Prikry forcing and its direct-extension order

Work in ZFC. Let $\kappa$ be an uncountable cardinal and let $U$ be a normal measure on $\kappa$ in the sense of Complete ultrafilters and measurable cardinals. A Prikry condition is a pair $p=(s_p,A_p)$ such that

• $s_p=⟨s_p(0),\dots ,s_p(n-1)⟩$ is a finite strictly increasing sequence of ordinals below $\kappa$;
• $A_p\in U$; and
• if $n>0$, then $s_p(n-1)<\min A_p$.

The empty stem imposes no maximum condition. The first coordinate $s_p$ is the stem, and $A_p$ is the upper part.

For $q=(s_q,A_q)$ and $p=(s_p,A_p)$, write $q\le p$ when $q$ is stronger than $p$, meaning that $s_q$ end-extends $s_p$, $A_q\subseteq A_p$, and every entry of $s_q$ after $s_p$ belongs to $A_p$. This is the stronger-below convention of Forcing preorders, compatibility and filters. Write

$q{\le }^{\ast }p$

and call $q$ a direct extension of $p$ when $q\le p$ and $s_q=s_p$.

Reflexivity is immediate. If $r\le q\le p$, then $s_r$ end-extends $s_p$ and $A_r\subseteq A_p$. An entry added by $r$ either was already added by $q$ and therefore lies in $A_p$, or lies in $A_q\subseteq A_p$. Thus $r\le p$, so this is a forcing preorder. Likewise, equality of stems and inclusion of upper parts show that ${\le }^{\ast }$ is reflexive and transitive. For a fixed stem, any two conditions are compatible: $(s,A\cap B)$ is a common extension, since a proper filter is closed under finite intersections.

No choice is needed to form a condition or compare two fixed conditions. The dependency The Axiom of Choice records the ambient ZFC hypothesis used for cardinal arithmetic and for the simultaneous measure-one selections in later results on this page; it is not being inferred from the existence of $U$.