---

Source. Theorem 13 and its proof, printed p. 1329 (published PDF).

Printed statement. Let $\mathcal A=(A_1,\ldots,A_n)$ and $\mathcal B=(B_1,\ldots,B_m)$ be finite indexed families, let $p_i\ge0$ be integers, and let $k\ge0$ be an integer. The source states that there is one subset $X\subseteq S$ which is both an $A$ $p$-transversal and a $B$ $k$-transversal if and only if

$$k|B(K)|\ge |K| \tag{15}$$

and

$$|A(J)\cap B(K)|\ge p(J)+|K|-m \tag{16}$$

for every $J\subseteq[n]$ and $K\subseteq[m]$.

This exact-common-support conclusion is false.

Counterexample. Take

$$S=\{a,b\},\qquad \mathcal A=(\{a\}),\qquad p_1=1,$$

and

$$\mathcal B=(\{a\},\{b\}),\qquad k=1.$$

Condition (15) is Hall's condition for the two singleton sets and holds for all $K\subseteq[2]$. For (16), when $J=\varnothing$ the right side is $|K|-2\le0$. When $J=\{1\}$, the required lower bound is $|K|-1$: it is at most zero for $|K|\le1$, and for $K=[2]$ the intersection has size one. Thus (16) also holds in every case.

The only $A$ $p$-transversal is $\{a\}$. The only $B$ $1$-transversal is $\{a,b\}$. No subset is both.

The proof's failure occurs after Theorem 5 produces a $B$ $k$-transversal $X$ of rank

$$N=\sum_{i=1}^n p_i$$

in the $p$-transversal matroid. Rank $N$ means that $X$ contains a base of that matroid; it does not mean that $X$ itself is a base. In the example, $\{a,b\}$ has rank one and contains the base $\{a\}$. $\square$

The printed proof also refers to “(1)” where it needs (15). Correcting that label does not repair the substantive error. The actual containment criterion is proved in the compilation-supplied corrected theorem.