Finite products preserve Knaster

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.