---

Compilation-supplied correction. The inequalities are (15)–(16) in Theorem 13, printed p. 1329 (published PDF). The exact-common-support conclusion there is false.

Statement. Let $\mathcal A=(A_1,\ldots,A_n)$ and $\mathcal B=(B_1,\ldots,B_m)$ be finite indexed families. Let $p_i\ge0$ and $k\ge0$ be integers. There are sets $Y\subseteq X\subseteq S$ such that $Y$ is an $A$ $p$-transversal and $X$ is a $B$ $k$-transversal if and only if

$$k|B(K)|\ge |K| \qquad(K\subseteq[m]), \tag{1}$$

and

$$|A(J)\cap B(K)|\ge p(J)+|K|-m \qquad(J\subseteq[n],\ K\subseteq[m]). \tag{2}$$

Proof. Put $N=p([n])$. Suppose first that $Y\subseteq X$ as in the statement. The set $Y$ is a base of the rank-$N$ transversal matroid $T_p(\mathcal A)$, so $r_p(X)=N$. Apply the necessary direction of Theorem 5 to the $B$ $k$-transversal $X$ in that matroid, with $t=N$. Its first condition is (1), and its second is

$$r_p(B(K))\ge |K|+N-m. \tag{3}$$

By the rank formula, (3) is equivalent to (2) for every $J$.

Conversely, assume (1) and (2). Taking $K=[m]$ in (2) gives

$$|A(J)|\ge |A(J)\cap B([m])|\ge p(J).$$

Thus an $A$ $p$-transversal exists by Theorem 7, and $T_p(\mathcal A)$ has rank $N$ with the $A$ $p$-transversals as its bases. The rank formula turns (2) into (3). Theorem 5, applied with $t=N$, now gives a $B$ $k$-transversal $X$ with $r_p(X)\ge N$. Since the whole matroid has rank $N$, the restriction to $X$ contains a base $Y\subseteq X$. That base is an $A$ $p$-transversal, as required.

The endpoint $m=0$ is included. Then (1) is vacuous, while (2) at $K=\varnothing$ forces $p(J)=0$ for every $J$ and hence $N=0$; take $X=Y=\varnothing$. The case $k=0$ with $m>0$ is excluded on both sides by (1). $\square$

The correction retains the result actually proved by the full-rank argument. It is not labeled as the printed theorem or as a published erratum.