---

Source. Ford–Fulkerson (1958), the paragraph following Theorem 1, printed p. 82 (published scan).

Let $D\subseteq A$ be a prescribed set of $q$ distinct elements. The finite indexed family $\mathcal S$ has an SDR whose range contains $D$ if and only if, for every $X\subseteq[n]$,

$$|X|\le |I_{\mathcal S}(X)|, \qquad |X|+|D\setminus\bigcup_{j\in X}S_j|\le n.$$

Proof. Set $\alpha_i=1$ for $a_i\in D$ and $\alpha_i=0$ otherwise, and set every $\beta_i=1$. An SRR for these bounds is exactly an SDR containing $D$. Here $a=q$, $\alpha(I_{\mathcal S}(X))=|D\cap\bigcup_{j\in X}S_j|$ and $\beta(I_{\mathcal S}(X))=|I_{\mathcal S}(X)|$. Substitution in Theorem 1 gives precisely the displayed inequalities, in both directions. For $X=\varnothing$, the second inequality includes $q\le n$; mandatory elements lying outside every family member also fail the appropriate test. $\square$

Attribution and scope. The source identifies this specialization with the Hoffman–Kuhn condition and cites their 1956 paper. The proof here is the local substitution into Ford–Fulkerson's theorem; it does not reproduce the separate Hoffman–Kuhn proof or the broader partition-quota result mentioned in the introduction.

Bears on. Mandatory-element variants of finite transversal constructions. No Erdős problem: the paper states no relation to a numbered Erdős problem.