Hall’s marriage theorem, proof of


We prove Hall’s marriage theoremMathworldPlanetmath by inductionMathworldPlanetmath on |S|, the size of S.

The theorem is trivially true for |S|=0.

Assuming the theorem true for all |S|<n, we prove it for |S|=n.

First suppose that we have the stronger condition

|∪T|≥|T|+1

for all ∅≠T⊂S. Pick any x∈Sn as the representative of Sn; we must choose an SDR from

S′={S1∖{x},…,Sn-1∖{x}}.

But if

{Sj1∖{x},…,Sjk∖{x}}=T′⊆S′

then, by our assumptionPlanetmathPlanetmath,

|∪T′|≥|∪i=1kSji|-1≥k.

By the already-proven case of the theorem for S′ we see that we can indeed pick an SDR for S′.

Otherwise, for some ∅≠T⊂S we have the “exact” size

|∪T|=|T|.

Inside T itself, for any T′⊆T⊂S we have

|∪T′|≥|T′|,

so by an already-proven case of the theorem we can pick an SDR for T.

It remains to pick an SDR for S∖T which avoids all elements of ∪T (these elements are in the SDR for T). To use the already-proven case of the theorem (again) and do this, we must show that for any T′⊆S∖T, even after discarding elements of ∪T there remain enough elements in ∪T′: we must prove

|∪T′∖∪T|≥|T′|.

But

|∪T′∖∪T| =|⋃(T∪T′)|-|∪T|≥ (1)
≥|T∪T′|-|T|= (2)
=|T|+|T′|-|T|=|T′|, (3)

using the disjointness of T and T′. So by an already-proven case of the theorem, S∖T does indeed have an SDR which avoids all elements of ∪T.

This the proof.

Title Hall’s marriage theorem, proof of
Canonical name HallsMarriageTheoremProofOf
Date of creation 2013-03-22 12:45:12
Last modified on 2013-03-22 12:45:12
Owner mps (409)
Last modified by mps (409)
Numerical id 10
Author mps (409)
Entry type Proof
Classification msc 05D15