[proofplan]
We prove the contrapositive form of the countable chain condition: every uncountable subset of $\mathbb P$ contains two compatible conditions. Starting from an uncountable family of finite supports, we first extract an uncountable subfamily whose supports form a delta-system with a common finite root. We then use the countability of possible root data over that root to thin once more so that all remaining conditions have the same root data. Any two distinct conditions in the final uncountable subfamily satisfy the compatibility hypothesis, so the original uncountable set cannot be an antichain.
[/proofplan]
[step:Extract an uncountable delta-system of supports]
Let $A \subset \mathbb P$ be uncountable. For each $p \in A$, define the finite set
\begin{align*}
S_p := \operatorname{supp}(p) \subset I.
\end{align*}
We first record the finite delta-system extraction used in this proof.
[claim:Every uncountable family of finite sets has an uncountable delta-subfamily]
Let $J$ be an uncountable set, and let $S:J \to [I]^{<\omega}$ be a map. Then there are an uncountable subset $J_0 \subset J$ and a finite set $r \subset I$ such that
\begin{align*}
S(j) \cap S(k)=r
\end{align*}
for all distinct $j,k \in J_0$.
[/claim]
[proof]
For each $n \in \mathbb N \cup \{0\}$, define
\begin{align*}
J_n := \{j \in J : |S(j)|=n\}.
\end{align*}
The sets $J_n$ form a countable partition of $J$. If every $J_n$ were countable, then for each $n$ choose an injection $e_n:J_n \to \mathbb N$ and define $e:J \to (\mathbb N \cup \{0\}) \times \mathbb N$ by sending $j \in J_n$ to $(n,e_n(j))$. This map is injective, and $(\mathbb N \cup \{0\}) \times \mathbb N$ is countable, so $J$ would be countable. Since $J$ is uncountable, there is $n \in \mathbb N \cup \{0\}$ such that $J_n$ is uncountable. Replacing $J$ by $J_n$, we may assume that all sets $S(j)$ have the same finite cardinality $n$.
We prove the assertion by induction on $n$. If $n=0$, then $S(j)=\varnothing$ for all $j \in J$, and the conclusion holds with $J_0=J$ and $r=\varnothing$.
Assume $n \geq 1$ and suppose the result has been proved for families of finite sets of cardinality $n-1$. For each $i \in I$, define
\begin{align*}
J(i):=\{j \in J : i \in S(j)\}.
\end{align*}
If there is $i_0 \in I$ such that $J(i_0)$ is uncountable, define the map $T:J(i_0) \to [I]^{<\omega}$ by
\begin{align*}
T(j):=S(j)\setminus \{i_0\}.
\end{align*}
Every set $T(j)$ has cardinality $n-1$. By the induction hypothesis, there are an uncountable subset $J_0 \subset J(i_0)$ and a finite set $r_0 \subset I$ such that
\begin{align*}
T(j) \cap T(k)=r_0
\end{align*}
for all distinct $j,k \in J_0$. Since $i_0 \in S(j) \cap S(k)$ for all $j,k \in J_0$, it follows that
\begin{align*}
S(j) \cap S(k)=r_0 \cup \{i_0\}
\end{align*}
for all distinct $j,k \in J_0$. Thus the conclusion holds with $r=r_0 \cup \{i_0\}$.
It remains to consider the case in which $J(i)$ is countable for every $i \in I$. Well-order $J$, and construct a subset $\mathcal M \subset J$ by transfinite recursion as follows: when the construction reaches $j \in J$, put $j$ into $\mathcal M$ exactly when $S(j)$ is disjoint from $S(k)$ for every previously chosen $k \in \mathcal M$. The resulting family $\{S(j):j \in \mathcal M\}$ is pairwise disjoint by construction. It is maximal with respect to this property: if $j \notin \mathcal M$, then at the stage when $j$ was considered, there was already some chosen $k \in \mathcal M$ with $S(j) \cap S(k) \neq \varnothing$, so adjoining $j$ would destroy pairwise disjointness.
If $\mathcal M$ were countable, then the set
\begin{align*}
U:=\bigcup_{j \in \mathcal M} S(j)
\end{align*}
would be countable, because it is a countable union of finite sets. For each $i \in U$, the set $J(i)$ is countable by the present case assumption, so
\begin{align*}
\{j \in J : S(j) \cap U \neq \varnothing\}=\bigcup_{i \in U} J(i)
\end{align*}
would be countable. By maximality of $\mathcal M$, every $j \in J$ satisfies $S(j) \cap U \neq \varnothing$; otherwise $S(j)$ would be disjoint from every $S(k)$ with $k \in \mathcal M$, contradicting maximality. Hence $J$ would be countable, contradicting the hypothesis. Therefore $\mathcal M$ is uncountable.
Taking $J_0=\mathcal M$, the sets $\{S(j):j \in J_0\}$ are pairwise disjoint, so the conclusion holds with $r=\varnothing$.
[/proof]
Applying the claim to the map $p \mapsto S_p$ on the uncountable set $A$, choose an uncountable subset $B \subset A$ and a finite set $r \subset I$ such that
\begin{align*}
\operatorname{supp}(p) \cap \operatorname{supp}(q)=r
\end{align*}
for all distinct $p,q \in B$.
Since $r=\operatorname{supp}(p) \cap \operatorname{supp}(q)$ for distinct $p,q \in B$, we have $r \subset \operatorname{supp}(p)$ for every $p \in B$.
[guided]
Let $A \subset \mathbb P$ be uncountable. For each condition $p \in A$, define the finite set
\begin{align*}
S_p := \operatorname{supp}(p) \subset I.
\end{align*}
The goal of this step is to thin the uncountable family $(S_p)_{p \in A}$ until all pairwise intersections are the same finite set. This is the point where the finiteness of supports is used.
We prove the finite delta-system extraction needed here. Let $J$ be an uncountable set, and let $S:J \to [I]^{<\omega}$ be a map. We show that there are an uncountable subset $J_0 \subset J$ and a finite set $r \subset I$ such that
\begin{align*}
S(j) \cap S(k)=r
\end{align*}
for all distinct $j,k \in J_0$.
For each $n \in \mathbb N \cup \{0\}$, define
\begin{align*}
J_n := \{j \in J : |S(j)|=n\}.
\end{align*}
The sets $J_n$ form a countable partition of $J$. If every $J_n$ were countable, then for each $n$ choose an injection $e_n:J_n \to \mathbb N$. Define $e:J \to (\mathbb N \cup \{0\}) \times \mathbb N$ by sending $j \in J_n$ to $(n,e_n(j))$. This is injective because the first coordinate records the unique part $J_n$ containing $j$, and the second coordinate separates points inside that part. Since $(\mathbb N \cup \{0\}) \times \mathbb N$ is countable, this would make $J$ countable. Since $J$ is uncountable, at least one set $J_n$ is uncountable. Replacing $J$ by that uncountable set $J_n$, we may assume that every set $S(j)$ has the same finite cardinality $n$.
We now prove the assertion by induction on $n$. If $n=0$, then $S(j)=\varnothing$ for every $j \in J$, so taking $J_0=J$ and $r=\varnothing$ gives
\begin{align*}
S(j) \cap S(k)=\varnothing=r
\end{align*}
for all distinct $j,k \in J_0$.
Assume $n \geq 1$ and suppose the assertion has already been proved for uncountable families of sets of cardinality $n-1$. For each $i \in I$, define
\begin{align*}
J(i):=\{j \in J : i \in S(j)\}.
\end{align*}
First suppose that there is $i_0 \in I$ such that $J(i_0)$ is uncountable. Define the map $T:J(i_0) \to [I]^{<\omega}$ by
\begin{align*}
T(j):=S(j)\setminus \{i_0\}.
\end{align*}
Because $i_0 \in S(j)$ for every $j \in J(i_0)$ and $|S(j)|=n$, every set $T(j)$ has cardinality $n-1$. The induction hypothesis applied to $T$ gives an uncountable subset $J_0 \subset J(i_0)$ and a finite set $r_0 \subset I$ such that
\begin{align*}
T(j) \cap T(k)=r_0
\end{align*}
for all distinct $j,k \in J_0$. Since $i_0$ belongs to both $S(j)$ and $S(k)$ for $j,k \in J_0$, and since $T(j)$ and $T(k)$ are obtained by deleting $i_0$, we obtain
\begin{align*}
S(j) \cap S(k)=r_0 \cup \{i_0\}
\end{align*}
for all distinct $j,k \in J_0$. Thus the desired conclusion holds in this case with $r=r_0 \cup \{i_0\}$.
It remains to handle the case in which $J(i)$ is countable for every $i \in I$. We need a maximal pairwise disjoint subfamily, and we construct it directly. Well-order $J$. Proceed through this well-order and build a subset $\mathcal M \subset J$ by the following rule: when the construction reaches $j \in J$, put $j$ into $\mathcal M$ exactly when $S(j)$ is disjoint from $S(k)$ for every previously chosen $k \in \mathcal M$. The family $\{S(j):j \in \mathcal M\}$ is pairwise disjoint by construction. It is also maximal for pairwise disjointness: if $j \notin \mathcal M$, then when $j$ was considered, some already chosen $k \in \mathcal M$ satisfied $S(j) \cap S(k) \neq \varnothing$, so adding $j$ later would break pairwise disjointness.
If $\mathcal M$ were countable, then the set
\begin{align*}
U:=\bigcup_{j \in \mathcal M} S(j)
\end{align*}
would be countable, because it is a countable union of finite sets. For each $i \in U$, the set $J(i)$ is countable by the present case assumption. Therefore
\begin{align*}
\{j \in J : S(j) \cap U \neq \varnothing\}=\bigcup_{i \in U} J(i)
\end{align*}
would be countable as a [countable union of countable sets](/theorems/755). However, maximality of $\mathcal M$ implies that every $j \in J$ satisfies $S(j) \cap U \neq \varnothing$: if some $j \in J$ had $S(j) \cap U=\varnothing$, then $S(j)$ would be disjoint from every $S(k)$ with $k \in \mathcal M$, so $\mathcal M \cup \{j\}$ would contradict maximality. Hence $J$ would be countable, contradicting the hypothesis that $J$ is uncountable. Thus $\mathcal M$ is uncountable.
Taking $J_0=\mathcal M$, the sets $\{S(j):j \in J_0\}$ are pairwise disjoint, so
\begin{align*}
S(j) \cap S(k)=\varnothing
\end{align*}
for all distinct $j,k \in J_0$. The desired conclusion holds with $r=\varnothing$. This completes the proof of the finite delta-system extraction.
We now apply this extraction to the forcing conditions. Use $J=A$ and define
\begin{align*}
S:A \to [I]^{<\omega}
\end{align*}
by
\begin{align*}
S(p):=\operatorname{supp}(p).
\end{align*}
The result just proved gives an uncountable set $B \subset A$ and a finite set $r \subset I$ such that
\begin{align*}
S(p) \cap S(q)=r
\end{align*}
whenever $p,q \in B$ are distinct. Substituting the definition of $S(p)$, this becomes
\begin{align*}
\operatorname{supp}(p) \cap \operatorname{supp}(q)=r
\end{align*}
for all distinct $p,q \in B$.
This is the exact form needed for the compatibility hypothesis later: the supports of any two distinct conditions in $B$ intersect in the same finite root $r$. Since $B$ is uncountable, for each fixed $p \in B$ we may choose $q \in B$ with $q \neq p$. The equality
\begin{align*}
\operatorname{supp}(p) \cap \operatorname{supp}(q)=r
\end{align*}
then implies $r \subset \operatorname{supp}(p)$. Therefore the root datum $p|_r$ is defined for every $p \in B$, which is what the next thinning step requires.
[/guided]
[/step]
[step:Thin the delta-system so all root data agree]
By the first hypothesis applied to the finite set $r$, the set
\begin{align*}
R_r:=\{p|_r : p \in \mathbb P \text{ and } r \subset \operatorname{supp}(p)\}
\end{align*}
is countable. For each $d \in R_r$, define
\begin{align*}
B_d:=\{p \in B : p|_r=d\}.
\end{align*}
The family $\{B_d:d \in R_r\}$ is a countable partition of $B$. Since $B$ is uncountable, at least one piece is uncountable. Choose $d_0 \in R_r$ such that
\begin{align*}
C:=B_{d_0}
\end{align*}
is uncountable. Then, for all $p \in C$,
\begin{align*}
p|_r=d_0.
\end{align*}
In particular, for all $p,q \in C$,
\begin{align*}
p|_r=q|_r.
\end{align*}
[guided]
The delta-system step gives a common finite root $r$, but compatibility also requires agreement of the data over that root. The first hypothesis says that the collection
\begin{align*}
R_r:=\{p|_r : p \in \mathbb P \text{ and } r \subset \operatorname{supp}(p)\}
\end{align*}
is countable. This hypothesis applies because $r$ is finite and, from the previous step, every $p \in B$ satisfies $r \subset \operatorname{supp}(p)$.
For each $d \in R_r$, define
\begin{align*}
B_d:=\{p \in B : p|_r=d\}.
\end{align*}
The sets $B_d$ partition $B$: every $p \in B$ has a defined root datum $p|_r \in R_r$, so $p$ belongs to exactly the piece indexed by that datum. Since $R_r$ is countable and $B$ is uncountable, at least one piece must be uncountable; otherwise the same injection argument for countable unions of countable sets would make $B$ countable.
Choose $d_0 \in R_r$ such that
\begin{align*}
C:=B_{d_0}
\end{align*}
is uncountable. Then every $p \in C$ satisfies
\begin{align*}
p|_r=d_0.
\end{align*}
Therefore, for all $p,q \in C$,
\begin{align*}
p|_r=q|_r.
\end{align*}
This is exactly the root-data agreement needed for the compatibility hypothesis.
[/guided]
[/step]
[step:Find two compatible conditions in the original uncountable set]
Since $C$ is uncountable, choose distinct conditions $p,q \in C$. Because $C \subset B$, the delta-system property gives
\begin{align*}
\operatorname{supp}(p) \cap \operatorname{supp}(q)=r.
\end{align*}
Because $p,q \in C$, the root-data thinning gives
\begin{align*}
p|_r=q|_r.
\end{align*}
The compatibility hypothesis therefore applies to this pair $p,q$, and yields that $p$ and $q$ are compatible in $\mathbb P$.
Thus every uncountable subset $A \subset \mathbb P$ contains two compatible conditions. Equivalently, no antichain in $\mathbb P$ is uncountable. Hence every antichain in $\mathbb P$ is countable, so $\mathbb P$ satisfies the countable chain condition.
[guided]
The set $C$ is uncountable, so it contains two distinct conditions; choose $p,q \in C$ with $p \neq q$. Because $C \subset B$, the delta-system property from the first step applies to this pair and gives
\begin{align*}
\operatorname{supp}(p) \cap \operatorname{supp}(q)=r.
\end{align*}
Because $p,q \in C$, the thinning step gives common root data:
\begin{align*}
p|_r=q|_r.
\end{align*}
Now both hypotheses required by the compatibility assumption are verified for this specific pair $p,q$: their supports meet exactly in $r$, and their restrictions to $r$ agree. Therefore the compatibility hypothesis yields that $p$ and $q$ are compatible in $\mathbb P$.
We started with an arbitrary uncountable subset $A \subset \mathbb P$ and found two compatible members of it. Hence no uncountable subset of $\mathbb P$ can be an antichain. Equivalently, every antichain in $\mathbb P$ is countable, which is precisely the countable chain condition.
[/guided]
[/step]
