Suslin Trees, Lines, Algebras, and Independence: Examples and Counterexamples

A branch maximal under inclusion must end on level $3$, so it is determined by one of the eight words $000,001,010,011,100,101,110,111$. Conversely each such word is terminal and hence maximal. For distinct terminal words $s,t$, the comparison rule is $s<t⟺s(k)<t(k)\text{ for }k=\min \{i<3:s(i)\mathrm{\ne }t(i)\}\mathrm{.}$ The displayed list in the statement follows by sorting first by coordinate $0$, then by coordinate $1$, then by coordinate $2$.

The first-difference coordinates of consecutive branches are explicit:

The adjacent pair $000<001$ has no branch strictly between it, so this order is not dense. The elements $000$ and $111$ are respectively a first and last element, so it is not endpoint-free. Moreover $(000,010)=\{001\}$ is a nonempty open interval with the countable dense subset $\{001\}$; hence the nowhere-separable conclusion fails. Since the whole order has only eight elements, every antichain of pairwise disjoint nonempty open intervals is finite, so its ccc assertion holds only trivially.

The failed order-density and endpoint conclusions in step 2.1 come from the local successor order $\{0<1\}$: it has neither an intermediate successor nor successors on both sides. In [F1], replacing each successor set by a dense order without endpoints supplies exactly those insertions. By contrast, the failure of nowhere separability reflects bounded height: the argument in [F2] first bounds an arbitrary countable candidate dense set by a height $\delta$ and then splits above $\delta$, which is impossible in $2^{\le 3}$. Thus local dense successor orders yield density and no endpoints, while unbounded normal splitting is essential to the later nowhere-separability construction.

The root is the empty word, but no empty terminal branch occurs; at height zero it has exactly the two successors $0,1$. Every top word has length exactly three, so there is no hidden extension beyond the listed eight. All lists, comparisons, and witnesses are explicit finite data in ZF, and no choice principle is used.

At stage $0$ there are no earlier endpoints. Choose $c_0<a_0<b_0<d_0\mathrm{.}$ The interval $I_0=[a_0,b_0]$ is nonempty and nondegenerate. The empty set of old endpoints creates no avoidance condition.

Fix any later stage $\alpha$. The set $E_{\alpha }=\{a_{\xi },b_{\xi }:\xi <\alpha \}$ is countable because $\alpha <{\omega }_1$. It cannot be dense in $L$: if it were, then for any $u<v$ the countable set $E_{\alpha }\cap (u,v)$ would be dense in the nonempty open interval $(u,v)$, contrary to nowhere-separability. Hence some nonempty open gap $(c,d)$ misses $E_{\alpha }$, and density lets the recursion choose $c<a_{\alpha }<b_{\alpha }<d$. Now fix $\xi <\alpha$ and write $I_{\xi }=[a_{\xi },b_{\xi }]$. There are exactly three relative positions. If $(c,d)$ meets $(a_{\xi },b_{\xi })$, order-convexity and endpoint avoidance force $(c,d)\subset (a_{\xi },b_{\xi })$, hence $I_{\alpha }\subset (a_{\xi },b_{\xi })$. If it lies to the left, then $b_{\alpha }<a_{\xi }$; if it lies to the right, then $b_{\xi }<a_{\alpha }$. In the last two cases the two closed intervals, and therefore their open interiors, are disjoint. Equality at a boundary cannot occur because $I_{\alpha }$ lies strictly inside $(c,d)$.

At stage $1$, use the nonempty open interval $(a_0,b_0)$, which contains neither of its endpoint witnesses. Density gives $a_0<c_1<a_1<b_1<d_1<b_0\mathrm{.}$ Hence $I_1\subset (a_0,b_0)$, so index $0$ is a predecessor of index $1$ in the reverse-nesting tree.

At stage $2$, the interval $(a_0,a_1)$ is nonempty and avoids all four earlier endpoints $a_0,b_0,a_1,b_1$. Choose $a_0<c_2<a_2<b_2<d_2<a_1<b_1<b_0\mathrm{.}$ It follows that $I_2\subset (a_0,b_0)$ but $I_2\cap I_1=\mathrm{\varnothing }$. Thus $0$ is a predecessor of $2$, whereas $1$ and $2$ are incomparable. This is the displayed three-stage configuration.

Apply step 1.2 to every earlier index. If two earlier indices $\xi ,\eta$ both contain a later $I_{\alpha }$, their own intervals intersect, so the disjoint alternative is impossible and one is nested inside the other according to index order. Therefore predecessor sets are linearly ordered. Conversely, incomparable indices must fall into the disjoint alternative; this is exactly what happens to indices $1$ and $2$ in step 3.1.

Stages zero, one, and two respectively treat the empty old-endpoint set, a single old interval, and two earlier intervals. Every chosen interval has two strictly ordered endpoints; no empty or singleton interval is admitted. The finite trace uses only finitely many existential witnesses, while [A1] is retained for the simultaneous ${\omega }_1$-stage recursion in the supplier. The construction uses order endpoints only as avoided boundary points and assumes that the ambient line itself has no first or last element.

First distinguish the two objects. For a ground condition $p$, a set $B\subseteq T_p$ is already a ground antichain and its membership is settled. By contrast, $\dot{A}$ is a name for a subset of the eventual union: $p_0$ need not decide its members, and its value need not be contained in $T_{p_0}$. Maximality of $\dot{A}$ says that every node of $\dot{T}$ is comparable with some named member, not that the intersection $\dot{A}\cap {\check{T}}_{p_0}$ is already maximal in $T_{p_0}$.

Put $p_0^{\ast }=r_0$. At outer round $n$, enumerate the countable tree $T_{p_n^{\ast }}$ as $(t_{n,m}{)}_{m<\omega }$. Starting with $p_{n,0}=p_n^{\ast }$, construct a descending sequence. Because $p_{n,m}$ still forces $\dot{A}$ maximal, it forces that some member of $\dot{A}$ is comparable with $t_{n,m}$. The existential forcing clause from [F7], dense decision from [F5], and [F4] let us strengthen to $p_{n,m+1}$ and decide such a member as a ground countable sequence $a_{n,m}$. The atomic clause [F6] permits a further strengthening whose tree actually contains $a_{n,m}$. Thus $p_{n,m+1}⊩{\check{a}}_{n,m}\in \dot{A}\text{ and }{\check{a}}_{n,m}\parallel {\check{t}}_{n,m}\mathrm{.}$ Countable closure gives a lower bound ${\bar{p}}_n$ for the inner sequence. Strengthen once more to $p_{n+1}^{\ast }\le {\bar{p}}_n$ with strictly larger top height. Use [F9] and [A1] for all $n,m<\omega$.

Let $U={\bigcup }_{n<\omega }{T}_{p_n^{\ast }},A^{\ast }=\{a_{n,m}:n,m<\omega \}\mathrm{.}$ The strictly increasing top heights make $U$ a normal splitting tree of nonzero countable limit height; $U$ itself has no top and is not yet a forcing condition. It is countable by [F8]. Every $t\in U$ lies in some $T_{p_n^{\ast }}$ and appears in that round's enumeration, so it is comparable with the corresponding $a_{n,m}\in A^{\ast }$. If two distinct elements of $A^{\ast }$ were comparable, a sufficiently late condition would contain them both and, by persistence in [F5], force both into the antichain $\dot{A}$, a contradiction. Hence $A^{\ast }$ is an antichain and the preceding coverage makes it maximal in $U$. Repeated decisions may yield the same $a_{n,m}$; set formation removes repetitions.

Apply [F3] to $(U,A^{\ast })$. Choose a countable covering family of cofinal branches of $U$, each meeting $A^{\ast }$, and add one distinct top node for each distinct branch. The result is a countable normal splitting tree $T_q$ with a genuine new top, hence a condition $q$. It end extends every fusion condition, so $q\le r_0$, and persistence gives $q⊩{\check{A}}^{\ast }\subseteq \dot{A}\mathrm{.}$ Every node of $T_q$ is comparable with a member of $A^{\ast }$, including each new top by construction.

Let $s\le q$. Literal end extension [F2] preserves $T_q$. Any node of $T_s$ already in $T_q$ is comparable with $A^{\ast }$ by step 4.1. Any new node has a unique predecessor on the top level of $T_q$; that top extends a member of $A^{\ast }$, so the new node does too. Thus every future node remains comparable with $A^{\ast }$, and density plus [F5] gives $q⊩{\check{A}}^{\ast }\text{ is maximal in }\dot{T}\mathrm{.}$ Since $q$ also forces that $\dot{A}$ is an antichain containing $A^{\ast }$, no distinct member can be added to $A^{\ast }$; therefore $q⊩\dot{A}={\check{A}}^{\ast }$.

The root handles a starting tree with only one node and shows that the forced maximal antichain cannot be empty. The indices $n=m=0$ give the first actual decision; zero outer rounds would not cover even the starting tree and are not used. The omega-union deliberately loses a top, and step 4.1 restores one. Empty or repeated branch/decision lists are not silently counted as new nodes: repetitions are removed, while nonemptiness follows from the root comparison. All names and assertions remain inside the forcing relation of [F7]; no universe-generic is claimed. Choice is spent in [A1] on the evolving tree enumerations and deciding extensions and through [F8], while the ground sealing operation [F3] itself is choice-free.

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.

Put $T_{mathrmMA}=\mathrm{Z}\mathrm{F}\mathrm{C}+\mathrm{M}\mathrm{A}+\mathrm{\neg }\mathrm{C}\mathrm{H}$ and $S_0=\mathrm{Z}\mathrm{F}\mathrm{C}+\mathrm{S}\mathrm{H}+\mathrm{\neg }\mathrm{C}\mathrm{H}$. By [F1], the given consistency hypothesis makes $T_{\mathrm{M}\mathrm{A}}$ externally consistent. Fix the finite $T_{\mathrm{M}\mathrm{A}}$ proof of SH supplied by [F2]. If $S_0$ had a certified finite refutation, replace every occurrence of its added SH axiom by a fresh copy of that fixed proof; its ZFC and $\mathrm{\neg }$CH axiom lines already belong to $T_{\mathrm{M}\mathrm{A}}$. The resulting finite derivation would refute $T_{\mathrm{M}\mathrm{A}}$, contrary to [F1] and [F11]. Hence $S_0$ is consistent. It asserts SH and $\mathrm{\neg }$CH together, so it refutes the implication SH$\to$CH. Zero, one, or repeated SH-axiom occurrences are handled by the same finite splice.

Build the other branch inside the verified $L$-interpretation. By [F4], there are fixed ZF proofs that internally $L$ satisfies ZFC, $V=L$, and AC. Translate [F5] and then [F6] inside $L$ to obtain one fixed certified ZF proof $d_{\mathrm{C}\mathrm{H}}$ of ${\mathrm{C}\mathrm{H}}^L$. Independently, translate [F7] and [F8] inside $L$ and use [F9] to obtain one fixed certified ZF proof $d_{\mathrm{\neg }\mathrm{S}\mathrm{H}}$ of $(\mathrm{\neg }\mathrm{S}\mathrm{H}{)}^L$. These are literal guarded relativizations in the fixed calculus, with capture-free substitutions. Object-level Choice is used inside $L$ by the diamond, tree and tree-to-line suppliers as recorded in [A1]; ambient ZF makes no family choice here.

Let $S_1=\mathrm{Z}\mathrm{F}\mathrm{C}+\mathrm{C}\mathrm{H}+\mathrm{\neg }\mathrm{S}\mathrm{H}$. Extend the verified dispatcher [F3] by two decidable axiom tags: on the CH tag return the constant block $d_{\mathrm{C}\mathrm{H}}$, and on the $\mathrm{\neg }$SH tag return $d_{\mathrm{\neg }\mathrm{S}\mathrm{H}}$. Retain the existing ZFC branches, capture guards, malformed-input tautology, and the terminal relativized-contradiction block supplied by [F12]. Two finite case branches and two constant certified blocks preserve primitive recursiveness, and PA verifies their line-prefix checker acceptance. Thus PA verifies a total map sending every certified $S_1$ refutation first to a ZF refutation and then, by identical axiom lines, to a ZFC refutation. Zero or repeated occurrences of either added axiom reuse the same dispatcher branches.

Apply [F10] to the map of step 2.1, with source ZFC and target $S_1$. It yields $\mathrm{P}\mathrm{A}⊢\mathrm{Con}(\mathrm{Z}\mathrm{F}\mathrm{C})⟶\mathrm{Con}(S_1),$ and therefore the required external consistency consequence. The theory $S_1$ asserts CH and $\mathrm{\neg }$SH, so it refutes CH$\to$SH. This is a syntactic consistency transfer through $L$, not an extraction of a transitive model from consistency.

By steps 1.1 and 3.1, under external $\mathrm{Con}(\mathrm{Z}\mathrm{F}\mathrm{C})$ there is a consistent extension refuting SH$\to$CH and a consistent extension refuting CH$\to$SH. A certified ZFC proof of either implication would also be a proof in the corresponding extension and, together with that extension's two added axioms, would give a fixed propositional refutation. Hence ZFC proves neither implication and cannot prove SH$\leftrightarrow$CH. The empty or malformed proof-code cases do not witness derivability; each joint theory has both advertised axioms, including CH and SH in opposite truth patterns. No actual generic or set model is inferred. The only Choice uses are already inside the two ZFC branches and internally in $L$; the final proof-code transformations use finite recursion only.