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Source. Theorem 3, printed p. 1324 (published PDF).

Printed statement. For a fixed vector $p$ of nonnegative integers, the source asserts without qualification that the $p$-transversals of $\mathcal A$ form the bases of a matroid on $S$ (p. 1324). As a matroid has at least one base, this fails when $\mathcal A$ has no $p$-transversal.

Corrected statement. Let $\mathcal A=(A_i)_{i\in I}$ be a finite indexed family and let $p_i\ge0$ be integers. The replicated family $\mathcal A^p$ defines a transversal matroid on $S$ in every case. If $\mathcal A$ has a $p$-transversal, then this matroid has rank

$$N=\sum_{i\in I}p_i,$$

and its bases are exactly the $p$-transversals of $\mathcal A$. If no $p$-transversal exists, the matroid still exists but has rank less than $N$; its bases cannot be the empty family asserted by the unqualified printed statement.

Proof. Use the replicated index set

$$I^p=\{(i,h):i\in I,\ 1\le h\le p_i\}, \qquad A^p_{i,h}=A_i.$$

By the finite transversal-matroid theorem, the ranges of partial transversals of $\mathcal A^p$ are the independent sets of a matroid $T_p(\mathcal A)$ on $S$.

A full transversal of $\mathcal A^p$ assigns $p_i$ distinct elements to the $p_i$ copies of index $i$. Grouping those elements by $i$ gives pairwise disjoint sets $X_i\subseteq A_i$ with $|X_i|=p_i$. Its range is therefore a $p$-transversal. Conversely, label the elements of each part $X_i$ by the $p_i$ copied indices. This turns any $p$-transversal into a full transversal of $\mathcal A^p$.

Suppose one full transversal exists. It has $N=|I^p|$ elements, and no partial transversal can have more than $N$ elements. Hence $T_p(\mathcal A)$ has rank $N$. Every full transversal is an independent $N$-set and hence a base. Conversely, every base has $N$ elements and is a partial-transversal range; its witnessing injection uses all $N$ replicated indices, so it is full and its range is a $p$-transversal.

If no full transversal exists, the same transversal matroid has rank below $N$. It has at least one base because its ground set is finite, whereas the family of $p$-transversals is empty. For example, $I=\{1\}$, $A_1=\varnothing$, and $p_1=1$ give exactly this obstruction. When $N=0$, the replicated family is empty and its unique full transversal and base are both $\varnothing$. $\square$

The existence qualification is explicit in source corrections.