---

Source. Theorem 9 and the preceding rank sentence, printed p. 1327 (published PDF).

Printed statement. Welsh states that $\mathcal A$ has a $p$-transversal containing $U\subseteq S$ if and only if

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

This statement is false. Condition (1) is instead the Hall criterion for a $p$-transversal contained in $U$. The prose immediately before the theorem uses the matroid with the single base $U$, of rank $r(X)=|X\cap U|$, which does express the “contains $U$” problem. Thus neither changing only the prose nor changing only the displayed inequality gives an unambiguous transcription repair.

Counterexamples. First, (1) is not necessary for a $p$-transversal containing $U$. Take

$$I=\{1\},\quad p_1=2,\quad S=A_1=\{u,v\},\quad U=\{u\}.$$

The unique $p$-transversal $\{u,v\}$ contains $U$, but at $J=\{1\}$ the printed inequality reads $1\ge2$.

Nor is the printed inequality sufficient for the printed conclusion. Take

$$I=\{1\},\quad p_1=1,\quad S=A_1=U=\{a,b\}.$$

Condition (1) holds for both subsets of $I$, but every $p$-transversal has one element and therefore cannot contain the two-element set $U$. $\square$

The valid contained-in criterion is proved in the corrected contained theorem. The valid contains-$U$ criterion needs ordinary $p$-Hall conditions and an additional defect inequality, and is proved in the corrected contains theorem. Both are compilation-supplied results, not claims about a published erratum.