Set-Theoretic Trees, Delta Systems, and Diamond

Let $\alpha =\mathrm{ht}(t)$ and let $e:\alpha \to P_t$ be its order isomorphism. For $\beta <\alpha$, transitivity and order reflection give $P_{e(\beta )}=e[\beta ]$. Thus $\mathrm{ht}(e(\beta ))=\beta$. Conversely every predecessor is $e(\gamma )$ for exactly one $\gamma <\alpha$ and has height $\gamma$, proving existence and uniqueness.

If $s,t{\le }_{T}u$ and either equals $u$, they are comparable. Otherwise both belong to the well-ordered set $P_u$, so its linear order compares them. This includes $s=t$.

If $s{<}_{T}t$, step 1.1 puts $s=e(\beta )$ at some $\beta <\mathrm{ht}(t)$, so $\mathrm{ht}(s)<\mathrm{ht}(t)$. Distinct nodes of one level therefore cannot be comparable; each level is an antichain. For a root the predecessor assertion has no indices, and empty levels satisfy the antichain assertion vacuously.

For each $t$, let $S_t$ be its immediate-successor set. With alphabet $A=T$, use the injection $e_t:S_t\to T$, $e_t(u)=u$. In the countable-successor case, the set $J_t$ of injections $S_t\to \omega$ is nonempty, including the empty map when $S_t$ is empty. All these maps lie in a set of relations contained in $T\times \omega$. Well-order that set and let $e_t$ be the least member of $J_t$. This is the sole use of AC; take $A=\omega$ in this case.

Define codes by recursion on levels. Give the root the empty code. At height $\beta +1$, the unique predecessor $t$ at height $\beta$ is the immediate predecessor of $u$; set $f(u)=f(t{)}^{⌢}⟨e_t(u)⟩$. At a nonzero limit height $\lambda$, put $f(u)={\bigcup }_{\beta <\lambda }f(u_{\beta })$, where $u_{\beta }$ is the unique height-$\beta$ predecessor. These are set-valued level operations, so transfinite recursion applies. On histories not satisfying the stated coherence condition the operation can be assigned the empty set; the next step proves that such histories never occur in the recursion.

By induction on the constructed level, $f(u)$ has domain $\mathrm{ht}(u)$ and restricts to $f(u_{\gamma })$ at every lower height $\gamma$. This is vacuous at the root. At a successor, appending one coordinate gives the domain and preserves all earlier restrictions. At a limit the earlier codes agree on overlaps by the induction assertion; their union is a function with domain the union of all smaller ordinals, namely that limit. Its restrictions are exactly the earlier codes.

The codes are injective on each level, again by induction. At zero there is just one root. At a successor, equality of codes gives equality of parent codes and hence of parents; equality of last coordinates then gives equality of the successors by injectivity of $e_t$. At a nonzero limit, equality of codes and step 3.1 give equal codes for each pair of predecessors; level injectivity below the limit makes all those predecessors equal. Thus the predecessor sets agree, and normality makes the two nodes equal.

Different levels give different code domains, so $f$ is injective on $T$. If $s{<}_{T}t$, step 3.1 identifies $f(s)$ with a proper restriction of $f(t)$. Conversely, if $f(s)$ is a proper initial segment of $f(t)$, take the predecessor $u$ of $t$ at height $\mathrm{ht}(s)$. Step 3.1 gives $f(u)=f(s)$, and step 4.1 gives $u=s$, hence $s{<}_{T}t$.

Every proper restriction of $f(t)$ is the code of its predecessor at that length, so the image is downward closed. The map onto its image is therefore the required order isomorphism and preserves heights by the domain computation. A height-one tree maps just to the empty sequence; no surjectivity onto all $A^{<\alpha }$ is needed.

Call $t$ good if the heights of nodes above or equal to $t$ are unbounded in $\omega$. The height assumption and F1 imply that every level is nonempty. Some root is good: otherwise, each of the finitely many roots has a finite height bound on its extensions; their maximum bounds all nodes because each node has a root predecessor (or is a root). That contradicts height $\omega$.

If a good node $t$ has height $n$, its immediate successors are precisely its extensions at height $n+1$, by F1. This is a finite set. Each higher extension of $t$ passes through one of these successors. If none were good, the maximum of their finitely many bounds, together with $n+1$, would bound all extensions of $t$. Thus a good immediate successor exists.

Fix a well-order of $T$ using AC. Let $t_0$ be the least good root and send each good node to its least good immediate successor. On the set of good nodes this is a self-map, so recursion gives $t_n$ at height $n$ with $t_n{<}_T{t}_{n+1}$ for every $n$.

The set $B=\{t_n:n<\omega \}$ is an infinite chain. If $u$ could be added to it, put $n=\mathrm{ht}(u)$. Comparability with $t_n$ and the level-antichain conclusion of F1 force $u=t_n$. Thus $B$ is already maximal, hence is an infinite branch.

Fix surjections $e:\omega \to T$ and $d:\omega \to \delta$. They exist by countability: $\delta$ is infinite, and $T$ has nodes at arbitrarily high levels below $\delta$, so it too is infinite. Only these two witnesses are fixed. Define ${\gamma }_0=d(0)$ and ${\gamma }_{n+1}=\max \{{\gamma }_n+1,d(n+1)\}$. The successor of every ordinal below the limit $\delta$ is still below $\delta$, so recursion gives a strictly increasing sequence in $\delta$. It is cofinal since ${\gamma }_n\ge d(n)$. The nonautonomous rule is a recursion on the state $(n,{\gamma }_n)$.

Starting at $t_0=t$, let ${\beta }_n=\max \{{\gamma }_n,\mathrm{ht}(t_n)+1\}<\delta$ and take $t_{n+1}=e(k)$ for the least $k$ for which $t_n{<}_{T}e(k)$ and $\mathrm{ht}(e(k))={\beta }_n$. The candidate set is nonempty by normality. Recursion on $(n,t_n)$ supplies this sequence, and its heights are cofinal because $\mathrm{ht}(t_{n+1})\ge {\gamma }_n$. Least natural indices require no choice function.

Put $b_t=\{s\in T:\mathrm{\exists }ns{\le }_T{t}_n\}$. It contains $t$ and is a chain: if $s{\le }_T{t}_n$ and $u{\le }_T{t}_m$, both lie below $t_{\max \{n,m\}}$, so they are comparable. Its heights are cofinal by step 2.1.

If $s$ can be adjoined to $b_t$ while retaining a chain, choose $n$ with $\mathrm{ht}(t_n)>\mathrm{ht}(s)$, possible by cofinality and the limit-height hypothesis. Comparability with $t_n$ and strict increase of height force $s{<}_T{t}_n$, hence $s\in b_t$. Thus $b_t$ is maximal and is a cofinal branch.

The same fixed $e,d$ determine $b_t$ uniquely for each $t\in T$. Replacement therefore forms $\mathcal{B}=\{b_t:t\in T\}$. The sequence $n\mapsto b_{e(n)}$ is onto $\mathcal{B}$, so it is countable, and $t\in b_t$ proves that it covers $T$. The construction includes the root and every prescribed node; no last level exists at the limit height.

For each $\alpha <{\omega }_1$, cofinality gives $v\in b$ of height at least $\alpha$. If its height is greater, let $u$ be its unique predecessor of height $\alpha$; otherwise set $u=v$. Every $w\in b$ is comparable with $u$: if $w{\le }_{T}v$, use common-predecessor comparability, and if $v{<}_{T}w$, use $u{\le }_{T}v{<}_{T}w$. Maximality of $b$ therefore puts $u$ in $b$. Distinct nodes of the same height cannot both belong to a chain. Thus there is a unique $b_{\alpha }\in b\cap T_{\alpha }$ for every $\alpha$. For $\alpha <\beta$, comparability and height give $b_{\alpha }{<}_T{b}_{\beta }$.

The node $b_{\alpha +1}$ is an immediate successor of $b_{\alpha }$, since an intermediate node would have height strictly between $\alpha$ and $\alpha +1$. Conversely every immediate successor $u$ of $b_{\alpha }$ has height $\alpha +1$: if its height were larger, its predecessor at $\alpha +1$ would lie strictly between $b_{\alpha }$ and $u$. Since $\alpha +1<{\omega }_1$, splitting makes $S_{\alpha }=\{u:u\text{ is an immediate successor of }{b}_{\alpha },u\mathrm{\ne }{b}_{\alpha +1}\}$ nonempty. AC gives $a_{\alpha }\in S_{\alpha }$ for every $\alpha <{\omega }_1$.

If $\alpha <\beta$, then $b_{\alpha +1}{\le }_T{b}_{\beta }{<}_T{a}_{\beta }$. Were $a_{\alpha }$ and $a_{\beta }$ comparable, their different heights $\alpha +1<\beta +1$ would force $a_{\alpha }{<}_T{a}_{\beta }$. The two distinct nodes $a_{\alpha },b_{\alpha +1}$ of height $\alpha +1$ would then be predecessors of $a_{\beta }$, contradicting uniqueness at that height. Thus $a_{\alpha },a_{\beta }$ are incomparable.

Consequently $\{a_{\alpha }:\alpha <{\omega }_1\}$ is an antichain. Its indexing is injective because $\mathrm{ht}(a_{\alpha })=\alpha +1$, so it has cardinality ${\mathrm{\aleph }}_1$. The argument includes $\alpha =0$ and adjacent levels; all successor levels used remain below ${\omega }_1$.

Fix enumerations of $T$ and $\delta$. From an enumeration $d:\omega \to \delta$, define ${\gamma }_0=d(0)$ and ${\gamma }_{k+1}=\max \{{\gamma }_k+1,d(k+1)\}$. All terms lie below the limit $\delta$ and ${\gamma }_k\ge d(k)$, so the sequence is cofinal. By F1 and F2, $T\times \mathbb{Q}$ has an enumeration. Keep precisely the entries $(x,r)$ with $\mathrm{\ell }(x)<r$ to enumerate all requests as $(x_n,r_n)$. There are infinitely many entries to keep, since for one fixed $x$ the distinct rationals $\mathrm{\ell }(x)+m+1$ give requests for all natural $m$. Retaining successive least valid indices is recursion, not countable choice.

Suppose branches $b_0,\dots ,b_{n-1}$ for the earlier requests have been defined. Set $q_n=(\mathrm{\ell }(x_n)+r_n)\mathrm{/}2$, so $\mathrm{\ell }(x_n)<q_n<r_n$. Infinitely many immediate successors of $x_n$ have labels below $q_n$. Each earlier branch contains at most one of them, since distinct immediate successors are incomparable by F5. Thus finitely many earlier branches exclude at most finitely many candidates. Choose the least enumerated remaining successor $z_0$; it has label below $q_n$ and belongs to none of the earlier branches.

Recursively, with $z_k$ already defined, put ${\eta }_k=\max \{{\gamma }_k,\mathrm{ht}(z_k)+1\}<\delta$. The first invariant gives an extension $z_{k+1}$ of height ${\eta }_k$ and label below $q_n$, since $\mathrm{\ell }(z_k)<q_n$. Take the least enumerated such extension. This defines a strictly increasing chain whose heights are cofinal and whose labels are all less than $q_n$. Recursion uses the state $(k,z_k)$, and the successor bound stays below $\delta$ because it is limit.

Let $b_n=\{s:\mathrm{\exists }ks{\le }_T{z}_k\}$. Common-predecessor comparability makes this a chain; its cofinality and unique predecessors give exactly one node of each height below $\delta$. A node comparable with all of $b_n$ lies below some $z_k$ of greater height and hence belongs to $b_n$, so it is a maximal chain. Every label on $b_n$ is less than $q_n$: for $s{\le }_T{z}_k$, strict increase gives $\mathrm{\ell }(s)\le \mathrm{\ell }(z_k)<q_n$. It contains $x_n$ and $z_0$, and $z_0$ belongs to no earlier branch, so $b_n$ differs from all of them. This construction determines $b_n$ uniquely from the finite list of previous branches; recursion on finite lists therefore produces all $b_n$ simultaneously.

Adjoin a distinct top $v_n$ above precisely $b_n$, with label $q_n$. Formally use the disjoint union of a tagged copy of $T$ and a tagged copy of $\omega$, ordered by the old order and $s<v_n$ iff $s\in b_n$. A branch is downward closed, so this order is transitive. The predecessor set of $v_n$ is ordered like $\delta$ by height, proving that these are exactly a new level at $\delta$. Their predecessor sets are distinct by step 4.1. The new tree is countable by interleaving its old enumeration with $n\mapsto v_n$. Strict increase holds on new comparisons because $\mathrm{\ell }(s)<q_n=\mathrm{\ell }(v_n)$ for $s\in b_n$; there are no comparisons between new tops.

For any old $x$ and rational $r>\mathrm{\ell }(x)$, its request occurs at some $n$. Then $x{<}_T{v}_n$ and $\mathrm{\ell }(v_n)=q_n<r$, proving the invariant at the new level. Earlier instances are unchanged and new tops have no higher level to check. If the old tree is normal, its root and old limit-level uniqueness persist, the request argument supplies extensions to the new level, and step 5.1 supplies new limit-level uniqueness. No immediate successors of old nodes are lost or changed, since their successor heights are strictly below $\delta$; new tops have no successor requirement. Every selection used least indices after finitely many fixed enumerations.

Fix an enumeration of $\mathbb{Q}$, a pairing enumeration of $\omega \times \omega$, and, by AC, surjections $d_{\alpha }:\omega \to \alpha$ for all $0<\alpha <{\omega }_1$. Each set of such surjections is nonempty by countability of $\alpha$; these sets form a set-indexed family. At every stage store a surjection onto the newly constructed level. Enumerated nodes can be renamed by their least enumeration indices, with level tags $(\alpha ,n)$, so all old nodes keep their names. All possible codes for a countable new level on the fixed ambient node set ${\omega }_1\times \omega$—its predecessor sets, rational labels and a surjection from $\omega$ onto the level—form a set $C$: the node sets, predecessor relations, label graphs and enumeration graphs are subsets of fixed sets built from ${\omega }_1\times \omega$, $\mathbb{Q}$ and $\omega$. By A1 fix a choice function $c$ on $\mathcal{P}(C)\setminus \{\mathrm{\varnothing }\}$. Applying this fixed function to the nonempty set of eligible limit-level codes makes every stage below a specified rule.

Start with the singleton root $T_0=\{(0,0)\}$ with label zero and its constant enumeration. At a successor stage, for each $x\in T_{\alpha }$ create a distinct immediate successor $(x,q)$ for every rational $q>\mathrm{\ell }(x)$, with label $q$, then rename these pairs by level-tagged least indices as in step 1.1. Its predecessors are $x$ and all predecessors of $x$. The level is nonempty and countable, since it is an enumerated subset of $T_{\alpha }\times \mathbb{Q}$. Strict labeling persists. There are infinitely many successors below any $r>\mathrm{\ell }(x)$: the distinct rationals $\mathrm{\ell }(x)+(r-\mathrm{\ell }(x))\mathrm{/}(m+2)$ for $m<\omega$ lie strictly between the two bounds. In particular every old last-level node splits.

At every partial stage the small-successor condition is required only for nodes whose successor level has already been constructed. Maintain also the invariant that for $\mathrm{ht}(x)<\beta$ and rational $r>\mathrm{\ell }(x)$ an extension at level $\beta$ has label less than $r$. It is vacuous at the root stage. At the successor level $\alpha +1$, if $x\in T_{\alpha }$, its successor of label $(\mathrm{\ell }(x)+r)\mathrm{/}2$ works. If $x$ lies below $\alpha$, use the earlier invariant with bound $s=(\mathrm{\ell }(x)+r)\mathrm{/}2$ to obtain $y\in T_{\alpha }$ above $x$ with $\mathrm{\ell }(y)<s$, and extend $y$ by the successor of label $(\mathrm{\ell }(y)+r)\mathrm{/}2<r$. This verifies every request at the new level; all earlier requests persist.

At a nonzero limit $\delta <{\omega }_1$, the union of earlier levels is countable: enumerate its nodes by pairing $d_{\delta }(n)$ with the stored enumeration of that level. Its height is $\delta$, and its order and labeling satisfy all previous invariants, since every pair of old nodes and every request involving a level below $\delta$ occurs in an earlier stage. Every old node has infinitely many small successors, supplied at its successor stage, which is below $\delta$. The old union is normal: roots agree, extensions persist, and predecessor sets at each old limit level were fixed when that level was added. F1 gives at least one nonempty countable new level preserving strict labeling, bounded extensions and normality, and retaining the small-successor condition at old nodes. New tops have no successor requirement until the next stage. Injectively rename its nodes by tags $(\delta ,n)$ and store a surjection from $\omega$ onto the renamed level. Let $E\subseteq C$ be the set of eligible codes for this history and take the new level record to be $c(E)$. Eligibility requires the old order and labels to be unchanged, the new nodes to have level tag $\delta$, and exactly the properties just obtained from F1. The set $E$ is nonempty by F1 and countability, so this selection is defined uniquely from the earlier history and the fixed $c$. All invariants, with the stage-relative successor requirement of step 3.1, and normality persist.

Steps 2.1–4.1 prescribe the next level and its enumeration from the earlier history and the fixed parameters. Extend the rule arbitrarily, say by the empty record, on histories not satisfying the invariants. F2 then defines all levels for $\alpha <{\omega }_1$. Their union is a set by Replacement and Union, with height ${\omega }_1$ and nonempty countable levels. Normality and splitting follow from their stagewise verification: any requested extension, limit-level comparison, or immediate successor appears at some stage. The labeling is strictly increasing because every comparison already appears at one stage.

On any chain the labeling is injective into $\mathbb{Q}$, because distinct comparable nodes have strictly different labels. F4 makes the chain countable. Its height image is countable and hence not cofinal in ${\omega }_1$ by F7, whose choice hypothesis follows from A1. No branch is cofinal. Thus the constructed ${\omega }_1$-tree is Aronszajn, and the increasing rational labeling makes it special by F8.

The tree is uncountable: its height map is onto ${\omega }_1$ since every level is nonempty, whereas the image of a countable set is countable. For each rational $q$, the fiber ${\mathrm{\ell }}^{-1}(\{q\})$ is an antichain by strict increase. If all these fibers were countable, F4 would index them countably and F3, using the countable choice supplied by A1, would make their union $T$ countable, a contradiction. At least one fiber is an uncountable antichain; by F8 the tree is not Suslin.

A union of fewer than $\kappa$ sets each of size less than $\kappa$ has size less than $\kappa$. Indeed the set of their cardinalities has size less than $\kappa$, so by regularity and F2 it is bounded below $\kappa$. Choose an infinite cardinal $\mu <\kappa$ bounding these sizes and the size of the index set. AC selects injections of the sets into $\mu$; assigning each element its least containing index in a fixed well-order injects the union into the product of the index set and $\mu$. Its cardinality is at most $\mu \cdot \mu =\mu <\kappa$. Empty index sets give empty union directly.

Write ${\mathcal{F}}_n=\{a\in \mathcal{F}:\mathrm{\mid }a\mathrm{\mid }=n\}$. If every ${\mathcal{F}}_n$ had size less than $\kappa$, step 1.1, applied to the countable index set and uncountable $\kappa$, would give $\mathrm{\mid }\mathcal{F}\mathrm{\mid }<\kappa$. Hence some ${\mathcal{F}}_n$ has size $\kappa$. It remains to prove the result for uniform size $n$, by induction on $n$. Size zero cannot occur with $\kappa$ distinct sets; for size one all members are pairwise disjoint, giving root $\mathrm{\varnothing }$.

Suppose the uniform-size assertion holds at $n$, and $\mathcal{G}$ consists of $\kappa$ distinct sets of size $n+1$. If some $x$ belongs to $\kappa$ members, delete $x$ from those members. Deletion is injective on sets containing $x$, since adjoining $x$ recovers the original set. The resulting $\kappa$ distinct $n$-element sets have a delta subsystem with root $r$ by the induction hypothesis. Reattach $x$; for distinct members $a,b$ the intersection is $(a\setminus \{x\})\cap (b\setminus \{x\})\cup \{x\}=r\cup \{x\}$.

In the remaining situation each $x$ belongs to fewer than $\kappa$ members of $\mathcal{G}$. Fix a bijective enumeration of $\mathcal{G}$ by $\kappa$. At stage $\gamma <\kappa$, let $U$ be the union of the previously selected sets. By step 1.1, $\mathrm{\mid }U\mathrm{\mid }<\kappa$. The sets intersecting $U$ form the union, over $x\in U$, of fewer-than-$\kappa$ sized subfamilies, so again fewer than $\kappa$ members are excluded. Also exclude all previously selected sets. Fewer than $\kappa$ candidates are excluded in total, leaving a candidate; select the least index. Recursion gives $\kappa$ distinct pairwise disjoint members, a delta system with empty root.

The two alternatives in steps 3.1 and 4.1 exhaust the possibilities and prove the uniform-size successor assertion. Induction with the zero and one cases from step 2.1 proves it at every finite size. Applying it to the subfamily found in step 2.1 proves the theorem.

If some finite set $a$ has uncountable fiber $J=\{\xi :a_{\xi }=a\}$, put $r=a$. For distinct $\xi ,\eta \in J$ one has $a_{\xi }\cap a_{\eta }=a\cap a=a$. This includes the case $a=\mathrm{\varnothing }$.

Otherwise every fiber is countable. Let $\mathcal{A}=\{a_{\xi }:\xi <{\omega }_1\}$. If $\mathcal{A}$ were countable, enumerate its values and apply F2 to their countable fibers; their union is all of ${\omega }_1$, which is uncountable. Hence $\mathcal{A}$ is uncountable. The map $a\mapsto m(a)=\min \{\xi :a_{\xi }=a\}$ injects $\mathcal{A}$ into ${\omega }_1$, so $\mathrm{\mid }\mathcal{A}\mathrm{\mid }={\mathrm{\aleph }}_1$. AC supplies the countable choice used in F2; the least-index map itself needs no choices.

Every ordinal below ${\omega }_1$ is countable; F3 therefore excludes every cofinal subset of cardinality below ${\mathrm{\aleph }}_1$. Thus ${\mathrm{\aleph }}_1$ is regular, and F1 gives an uncountable delta subsystem $\mathcal{B}\subseteq \mathcal{A}$ with finite root $r$. Set $J=\{m(a):a\in \mathcal{B}\}$. The map $m$ is injective, so $J$ is uncountable, and distinct indices in $J$ represent distinct members of $\mathcal{B}$, whose intersection is $r$. Together with the repeated-value alternative this proves the indexed assertion.

For any Knaster poset $Q$ and family $(q_{\xi }{)}_{\xi \in J}$ with uncountable $J\subseteq {\omega }_1$, first suppose the range is countable. Some fiber is uncountable: otherwise an enumeration of the range and F2 would express $J$ as a countable union of countable fibers. Retain such a fiber; all its values coincide and hence are pairwise compatible. If the range is uncountable, F1 gives an uncountable pairwise compatible subset $B$ of the range. Retain $\{\xi \in J:q_{\xi }\in B\}$. Its image is $B$, so it is uncountable, and its values are pairwise compatible, even when repeated. The only infinite choice used here is the countable choice for F2, supplied by A1.

Start with the given indexed family in ${\prod }_{i<n}{P}_i$ and set $J_0=J$. For each $i<n$ apply step 1.1 to the $i$th coordinate on $J_i$, obtaining an uncountable $J_{i+1}\subseteq J_i$ on which that coordinate is pairwise compatible. Earlier coordinate compatibility persists under restriction. Finite iteration gives uncountable $K=J_n$ with every coordinate pair compatible. If $n=0$, set $K=J$ and there are no coordinate conditions to impose.

For distinct $\xi ,\eta \in K$, select for each $i<n$ a common lower bound $r_i{\le }_i{p}_{\xi }(i),p_{\eta }(i)$. There are only finitely many selections, so $r=(r_i{)}_{i<n}$ is a tuple in the product and a lower bound for both. At $n=0$ it is the empty tuple. Thus step 2.1 proves the indexed assertion without any greatest-element hypothesis.

Given an uncountable subset $X$ of the product, F3 and A1 give a well-order of $X$ and hence an injection ${\omega }_1\to X$: enumerate the first ${\omega }_1$ elements of its order type, which must be at least ${\omega }_1$ since $X$ is uncountable. Apply steps 2.1 and 3.1 to this injective enumeration. Its restriction to $K$ remains injective, so its image is an uncountable compatible subset of $X$, as F1 requires. If the product is empty or a singleton (including the empty product), it has no uncountable subset and the Knaster assertion is vacuous.

Well-order $X$ using F5 and A1 and take an injective family $(p_{\xi }{)}_{\xi <{\omega }_1}$ from it, possible since $X$ is uncountable. By F3 the supports $a_{\xi }$ are finite. Apply F1 to obtain an uncountable $J\subseteq {\omega }_1$ and finite $r$ with $a_{\xi }\cap a_{\eta }=r$ for distinct $\xi ,\eta \in J$. In particular $r\subseteq a_{\xi }$ for every $\xi \in J$, since each index has a distinct partner in $J$.

Each restriction $p_{\xi }\upharpoonright r$ belongs to the finite product ${\prod }_{i\in r}{P}_i$. Enumerate the finite set $r$ to apply F2 and obtain uncountable $K\subseteq J$ with pairwise compatible restrictions. Repeated restrictions cause no loss of indices because F2 states its indexed version. If $r=\mathrm{\varnothing }$, all restrictions are the empty tuple and one may take $K=J$.

Fix distinct $\xi ,\eta \in K$. On $r$ take a tuple $q$ below both restrictions. On $a_{\xi }\setminus r$ use $p_{\xi }(i)$, and on $a_{\eta }\setminus r$ use $p_{\eta }(i)$. These two sets are disjoint by step 1.1. Put $s(i)=1_i$ outside $a_{\xi }\cup a_{\eta }$. On a petal the other condition has value $1_i$, so the chosen value is below both; on $r$ use $q$; elsewhere both original values are $1_i$. Thus $s\le p_{\xi },p_{\eta }$ and its support is contained in the finite union $a_{\xi }\cup a_{\eta }$, so F3 puts $s$ in the product. The family $(p_{\xi }{)}_{\xi \in K}$ remains injective, giving an uncountable compatible subset of $X$. Hence the product is Knaster and is ccc by F4.

For the partial-function assertion, take a disjoint tagged copy $\widehat{C}$ of $C$ and a new element $\mathrm{\top }$. Let $Q=\widehat{C}\cup \{\mathrm{\top }\}$, with $u\le v$ iff $u=v$ or $v=\mathrm{\top }$. This is a partial order: reflexivity holds by equality, distinct tagged values are unrelated, and any strict comparison ends at $\mathrm{\top }$, which verifies antisymmetry and transitivity. It is countable and has greatest element $\mathrm{\top }$, so F4 makes it Knaster. A finite partial function $f$ corresponds to the tuple equal to the tagged $f(i)$ on its domain and $\mathrm{\top }$ elsewhere. The support is exactly $\mathrm{dom}(f)$; conversely any finite-support tuple gives exactly that finite partial function. Moreover, the tuple of $g$ is below that of $f$ precisely when $g$ extends $f$, because a tagged value has no smaller element other than itself. This order isomorphism transfers the Knaster conclusion of step 3.1 to finite partial functions. The empty domain maps to the greatest tuple, and for $I=\mathrm{\varnothing }$ both posets are singletons.

Set $\mathcal{F}=\{A\subseteq X:X\setminus A\text{ is countable}\}$. It contains $X$ because the empty set is countable, and omits $\mathrm{\varnothing }$ because $X$ is uncountable. If $A\in \mathcal{F}$ and $A\subseteq B\subseteq X$, then $X\setminus B\subseteq X\setminus A$, so F4 gives $B\in \mathcal{F}$.

If $A,B\in \mathcal{F}$, then $X\setminus (A\cap B)=(X\setminus A)\cup (X\setminus B)$ is countable: apply F3 to the sequence with these first two terms and empty remaining terms, using the countable choice supplied by A1. Therefore $A\cap B\in \mathcal{F}$. Together with step 1.1 this verifies every filter axiom in F1.

Apply F2 to this proper filter, using A1, to obtain an ultrafilter $\mathcal{U}\supseteq \mathcal{F}$. If countable $D\subseteq X$ belonged to $\mathcal{U}$, its complement would belong to $\mathcal{F}\subseteq \mathcal{U}$; F1 would put $D\cap (X\setminus D)=\mathrm{\varnothing }$ in $\mathcal{U}$, contradicting properness. Thus every member of $\mathcal{U}$ is uncountable. Only pairwise intersections were used.