\alpha ,a){<}_T(\beta ,b)$ exactly when $\alpha <\beta$ and $b{\le }_{B}a$. By step 3.1, every $b\in A_{\beta }$ lies below exactly one member of each $A_{\alpha }$ for $\alpha <\beta$: existence is refinement and uniqueness is disjointness. Consequently the strict predecessors of $(\beta ,b)$ are well-ordered with one node at each height below $\beta$, so $T$ is a tree, its $\alpha$-th level is the tagged copy of $A_{\alpha }$, and its height is ${\omega }_1$.

The node $(0,1)$ is the unique root. If $(\alpha ,a)\in T$ and $\alpha <\beta <{\omega }_1$, some member of $A_{\beta }$ lies below $a$: otherwise refinement would put every member of $A_{\beta }$ below an $A_{\alpha }$-member disjoint from $a$. In a complete Boolean algebra, fixed meet distributes over an arbitrary join (if $y$ bounds every $a\wedge x$, then $\mathrm{\neg }a\vee y$ bounds every $x$), so this would give $a=a\wedge \bigvee A_{\beta }={\bigvee }_{b\in A_{\beta }}(a\wedge b)=0$, a contradiction. Thus every node extends to every higher level. If two nodes on a nonzero limit level have the same strict predecessors, their coherent earlier choices agree, and step 2.1 makes both Boolean values the meet of that same branch, so the nodes coincide. At a successor level, step 1.1 gives exactly the two immediate successors $(\alpha +1,a^0)$ and $(\alpha +1,a^1)$ of $(\alpha ,a)$. Hence $T$ is normal and splitting in the exact sense of F2.

If two nodes are incomparable in $T$, their Boolean values are disjoint: for nodes on different levels, the later value lies below a unique member of the earlier antichain, and incomparability says that member is not the earlier node. Thus a tree antichain maps injectively to a Boolean antichain, which is countable by the ccc of $B$. In particular every level is countable, as was also proved in step 3.1.

Suppose that $C$ were a cofinal branch. Maximality of a branch together with the unique-predecessor description in step 4.1 puts exactly one node $(\alpha ,a_{\alpha })$ of $C$ on every level. At each successor, $0<a_{\alpha +1}<a_{\alpha }$ by the strict split, so $d_{\alpha }=a_{\alpha }\wedge \mathrm{\neg }{a}_{\alpha +1}$ is nonzero. If $\alpha <\beta$, then $d_{\beta }\le a_{\beta }\le a_{\alpha +1}$ while $d_{\alpha }\wedge a_{\alpha +1}=0$; hence $(d_{\alpha }{)}_{\alpha <{\omega }_1}$ is an uncountable Boolean antichain, contradicting ccc. Therefore $T$ has no cofinal branch.

Steps 4.1, 5.1, 5.2, and 6.1 verify height ${\omega }_1$, countable levels, no cofinal branch, no uncountable antichain, normality, and splitting. By F3, $T$ is a normal splitting Suslin tree, and step 3.1 supplies the promised continuous refining antichain sequence. AC was used only to fix the simultaneous split selector and the countable enumerations/cofinal sequences; no Boolean prime ideal theorem or maximal-antichain extension is used.

Write $\mathsf{S}\mathsf{T}$, $\mathsf{S}\mathsf{L}$, and $\mathsf{S}\mathsf{A}$ for the respective existence assertions in clauses 1-3. These are genuine existence statements under the fixed nonempty, nontrivial conventions in F1 and the cited tree interfaces.

If $\mathsf{S}\mathsf{T}$ holds, F2 constructs a Suslin line, so $\mathsf{S}\mathsf{T}\Rightarrow \mathsf{S}\mathsf{L}$. Conversely, if $\mathsf{S}\mathsf{L}$ holds, F3 constructs a Suslin tree, so $\mathsf{S}\mathsf{L}\Rightarrow \mathsf{S}\mathsf{T}$. Hence $\mathsf{S}\mathsf{T}⟺\mathsf{S}\mathsf{L}$.

If $\mathsf{S}\mathsf{T}$ holds, first apply F4 to obtain a normal splitting Suslin tree, then apply F5 to its reverse-order regular-open completion. The output is a Suslin algebra, so $\mathsf{S}\mathsf{T}\Rightarrow \mathsf{S}\mathsf{A}$.

Conversely, F6 sends any Suslin algebra to a normal splitting Suslin tree, so $\mathsf{S}\mathsf{A}\Rightarrow \mathsf{S}\mathsf{T}$. Together with step 2.2 this gives $\mathsf{S}\mathsf{T}⟺\mathsf{S}\mathsf{A}$.