A graph-theoretic version of the union-closed sets conjecture

A Graph-Theoretic Versionof the Union-Closed SetsConjecture

Mohamed H. El-ZaharMATHEMATICS DEPARTMENT

FACULTY OF SCIENCEAIN SHAMS UNIVERSITY

ABBASEIA, CAIRO, EGYPTE-mail: [email protected]

Received November 10, 1995

Abstract: An induced subgraph S of a graph G is called a derived subgraph of G if Scontains no isolated vertices. An edge e of G is said to be residual if e occurs in morethan half of the derived subgraphs of G. We introduce the conjecture: Every non-emptygraph contains a non-residual edge. This conjecture is implied by, but weaker than, theunion-closed sets conjecture. We prove that a graphG of ordern satisfies this conjecturewhenever G satisfies any one of the conditions: δ(G) ≤ 2, log2 n ≤ δ(G), n ≤ 10, or thegirth of G is at least 6. Finally, we show that the union-closed sets conjecture, in itsfull generality, is equivalent to a similar conjecture about hypergraphs. c© 1997 John Wiley &

Sons, Inc. J Graph Theory 26: 155–163, 1997

Keywords: derived subgraphs, derived subhypergraphs, union-closed sets conjecture

1. INTRODUCTION

A union-closed family of sets A is a finite collection of sets not all empty such that the union ofany two members of A is also a member of A. The following Conjecture is due to Peter Frankl[cf. 1, 2, 5].

Conjecture 1. Let A = {A1, A2, . . . , An} be a union-closed family of n distinct sets. Thenthere exists an element which belongs to at least n/2 of the sets in A

c© 1997 John Wiley & Sons, Inc. CCC 0364-9024/97/030155-09

156 JOURNAL OF GRAPH THEORY

Some results on this intriguing problem appear in [3, 4, 5].Let A = ∪Ai. If we replace each set Ai by Bi = A− Ai then we get an intersection-closed

family of sets, which we call the dual family of A. Therefore Conjecture 1 is equivalent to thefollowing.

Conjecture 2. Let B = {B1, B2, . . . , Bn} be an intersection-closed family of n distinct sets.Then there exists an element which belongs to at most n/2 of the sets in B.

In Section 2, we shall formulate a weaker version of Conjecture 2 specialized for graphs. InSections 3 and 4, we investigate which graphs satisfy this weaker version. Finally, in Section 5,we show that Conjecture 1 is equivalent to a similar conjecture on hypergraphs.

2. THE DERIVED SUBGRAPHS

Let G = (V,E) be a simple graph with neither loops nor multiple edges. For each subset X ⊆ V ,let EX = {e ∈ E(G) : e is incident with some vertex in X}. Now EX ∪EY = EX∪Y so that thefamily E = {EX : X ⊆ V (G)} is union-closed. Therefore, Conjecture 1 implies that G containsan edge which belongs to at least half of the distinct sets in the family E . This can be reformulatedin another way as follows. For each subset X ⊆ V (G), let FX = E(G)−E(V (G)−X). Note thatFX is exactly the edge set of the subgraph 〈X〉 induced by X . Again by Conjecture 2, G containsan edge which belongs to at most half of the distinct sets in the familyF = {FX : X ⊆ V (G)}. Tostudy this special case of the union-closed sets conjecture, we need a formulation that is more directin terms of graph-theoretical concepts. Define an equivalence relation ∼ on the set of inducedsubgraphs of G as follows. For X,Y ⊆ V (G), let 〈X〉 ∼ 〈Y 〉 whenever the two subgraphs 〈X〉and 〈Y 〉 have the same edge set, i.e., whenever FX = FY . If 〈X〉 has no isolated vertices then itis chosen as the representative of the equivalence class of ∼ to which it belongs. We call theseclass representatives the derived subgraphs of G. To summarize, the derived subgraphs of G arethe induced subgraphs of G having no isolated vertices, or equivalently, those subgraphs whichare vertex-induced and edge-induced at the same time. We adopt the convention that 〈∅〉 is aderived subgraph of G so that it is the representative of the equivalence class containing the emptyinduced subgraphs. Let D(G) denote the set of derived subgraphs of G and put nd(G) = |D(G)|.Finally, let us call an edge e of G residual if it belongs to more than half of the derived subgraphsof G, otherwise e is non-residual. We arrive at the following conjecture which is implied by, butweaker than, the union-closed sets conjecture.

Conjecture 3. Every non-empty graph contains a non-residual edge.

Let us give some examples to illustrate the above definitions. Consider the 6-cycleC6. The derivedsubgraphs of C6 are ∅, C6 and the subgraphs S1, . . . , S5 together with their cyclic permutations,see Figure 1. In all, we have nd(C6) = 29 (compare to 64 induced subgraphs of C6). Each edgeof C6 is contained in exactly 12 derived subgraphs and, therefore, is non-residual.

Another example is the graph G1 of Figure 2. This graph has nd(G1) = 34. Each of theedges e1, e2, e3 occur in 18 derived subgraphs so that it is residual. The remaining edges arenon-residual belonging only to 13 derived subgraphs. How about the average size of derivedsubgraphs? The star graph K(1, n) has 2n derived subgraphs and each edge occurs in 2n−1

(exactly half) of them. So, the average derived subgraphs of K(1, n) contains exactly 12 of its

edges. The average derived subgraph of C6 contains less than 3 edges. On the other hand, thegraph G2 of Figure 3 has 79 derived subgraphs. The average size of a derived subgraph of G2

UNION-CLOSED SETS CONJECTURE 157

FIGURE 1. Derived Subgraphs of C6

is 280/79 which is slightly larger than |E(G2)|/2. This may mean that there is no hope to proveConjecture 3 by considering the average size of derived subgraphs.

3. GRAPHS SATISFYING THE DERIVED SUBGRAPHSCONJECTURE

In this section we investigate which graphs satisfy Conjecture 3.

Lemma 1. Let V = V1 ∪ V2 be a partition of the vertex set of the graph G = (V,E). Let G1

and G2 respectively denote the subgraphs induced by V1 and V2. Then nd(G) ≥ nd(G1)nd(G2).

Proof. Obvious.

Lemma 2. Suppose that the graph G is the disjoint union of the two graphs G1 and G2. Thennd(G) = nd(G1)nd(G2). Moreover an edge e of G1 (resp. G2) is residual in G if and only if itis residual in G1 (resp. G2).

Proof. Let S ∈ D(G) and denote by Si its restriction to Gi for i = 1, 2. Then each Si hasno isolated vertices so that Si ∈ D(Gi), i = 1, 2. Conversely if Si ∈ D(Gi), for i = 1, 2, thenS1 ∪ S2 ∈ D(G). It follows that D(G) = {S1 ∪ S2 : Si ∈ D(Gi), i = 1, 2}. The two assertionsof the lemma are now obvious.

The previous lemma shows us that it suffices to consider only connected graphs.

Theorem 1. Let G be a graph on n vertices. If∑

v∈V (G) 2−deg(v) ≤ 1 then every edge of G isnon-residual. In particular if δ(G) ≥ log2 n then G has no residual edges.

Proof. If v ∈ V (G) then the number of induced subgraphs of G that contain v as an isolatedvertex is 2n−1−deg(v). Hence, nd(G) ≥ 2n −∑

v∈V (G) 2n−1−deg(v). On the other hand, everyedge of G is contained in exactly 2n−2 induced subgraphs of G and, therefore, occurs in at most

FIGURE 2. Residual edges


FIGURE 3. Average density larger than |E(G)|/2

2n−2 derived subgraphs of G. Hence if G has a residual edge then we must have nd(G) < 2n−1.Thus if G has a residual edge then 2n−1 > 2n − ∑

v∈V (G) 2n−1−deg(v) which implies that

1 <∑

v∈V (G) 2−deg(v). This proves the first assertion of the theorem. Now δ(G) ≥ log2 n

implies that∑

v∈V (G) 2−deg(v) ≤ 1, which proves the second assertion.

Theorem 1 shows that a graph not satisfying conjecture 3 cannot have a large minimal degree.On the other hand, it is known [3] that if a union-closed family A contains a non-empty set of atmost two elements then an element of this set occurs in at least half of the members of A. Thisimplies that if G has a vertex v of degree 1 or 2 then v is incident with a non-residual edge. Weprefer to give graph-theoretic proofs for this result.

Theorem 2. Let the graph G have a vertex v with deg(v) = 1. Then the edge incident with v isnon-residual.

Proof. Denote the vertex adjacent to v by w. Put G1 = G− v. Let D(G) = S1 ∪ S2 whereS1 = {S ∈ D(G) : v /∈ S} and S2 = {S ∈ D(G) : v ∈ S}. If S ∈ S1 then S is also a derivedsubgraph of G1, so that |S1| = nd(G1). If S ∈ S2 then S − v either is a derived subgraph of G1,or else contains w as an isolated vertex. Define a map φ : S2 → D(G1) by

φ(S) = S − v − w w is isolated in S − v,

= S − v otherwise.

The map φ is 1-1 so that |S2| ≤ nd(G1) and hence |S2| ≤ |S1|. This implies that the edge vw isnon-residual since it belongs to every member of S2 and to no member of S1.

Let us remark here that the map φ above is not necessarily onto. A derived subgraph S1 of G1

such that w /∈ S1 but u ∈ S1 for some vertex u adjacent to w, is not the image under φ of anyS ∈ S2.

Theorem 3. Let the graph G have a vertex v with deg(v) = 2. Then at least one of the edgesincident with v is non-residual.

Proof. Denote the vertices adjacent to v by w1 and w2 and let ei = vwi, i = 1, 2. PutG1 = G− v. Partition D(G) into S0 ∪ S1 ∪ S2 ∪ S3 where

S0 = {S ∈ D(G) : v /∈ S}, S1 = {S ∈ D(G) : e1 ∈ S, e2 /∈ S},

S2 = {S ∈ D(G) : e1 /∈ S, e2 ∈ S}, S3 = {S ∈ D(G) : e1 ∈ S, e2 ∈ S}.Clearly, |S0| = nd(G1). For each S ∈ S3 denote by φ(S) the subgraph obtained from S

by deleting the vertex v and any other vertex (from among w1, w2) that becomes isolated. Thenφ(S) ∈ D(G1). This shows that |S3| ≤ nd(G1) = |S0|. Now if |S1| ≤ |S2| then e1 is non-residual, otherwise e2 is non-residual.


For a vertex of degree 3, it could happen that all of the edges incident with it are residual. Anexample is the vertex v in the graph G1 of Figure 2. However, we still can say something.

Theorem 4. Suppose that the graphGhas two adjacent verticesv1, v2 withdeg(v1) = deg(v2) =3. Then one of v1 and v2 is incident with a non-residual edge.

Proof. Put e = v1v2 and let G1 = G− e. Let S ∈ D(G). If e /∈ S then S ∈ D(G1). And ife ∈ S then either S − e ∈ D(G1) or else S − e contains v1 or v2 as an isolated vertex. PartitionD(G) into S0 ∪ S1 ∪ S2 ∪ S3, where S0 = {S ∈ D(G) : S − e ∈ D(G1)}, and S1,S2,S3,respectively, contain thoseS ∈ D(G) such that the set of isolated vertices ofS−e is, respectively,{v1}, {v2}, {v1, v2}. In G1, we have deg(v1) = deg(v2) = 2 so that, by Theorem 3, there aretwo edges e1, e2 incident respectively with v1, v2 such that e1, e2 are non-residual in G1. Thuseach of e1 and e2 occurs in at most half of the members of S0. Assume that |S1| ≤ |S2|. Then e2

is non-residual in G since e2 occurs in no member of S2,S3. Similarly |S2| ≤ |S1| implies thate1 is non-residual in G. This completes the proof of the theorem.

Theorem 5. Every graph on n ≤ 10 vertices satisfies Conjecture 3.

Proof. Assume there is a graph G of order n which violates Conjecture 3. From Theorems 2and 3, δ(G) ≥ 3, and by Theorem 1, δ(G) < log2 n. This implies that n ≥ 9. Suppose first thatn = 9. By Theorem 1, we must have

∑v∈V (G) 2−deg(v) > 1. This implies that G has at least

8 vertices of degree 3. By Theorem 4, these vertices of degree 3 must be pairwise non-adjacentwhich is impossible. Now let n = 10. We can similarly deduce that G has at least 7 pairwisenon-adjacent vertices of degree 3. This forces G to have 3 vertices of degree at least 7. But thenthe condition

∑v∈V (G) 2−deg(v) > 1 cannot be satisfied. This contradiction completes the proof

of the theorem.

Most of the familiar graphs, such as complete graphs, trees, paths and cycles, among othergraphs are now known to satisfy Conjecture 3.

4. GRAPHS WITH GIRTH ≥

In this section we shall prove the following theorem.

Theorem 6. Every graph with girth ≥ 6 satisfies Conjecture 3.

The rest of this section is devoted to proving this theorem. But first a definition and a technicallemma. Let v1, v2, . . . , vm be pairwise non-adjacent vertices in a graph G and let S ∈ D(G).The vertices v1, v2, . . . , vm are said to be free for S whenever S contains none of these verticesand none of their neighbors. Note that, in this case, the vertices v1, v2, . . . , vm are exactly theisolated vertices of the subgraph 〈V (S) ∪ {v1, v2, . . . , vm}〉.Lemma 3. Let v1, v2, . . . , vm be pairwise non-adjacent vertices in the graph G with degreesd1, d2, . . . , dm respectively. Assume further that no pair of these vertices has a common neighbor.Let n0 denote the number of derived subgraphs of G for which these vertices are free. Thenn0 ≤ 2−(d1+···+dm)nd(G).

Proof. LetG1 denote the subgraph ofG spanned by v1, v2, . . . , vm and their neighbors and putG2 = 〈V (G)− V (G1)〉. From Lemma 1, we have nd(G) ≥ nd(G1)nd(G2). But nd(G2) = n0

and G1 is the union of disjoint starts so that nd(G1) = 2d1+···+dm . Substituting, we get therequired result.


Proof of Theorem 6. Let G be a graph of girth ≥ 6. By Theorems 1 and 2, we can assumethat δ(G) ≥ 3. Choose two adjacent vertices u, v ∈ G such that deg(u) + deg(v) is minimum.We shall prove that the edge uv is non-residual. Let deg(u) = α and deg(v) = β, and supposethat u is adjacent to v, x1, . . . , xα−1 and v is adjacent to u, y1, . . . , yβ−1. Let G0 = G− {u, v}and put N0 = nd(G0).

Consider a derived subgraph S ∈ D(G). If u, v /∈ S then S ∈ D(G0). If S contains either uor v, then put S0 = S − {u, v}. For the subgraph S0, we have either S0 ∈ D(G0), or else S0

contains some of the vertices x1, . . . , xα−1, y1, . . . , yβ−1 as isolated vertices. In the latter case,of course, S0 /∈ D(G0). Thus we partition D(G) into S1 ∪ S2 ∪ S3 where

S1 = D(G0),

S2 = {S ∈ D(G) : S /∈ D(G0) but (S − {u, v}) ∈ D(G0)},S3 = {S ∈ D(G) : S /∈ D(G0) and (S − {u, v}) /∈ D(G0)}.

Let

n = |{S ∈ S3 : uv ∈ S}| − |{S ∈ S3 : uv /∈ S}|,and

m = |{S ∈ S1 ∪ S2 : uv /∈ S}| − |{S ∈ S2 : uv ∈ S}|.We shall prove that n−m ≤ 0, thereby proving that the edge uv is non-residual.

Lemma 4. n ≤ 9N0/16.

Proof. Let S ∈ S3. Put S0 = S − {u, v}. Then S0 contains some of the verticesx1, . . . , xα−1, y1, . . . , yβ−1 as isolated vertices. Denote by S∗ the subgraph obtained fromS0 by deleting all its isolated vertices. Then S∗ ∈ D(G0). Note that the vertices of the setV (S0) − V (S∗) are free for S∗. The lemma will be proved through calculating an upper boundon the number of ways by which S∗ can be chosen and the number of ways by which S can be re-built fromS∗. Suppose first thatS∗ has the vertices xi1 , . . . , xik , yj1 , . . . , yjh as free vertices. Wecan choose xi1 , . . . , xik , yj1 , . . . , yjh by (α−1

k )(β−1h ) ways. Next the number of derived subgraphs

of G0 for which xi1 , . . . , xik , yj1 , . . . , yjh are free is less than or equal to 2−k(β−1)2−h(α−1)N0.This follows from Lemma 3 by noting that in G0 each xi has degree ≥ β − 1 and each yj hasdegree ≥ α− 1 by the minimality of deg(u) + deg(v). Also note that the condition on the girthof G is needed here to ensure that the xi’s and the yj’s have distinct neighbors. Now the verticesof S0 − S∗ form a subset of xi1 , . . . , xik , yj1 , . . . , yjh which is non-empty since S ∈ S3. In thecase u, v ∈ S we can choose S0 − S∗ by 2k2h − 1 ways. If u ∈ S and v /∈ S then S0 − S∗ is anon-empty subset of xi1 , . . . , xik which can be chosen in 2k − 1 ways. Similarly for u /∈ S andv ∈ S, S0 − S∗ can be chosen from yj1 , . . . , yjh in 2h − 1 ways. Therefore we have

n ≤α−1∑k=0

β−1∑h=0

N0(α−1k )(β−1

h )2−k(β−1)2−h(α−1)[(2k+h − 1) − (2k − 1) − (2h − 1)]

= N0[(1 + 4/2β)α−1 − (1 + 2/2β)α−1][(1 + 4/2α)β−1 − (1 + 2/2α)β−1]

≤ N0(1/2β−1)(α− 1)(1 + 4/2β)α−2(1/2α−1)(β − 1)(1 + 4/2α)β−2.

Let us re-write this inequality as

n ≤ N0F (α, β), (1)


where

F (α, β) = 22−α−β(α− 1)(β − 1)(1 + 4/2α)β−2(1 + 4/2β)α−2.

Now for α, β ≥ 3 we have

F (α + 1, β)

F (α, β)=

α

2(α− 1)

(2α−1 + 1

2α−1 + 2

)β−2

(1 + 4/2β)

≤ 3α

4(α− 1)

(2α−1 + 1

2α−1 + 2

)< 1.

This inequality together with the symmetry of F (α, β) implies that

F (α, β) ≤ F (3, 3) = 9/16 α, β ≥ 3.

Substituting in (1), we get the required result.

Next we put G1 = G0 − {x1, . . . , xα−1}, G2 = G0 − {y1, . . . , yβ−1}, and let Ni =nd(Gi), i = 1, 2.

Lemma 5.

m = 2N0 −N1 −N2. (2)

Proof. We have

m = |{S ∈ S1 ∪ S2 : uv /∈ S}| − |{S ∈ S2 : uv ∈ S}|= |S1| + |{S ∈ S2 : u ∈ S, v /∈ S}| + |{S ∈ S2 : u /∈ S, v ∈ S}| − |{S ∈ S2 : uv ∈ S}|.

But,

|S1| = N0 (3)

|{S ∈ S2 : uv ∈ S}| = |{〈V (S0) ∪ {u, v}〉 : S0 ∈ D(G0)}| = |D(G0)| = N0. (4)

Let S ∈ S2 be such that u ∈ S and v /∈ S. Put S0 = S − u. Then S0 ∈ D(G0) andS0 ∩ {x1, . . . , xα−1} /= ∅ since u is not isolated in S. Hence S0 /∈ D(G1). Therefore,

|{S ∈ S2 : u ∈ S, v /∈ S}| = |{S0 ∈ D(G0) : S0 /∈ D(G1)}| = N0 −N1. (5)

Similarly,

|{S ∈ S2 : u /∈ S, v ∈ S}| = N0 −N2. (6)

From (3), . . ., (6), we obtain m = 2N0 −N1 −N2.

Lemma 6. N1 and N2 satisfy

N1 ≤ N0[1 + (1 + 2/2β)α−1]/(1 + 2α−1) (7)

N2 ≤ N0[1 + (1 + 2/2α)β−1]/(1 + 2β−1). (8)

Proof. We shall prove (7) and the proof of (8) is similar. PutG′ = G−v and letN ′ = nd(G′).

Suppose that S ∈ D(G′). If u /∈ S then S ∈ D(G0). If u ∈ S then either (S − u) ∈ D(G0) or(S − u) /∈ D(G0) in which case S − u contains some of the vertices x1, . . . , xα−1 as isolatedvertices. Partition D(G′) into S ′

1 ∪ S ′2 ∪ S ′

3, where

S ′1 = D(G0),


S ′2 = {S ∈ D(G′) : u ∈ S, (S − u) ∈ D(G0)},

S ′3 = {S ∈ D(G′) : u ∈ S, (S − u) /∈ D(G0)}.

Then

|S ′1| = N0. (9)

Also if S ∈ D(G′) with u ∈ S then S ∩ {x1, . . . , xα−1} /= ∅, hence (S − u) ∈ D(G0)− D(G1). Therefore,

|S ′2| = N0 −N1. (10)

We calculate an upper bound for |S ′3| as follows. Suppose S ∈ S ′

3. Let S0 = S−u, and denoteby S∗ the subgraph obtained from S0 by deleting all of its isolated vertices. Assume S0 containsthe vertices xi1 , . . . , xik as isolated vertices. These vertices can be chosen from {x1, . . . , xα−1}in (α−1

k ) ways. We note that the vertices xi1 , . . . , xik are free for S∗. From Lemma 3, the numberof such S∗ is less than or equal to 2−k(β−1)N0. Hence

|S ′3| ≤ N0

α−1∑k=1

(α− 1k

)2−k(β−1),

i.e.,

|S ′3| ≤ N0[(1 + 2/2β)α−1 − 1]. (11)

By (9), (10) and (11) we obtain

N ′ = |S ′1| + |S ′

2| + |S ′3|,

N ′ ≤ N0[1 + (1 + 2/2β)α−1] −N1. (12)

On the other hand N1 is the number of derived subgraphs of G′ for which vertex u is free andso by Lemma 3 we have

N1 ≤ 2−(α−1)N ′. (13)

From (12) and (13) we get 2α−1N1 ≤ N0[1+(1+2/2β)α−1]−N1 which implies the requiredresult.

Lemma 7.

m ≥ 39N0/40. (14)

Proof. Since α ≥ 3 and β ≥ 3, then we can deduce from (7) and (8) that

N1 ≤ [1 + (5/4)α−1]N0/(1 + 2α−1) (15)

N2 ≤ [1 + (5/4)β−1]N0/(1 + 2β−1). (16)

Substituting from (15) and (16) in (2), we get

m ≥ 2N0 − [1 + (5/4)α−1]N0/(1 + 2α−1) − [1 + (5/4)β−1]N0/(1 + 2β−1)

= [1 − (5/8)α−1]N0/(1 + 2−α+1) + [1 − (5/8)β−1]N0/(1 + 2−β+1)

≥ [1 − (5/8)2]N0/(5/4) + [1 − (5/8)2]N0/(5/4) = 39N0/40.


which is the required result.

From Lemmas 4 and 7 we see that n −m ≤ 9N0/16 − 39N0/40 < 0. This shows that theedge uv is well non-residual in G which completes the proof of Theorem 6.

The proof of Theorem 6 suggests the following conjecture.

Conjecture 4. Let u, v be a pair of adjacent vertices in a graph G such that deg(u) + deg(v) isminimal. Then the edge uv is non-residual in G.

5. GENERALIZATION TO HYPERGRAPHS

The concepts of derived subgraphs and residual edges have a straightforward generalization tohypergraphs. So it is appropriate to use the ‘generalize and conjecture' technique.

Let H be a hypergraph. Let S be a subhypergraph of H and denote by E(S) the set of itshyperedges. Call S a derived subhypergraph of H if every vertex v ∈ S belongs to some hyper-edge e ∈ E(S) (i.e., v is not isolated in S). Let D(H) denote the set of derived subhypergraphs ofH. A hyperedge e is called residual if it belongs to more than half of the members of D(H). Nowif S1, S2 ∈ D(H) then E(S1)∩E(S2) = E(S′), where S′ ∈ D(H) is obtained from S1 ∩S2 bydeleting its isolated vertices, if any. This shows that the family B(H) = {E(S) : S ∈ D(H)} isintersection-closed. Therefore Conjecture 2 implies the following.

Conjecture 5. Every non-empty hypergraph contains a non-residual hyperedge.

We outline a proof that Conjecture 5 implies Conjecture 1. Let A = {A1, . . . , An} be a union-closed family of sets and assume that ∅ ∈ A. Let A = ∪Ai = {e1, . . . , em}. Let J denote theset of irreducible members of A, that is, J = {Ai ∈ A : Ai /= Aj ∪ Ak for j, k /= i}. Notethat A is the closure, under taking unions, of J . We define a hypergraph H = (V, E) as follows.For each X ∈ J we associate a vertex vx and put V = {vX : X ∈ J }. The set of hyperedgesof H is defined by E = {e1, . . . , em}, where the hyperedge ei is incident with vertex vX if andonly if ei ∈ X . It can be shown that the family B(H) = {E(S) : S ∈ D(H)} is the dual familyof A. Moreover a hyperedge ei is non-residual in H if and only if ei belongs to at least half ofthe members of A. This follows by generalizing the arguments of Section 2 to hypergraphs. Weomit the details.

References

[1] I. Rival (Ed.), Graphs and order, Reidel, Dordrecht-Boston, (1985), p. 25.

[2] R. P. Stanley, Enumerative combinatorics, vol. I, Wadsworth & Brooks/Cole, Belmont, CA, (1986).

[3] D. G. Sarvate and J-C Renaud, On the union-closed sets conjecture, Ars Combin. 27 (1989), 149–154.

[4] D. G. Sarvate and J-C Renaud, Improved bounds for the union-closed sets conjecture, Ars Combin. 29(1990), 181–185.

[5] B. Poonen, Union-closed families, J. Combin. Theory, A 59 (1992), 253–268.

Documents

A graph-theoretic version of the union-closed sets conjecture