[proofplan] We show that every uncountable family of Cohen conditions contains two compatible conditions. The key combinatorial input is the finite delta-system argument: an uncountable family of finite domains can be thinned to an uncountable subfamily whose pairwise intersections are a fixed finite root. After thinning once more so that all conditions agree on that root, the union of any two distinct conditions is again a finite partial function extending both, so the two conditions are compatible. Hence no uncountable antichain exists. [/proofplan] [step:Prove the finite delta-system thinning needed for domains] Let $\mathbb{N}$ denote the set of positive integers, let $\omega_1$ denote the first uncountable ordinal, and let $\operatorname{dom}(f)$ denote the domain of a function $f$. Let $X$ be a set, and let $\mathcal{F}$ be an uncountable family of finite subsets of $X$. We prove that there are an uncountable subfamily $\mathcal{G} \subseteq \mathcal{F}$ and a finite set $r \subseteq X$ such that \begin{align*} a \cap b = r \end{align*} for all distinct $a,b \in \mathcal{G}$. First thin $\mathcal{F}$ to an uncountable subfamily all of whose elements have the same finite cardinality. This is possible because \begin{align*} \mathcal{F} = \bigcup_{n \in \mathbb{N} \cup \{0\}} \{a \in \mathcal{F} : |a| = n\} \end{align*} is a countable union, and a [countable union of countable sets](/theorems/755) is countable. We prove the assertion by induction on this common cardinality $n$. If $n = 0$, then every member is $\varnothing$, and the assertion holds with root $r = \varnothing$. Assume $n \geq 1$, and suppose the assertion is known for families of sets of cardinality $n-1$. If there exists $x \in X$ such that \begin{align*} \mathcal{F}_x := \{a \in \mathcal{F} : x \in a\} \end{align*} is uncountable, define \begin{align*} \mathcal{F}_x' := \{a \setminus \{x\} : a \in \mathcal{F}_x\}. \end{align*} This is an uncountable family of sets of cardinality $n-1$ after identifying duplicate sets by retaining one preimage for each member if necessary; applying the induction hypothesis to an uncountable subfamily gives an uncountable $\mathcal{H}' \subseteq \mathcal{F}_x'$ and a finite root $r' \subseteq X \setminus \{x\}$. Pulling back to the corresponding members of $\mathcal{F}_x$, we obtain an uncountable $\mathcal{H} \subseteq \mathcal{F}$ with common root $r := r' \cup \{x\}$. It remains to handle the case in which each point $x \in X$ belongs to at most countably many members of $\mathcal{F}$. We construct an uncountable pairwise disjoint subfamily by recursion of length $\omega_1$. At every stage $\alpha < \omega_1$, the ordinal $\alpha$ is countable by the definition of $\omega_1$, so the set of earlier stages is countable. Suppose $\{a_\beta : \beta < \alpha\}$ has been chosen for some countable ordinal $\alpha < \omega_1$. Define the finite-union support already used by \begin{align*} S_\alpha := \bigcup_{\beta < \alpha} a_\beta. \end{align*} Since $\alpha$ is countable and each $a_\beta$ is finite, the set $S_\alpha$ is countable. For each $x \in S_\alpha$, the family $\{a \in \mathcal{F} : x \in a\}$ is countable by the present case assumption. Therefore the family of sets in $\mathcal{F}$ meeting $S_\alpha$ is countable. Since $\mathcal{F}$ is uncountable, choose $a_\alpha \in \mathcal{F}$ disjoint from $S_\alpha$. This recursion produces an uncountable pairwise disjoint subfamily, which is a delta-system with root $r = \varnothing$. [guided] Let $\mathbb{N}$ denote the set of positive integers, let $\omega_1$ denote the first uncountable ordinal, and let $\operatorname{dom}(f)$ denote the domain of a function $f$. We need a precise version of the familiar delta-system lemma only for finite sets. The goal is to take an uncountable family $\mathcal{F}$ of finite subsets of a set $X$ and find an uncountable subfamily whose intersections are all the same finite set. First, we reduce to the case where all sets have one fixed size. For each $n \in \mathbb{N} \cup \{0\}$, let \begin{align*} \mathcal{F}_n := \{a \in \mathcal{F} : |a| = n\}. \end{align*} Then \begin{align*} \mathcal{F} = \bigcup_{n \in \mathbb{N} \cup \{0\}} \mathcal{F}_n. \end{align*} If every $\mathcal{F}_n$ were countable, then $\mathcal{F}$ would be countable as a countable union of countable sets. This contradicts the hypothesis that $\mathcal{F}$ is uncountable. Hence some $\mathcal{F}_n$ is uncountable, and we replace $\mathcal{F}$ by that subfamily. We now argue by induction on $n$. If $n = 0$, every member is the empty set, so any uncountable subfamily has pairwise intersection $\varnothing$. Thus the root is $r = \varnothing$. Assume $n \geq 1$ and the result has already been proved for uncountable families of sets of size $n-1$. There are two possibilities. The first possibility is that some point $x \in X$ occurs in uncountably many members of $\mathcal{F}$. Formally, define \begin{align*} \mathcal{F}_x := \{a \in \mathcal{F} : x \in a\}. \end{align*} If $\mathcal{F}_x$ is uncountable, remove the common point $x$ from each member and consider \begin{align*} \mathcal{F}_x' := \{a \setminus \{x\} : a \in \mathcal{F}_x\}. \end{align*} Each member of $\mathcal{F}_x'$ has cardinality $n-1$. Applying the induction hypothesis to an uncountable subfamily of these reduced sets gives an uncountable subfamily with common root $r'$. Restoring the point $x$ gives pairwise intersections exactly $r' \cup \{x\}$ among the original sets. Thus the original family has an uncountable delta-system with root $r := r' \cup \{x\}$. The second possibility is that no point occurs uncountably often. Thus, for every $x \in X$, the family \begin{align*} \{a \in \mathcal{F} : x \in a\} \end{align*} is countable. In this case we build a pairwise disjoint uncountable subfamily. The recursion has length $\omega_1$, and for every stage $\alpha < \omega_1$, the set of earlier stages $\{\beta : \beta < \alpha\}$ is countable because $\omega_1$ is the first uncountable ordinal. Suppose that for some countable ordinal $\alpha < \omega_1$ we have already chosen sets $a_\beta \in \mathcal{F}$ for all $\beta < \alpha$, pairwise disjoint. Let \begin{align*} S_\alpha := \bigcup_{\beta < \alpha} a_\beta. \end{align*} Because $\alpha$ is countable and each $a_\beta$ is finite, $S_\alpha$ is countable. A set in $\mathcal{F}$ can meet $S_\alpha$ only by containing some $x \in S_\alpha$. For each such $x$, only countably many members of $\mathcal{F}$ contain $x$, and $S_\alpha$ itself is countable. Therefore only countably many members of $\mathcal{F}$ meet $S_\alpha$. Since $\mathcal{F}$ is uncountable, some member $a_\alpha \in \mathcal{F}$ is disjoint from $S_\alpha$. Choosing this $a_\alpha$ continues the recursion. Carrying out the recursion for all $\alpha < \omega_1$ gives an uncountable family $\{a_\alpha : \alpha < \omega_1\}$ of pairwise disjoint finite sets. Pairwise disjointness means \begin{align*} a_\alpha \cap a_\beta = \varnothing \end{align*} whenever $\alpha \neq \beta$, so this is a delta-system with root $r = \varnothing$. [/guided] [/step] [step:Thin an uncountable family of Cohen conditions to common domain root and common root values] Let $P := \operatorname{Add}(\omega,\kappa)$, and let $A \subseteq P$ be uncountable. For each $p \in A$, let \begin{align*} d_p := \operatorname{dom}(p) \subseteq \kappa \times \omega. \end{align*} Each $d_p$ is finite by the definition of $P$. The family $\{d_p : p \in A\}$ is uncountable: if it were countable, then each fixed finite domain $d \subseteq \kappa \times \omega$ would support exactly $2^{|d|}$ functions $d \to \{0,1\}$, so $A$ would be a countable union of finite sets and therefore countable, contradicting the choice of $A$. Applying the finite delta-system thinning from the previous step to the uncountable family $\{d_p : p \in A\}$, and retaining one corresponding condition for each chosen domain when necessary, gives an uncountable subfamily $A_0 \subseteq A$ and a finite set $r \subseteq \kappa \times \omega$ such that \begin{align*} \operatorname{dom}(p) \cap \operatorname{dom}(q) = r \end{align*} for all distinct $p,q \in A_0$. For each $p \in A_0$, the restriction $p|_r$ is a function $r \to \{0,1\}$. Since $r$ is finite, there are exactly $2^{|r|}$ such functions. Therefore the map \begin{align*} A_0 &\to \{f : r \to \{0,1\}\} \end{align*} given by $p \mapsto p|_r$ has finite range. Since $A_0$ is uncountable, at least one fiber is uncountable. Choose an uncountable subfamily $A_1 \subseteq A_0$ such that \begin{align*} p|_r = q|_r \end{align*} for all $p,q \in A_1$. [/step] [step:Unite two thinned conditions to obtain a common extension] Choose distinct $p,q \in A_1$. Define \begin{align*} u: \operatorname{dom}(p) \cup \operatorname{dom}(q) &\to \{0,1\} \end{align*} by requiring $u|_{\operatorname{dom}(p)} = p$ and $u|_{\operatorname{dom}(q)} = q$. This definition is well-defined. Indeed, if $(\alpha,m) \in \operatorname{dom}(p) \cap \operatorname{dom}(q)$, then $(\alpha,m) \in r$, and the choice of $A_1$ gives \begin{align*} p(\alpha,m) = p|_r(\alpha,m) = q|_r(\alpha,m) = q(\alpha,m). \end{align*} The domain $\operatorname{dom}(p) \cup \operatorname{dom}(q)$ is finite because it is a union of two finite sets, so $u \in P$. Since $u \supseteq p$ and $u \supseteq q$, the reverse-inclusion order on $P$ gives \begin{align*} u \leq p \end{align*} and \begin{align*} u \leq q. \end{align*} Thus $p$ and $q$ are compatible conditions in $P$. [/step] [step:Conclude that no uncountable antichain exists] We have shown that every uncountable subset $A \subseteq P$ contains two distinct compatible conditions. Therefore no uncountable subset of $P$ can be an antichain, since an antichain is a set of pairwise incompatible conditions. Hence every antichain in $P = \operatorname{Add}(\omega,\kappa)$ is countable. This is exactly the countable chain condition for $\operatorname{Add}(\omega,\kappa)$. [/step]