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Compilation-supplied correction. The formula is printed in Theorem 9 on p. 1327 (published PDF), but the source attaches it to the false “contains $U$” prose.

Statement. Let $U\subseteq S$. There is a $p$-transversal $X$ of $\mathcal A$ satisfying $X\subseteq U$ if and only if

$$|A(J)\cap U|\ge p(J) \qquad(J\subseteq I). \tag{1}$$

Proof. Put $C_i=A_i\cap U$. For every $J\subseteq I$,

$$C(J)=\bigcup_{i\in J}(A_i\cap U)=A(J)\cap U.$$

A $p$-transversal of $(C_i)$ is exactly a $p$-transversal of $\mathcal A$ contained in $U$. Applying Theorem 7 to $(C_i)$ gives precisely (1). This includes $U=\varnothing$, zero coordinates, and the empty family. $\square$

This proof identifies the theorem expressed by the printed inequality. It does not change the source's prose silently.