The indexed delta-system lemma

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.