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I am looking for a citeable reference to the following generalization of Hall's Marriage Theorem:

  • Given a bipartite graph of boys and girls. In addition to gender difference, they are divided into 1st and 2nd class citizens. Suppose that Hall's condition is satisfied for 1st class citizens. That is, for every set $M\subset\{\text{1st class boys}\}$ there are at least $|M|$ girls (from both classes) adjacent to $M$. And similarly for every $W\subset\{\text{1st class girls}\}$ there are at least $|W|$ boys adjacent to $W$. Then there exists a matching covering all 1st class citizens.

I need this fact as a lemma in a paper on geometric analysis. The proof is more or less straightforward but it occupies some space when written down. And I suspect that the fact may be well-known to specialists. Is this indeed the case and what are relevant references?

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  • $\begingroup$ It's a really nice theorem and you should write down your proof. Should it go into your paper on geometric analysis? I don't know, but I wouldn't see a reason why not. $\endgroup$ Apr 23, 2015 at 19:29
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    $\begingroup$ @darij: Yes, of course we will prove it in the paper if there is no reference. It just looks too natural to be unknown. $\endgroup$ Apr 23, 2015 at 20:14
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    $\begingroup$ @Dominic van der Zypen: Yes we have just finished it. It is on arxiv.org/abs/1506.06781 $\endgroup$ Jun 24, 2015 at 13:54
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    $\begingroup$ This question is unfair and politically incorrect! I like it very much nevertheless! $\endgroup$ May 12, 2023 at 7:35
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    $\begingroup$ At least it is not gender-biased :) $\endgroup$ May 14, 2023 at 0:01

2 Answers 2

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In this answer I sketch an easy proof of your lemma and then give some references.

The easy proof uses Knaster's fixed point theorem:

THEOREM. Let $S$ be any set (finite or infinite) and let $\varphi:\mathcal P(S)\to\mathcal P(S)$ be an order-preserving map, i.e., $X\subseteq Y\implies\varphi(X)\subseteq\varphi(Y).$ Then $\varphi$ has a fixed point.

PROOF. Let $X_0=\bigcup\{X\in\mathcal P(S):X\subseteq\varphi(X)\}.$ It is easy to see that $\varphi(X_0)=X_0.$

(Knaster's fixed point theorem was set as a Putnam Problem in 1957. The generalization to complete lattices is called the Knaster-Tarski Theorem.)

Now let $B_1,B_2,G_1,G_2$ be the set of all first-class boys, second-class boys, first-class girls, and second-class girls, respectively; $B=B_1\cup B_2,\ G=G_1\cup G_2,\ B_1\cap B_2=G_1\cap G_2=\emptyset.$ By Hall's theorem there are matchings $f:B_1\to G$ and $g:G_1\to B.$ Define an order-preserving map $\varphi:\mathcal P(B_1)\to\mathcal P(B_1)$ by setting, for $X\subseteq B_1,$ $$\varphi(X)=B_1\setminus g[G_1\setminus f[X]].$$ By Knaster's fixed point theorem we have $\varphi(X_0)=X_0$ for some $X_0\subseteq B_1$. Now we can match the boys in $X_0$ with the girls in $f[X_0]$ and the girls in $G_1\setminus f[X_0]$ with the boys in $g[G_1\setminus f[X_0]].$

Here are some references related to your lemma:

L. Mirsky, Transversal Theory, Academic Press, New York and London, 1971.

L. Mirsky and Hazel Perfect, Systems of representatives, J. Math. Analysis Appl. 15 (1966), 520-568.

O. Ore, Theory of Graphs, Amer. Math. Soc. Colloquium Publications No. 38, Providence, 1962 [Theorem 7.4.1].

Here is how your lemma is stated on p. 36 of Mirsky's book:

THEOREM 2.3.1. Let $(X,\Delta,Y)$ be a deltoid and let $X',Y'$ be admissible subsets of $X,Y$ respectively. Then there exist linked sets $X_0,Y_0$ such that $X'\subseteq X_0\subseteq X,\ Y'\subseteq Y_0\subseteq Y.$

The jargon is defined on pp. 33-34. Namely, a deltoid $(X,\Delta,Y)$ is a bipartite graph with partite sets $X,Y$ and edge set $\Delta$; a set $X'\subseteq X$ is admissible if there is an injective matching of $X'$ to $Y$; a set $Y'\subseteq Y$ is admissible if there is an injective matching of $Y'$ to $X$; two sets $X_0\subseteq X,Y_0\subseteq Y$ are linked if there is a bijective matching of $X_0$ to $Y_0.$

P.S. Sergei Ivanov has commented that the result, as stated in the question, can be traced back to Dulmage & Mendelsohn, Some generalizations of the problem of distinct representatives, Canad. J. Math. 10 (1958) 230–241, Theorem 1.

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    $\begingroup$ Very nice proof! I like how it weaponizes one of the most trivial facts in mathematics. $\endgroup$ Apr 24, 2015 at 0:36
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    $\begingroup$ Thanks! I tracked it from Mirski & Perfect's paper down to Dulmage & Mendelsohn, Coverings of bipartite graphs, Canad. J. Math. 10 (1958) 517-534, Theorem 1. And there (unlike the other sources) it is exactly it, not something that one has to combine with Hall's Theorem. $\endgroup$ Apr 24, 2015 at 1:12
  • $\begingroup$ @bof: I think we are going to thank you in acknowledgements. Do you prefer to be mentioned as an anonym or by some real name? $\endgroup$ Apr 29, 2015 at 13:12
  • $\begingroup$ The Dulmage/Mendelsohn reference is wrong; this is certainly not Theorem 1 in that paper. $\endgroup$ May 12, 2023 at 2:03
  • $\begingroup$ Oops, yes, wrong paper. It is Theorem 1 in the paper "Some Generalizations of the Problem of Distinct Representatives" same authors, same volume, pages 230-241. doi.org/10.4153/CJM-1958-027-8 . $\endgroup$ May 13, 2023 at 10:26
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I have not seen it stated anywhere, but I would call it a corollary of Hall. It is in fact very natural, so it would not surprise me if it has been stated before.

Simplest proof I can come up with using Hall:

Hall gives you a matching from the first class boys into the girls, and a matching from the first class girls into the boys, record all these edges with this direction to get a directed graph, which consists of directed paths and even cycles. Delete every other edge in every cycle, and delete the second, fourth, etc. edge of every path. Note that the last vertex in every path is 2nd class, so this yields a matching covering all first class vertices.

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    $\begingroup$ Yes, we have the very same argument. It would be 5 lines long if we were aiming at combinatorialists. But it took more than a page to make the text consumable by analysts. $\endgroup$ Apr 23, 2015 at 21:43
  • $\begingroup$ Why do you assume that analysts can't understand the five-line version? $\endgroup$ Apr 24, 2015 at 14:40
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    $\begingroup$ @David: Because we tested it on live analysts. It turns out that the very language of graph theory is not as widely understood as one might hope. $\endgroup$ Apr 29, 2015 at 13:19

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