A proof of a Graph-theoretic version of the Union-Closed Sets Conjecture Bernardo LLano, Juan Jose Montellano-Ballesteros, Eduardo Rivera-Campo, Ricardo Strausz 14 julio 2005

Abstract An induced subgraph S of a graph G is called a derived subgraph of G if S contains no isolated vertices. An edge e of G is said to be residual if e occurs in more than half of the derived subgraphs of G. In this paper we proved that every non-empty graph contains a non-residual edge.

1

Introduction

A union-closed family of sets A is a finite collection of sets not all empty such that the union of two members of A is also a member of A. The following conjecture is due to P. Frankl (see [2, 3]).

Conjecture 1 Let A = {A1 , . . . An } be a union-close family of of n distinct sets. Then there exists an element which belongs to at least

n 2

of the sets in A.

In [1] M. H. El-Zahar proposed a weaker version of this conjecture (Conjecture 2), and it is state in the following way: Let G = (V (G), E(G)) be a simple graph with neither loops nor multiple edges. A subgraph S of G is called a derived subgraph of G if S contains no isolated vertices, and an edge e of G is said to be residual if e occurs in more than half of the derived subgraphs of G. 1

Conjecture 2 Every non-empty graph contains a non-residual edge. In [1] it is proof that a graph G of order n satisfies Conjecture 2 whenever one of the following conditions holds: δ(G) ≤ 2; log2 n ≤ δ(G); or n ≤ 10 (where δ(G) denotes the minimum degree of G). For each u ∈ V (G) let dG (u) denotes the degree of u in G. In [1] it is also conjectures that Conjecture 3 Let u, v be a pair of adjacent vertices in a graph G such that dG (u) + dG (v) is minimal. Then the edge uv is non-residual in G. We will said that an edge xy of G is local minimum if for every z ∈ NG (x), dG (z) ≥ dG (y), and for every z ∈ NG (y), dG (z) ≥ dG (x), where NG (x) and NG (y) denotes the set of neighbors of x and y respectively. In this paper we proved the following theorem. Theorem 4 Let G be a graph. Every local minimum edge is non-residual. Since for every graph G, an edge xy ∈ E(G) such that dG (x) + dG (y) = min{dG (z) + dG (w) : zw ∈ E(G)} is local minimum, Theorem 4 implies the veracity of Conjectures 2 and 3.

2

Notation

Let G = (V (G), E(G)) be a simple graph with neither loops nor multiple edges. Given Z ⊆ V (G), G − Z will denote the graph obtained from G by deleting the vertices in Z. Given an edge e = xy ∈ E(G), we will said that e is an x-pendant edge if dG (y) = 1 and we will said that e = xy is an isolated edge if dG (x) = dG (y) = 1. Let D(G) denotes the set of derived subgraphs of G. Given e = xy ∈ E(G), De (G) will denote the set of derived subgraphs of G which contains e as an edge, and let D∗e (G) denotes the set of elements of De (G) such that e is not an isolated edge. Also, let T (e) denotes the set of vertices z ∈ V (G) such that {x, y, z} induces a triangle in G. Given S ∈ D∗e (G), let Px (S) be the set of vertices z ∈ V (S) such that zx is a x- pendant edge of S. In an analogous way is defined Py (S). Observe that for every S ∈ D∗e (G), Px (S) ⊆ NG (x) \ T (e) and Py (S) ⊆ NG (y) \ T (e). 2

Given a set H of derived subgraphs of G and a set of properties Q1 , . . . , Qr , we denote as H[Q1 , . . . , Qr ] the set of derived subgraphs in H that fulfills the properties Q1 , . . . , Qr . For instance, if e = xy ∈ E(G) and {z, w} ⊆ V (G), De (G)[w 6∈ V, NG (z) ∩ V = ∅, Py = ∅, d(x) ≥ 3] will denote the set of derived subgraphs S in De (G) such that w 6∈ V (S), NG (z) ∩ V (S) = ∅, Py (S) = ∅ and dS (x) ≥ 3.

3

The Results

Before proving Theorem 4, we need some previous results and definitions.

Lemma 5 Let G be a graph an e = xy ∈ E(G) be a residual edge. Thus |D∗e (G)[Px 6= ∅, Py 6= ∅]| > |D∗e (G)[Px = Py = ∅]|. Moreover, if T (e) = ∅, then |D∗e (G)[Px 6= ∅, Py 6= ∅]| > 2|D∗e (G)[Px = Py = ∅]|. Proof. Let e = xy ∈ E(G) be a residual edge. Since e is residual 2|De (G)| > |D(G)| which implies that |De (G)| > |D(G) \ De (G)|.

(1)

Clearly |D(G) \ De (G)| = |D(G)[x ∈ V, y 6∈ V ]|+ |D(G)[x 6∈ V, y ∈ V ]| + |D(G)[x 6∈ V, y 6∈ V ]|,

(2)

and |De (G)| = |De (G) \ D∗e (G)| + |D∗e (G)| = |De (G) \ D∗e (G)| + |D∗e (G)[Px 6= ∅, Py 6= ∅]| + |D∗e (G)[Px = ∅]|+ |D∗e (G)[Py = ∅]| − |D∗e (G)[Px = Py = ∅]|.

3

(3)

Given S ∈ D∗e (G)[Px = ∅], since S has no x-pendant edges and xy ∈ E(S), it follows that S − {x} has no isolated vertices and y ∈ V (S − {x}). Therefore (S − {x}) ∈ D(G)[x 6∈ V, y ∈ V ]. From here it is not hard to see that the function fx : D∗e (G)[Px = ∅] → D(G)[x 6∈ V, y ∈ V ] defined as fx (S) = (S − {x}) is an injective function and therefore |D(G)[x 6∈ V, y ∈ V ]| ≥ |D∗e (G)[Px = ∅]|. In an analogous way we see that |D(G)[x ∈ V, y 6∈ V ]| ≥ |D∗e (G)[Py = ∅]|. Given S ∈ De (G) \ D∗e (G) it follows that S 0 = S − {x, y} has no isolated vertices. Moreover, NG (x) ∩ V (S 0 ) = NG (y) ∩ V (S 0 ) = ∅ and therefore S 0 ∈ D(G)[x 6∈ V, y 6∈ V, NG (x) ∩ V = NG (y) ∩ V = ∅] (observe that the empty graph is an element of D(G)[x 6∈ V, y 6∈ V, NG (x) ∩ V = NG (y) ∩ V = ∅]). As before, we can see that |De (G) \ D∗e (G)| ≤ |D(G)[x 6∈ V, y 6∈ V, NG (x) ∩ V = NG (y) ∩ V = ∅]|. Therefore, from (1), (2) and (3), and since D(G)[x 6∈ V, y 6∈ V ] = D(G)[x 6∈ V, y 6∈ V, NG (x) ∩ V = NG (y) ∩ V = ∅]∪ D(G)[x 6∈ V, y 6∈ V, NG (x) ∩ V 6= ∅ or NG (y) ∩ V 6= ∅] it follows that |D∗e (G)[Px 6= ∅, Py 6= ∅]| > |D∗e (G)[Px = Py = ∅]| + |D(G)[x 6∈ V, y 6∈ V, NG (x) ∩ V 6= ∅ or NG (y) ∩ V 6= ∅]|

(4)

and so |D∗e (G)[Px 6= ∅, Py 6= ∅]| > |D∗e (G)[Px = Py = ∅]|, proving the first part of this lemma. If T (e) = ∅, given S ∈ D∗e (G)[Px = Py = ∅] neither x nor y has pendant edges in S and there is no triangle with e as an edge in S. So the graph S 0 = S − {x, y} has no isolated vertices and, since xy is not an isolated edge of S, NG (x) ∩ V (S 0 ) 6= ∅ or NG (y) ∩ V (S 0 ) 6= ∅ and therefore (S − {x, y}) ∈ D(G)[x 6∈ V, y 6∈ V, NG (x) ∩ V 6= ∅ or NG (y) ∩ V 6= ∅]. From here it is not hard to see that |D∗e (G)[Px = Py = ∅]| ≤ |D(G)[x 6∈ V, y 6∈ V, NG (x) ∩ V 6= ∅ or NG (y) ∩ V 6= ∅]|, and therefore, from (4) |D∗e (G)[Px 6= ∅, Py 6= ∅]| > 2|D∗e (G)[Px = Py = ∅]|.

4

Let G be a graph, e = xy ∈ E(G) and let S ∈ D∗e (G)[Px 6= ∅]. We will said that Z ⊆ V (G) is an extension of Px (S) if i) Z ∩ V (S) = ∅. ii) For every w ∈ Px (S), NG (w) ∩ Z 6= ∅. iii) For every z ∈ Z, NG (z) ∩ Px (S) 6= ∅. Given an extension Z of Px (S), the graph induced by V (S) ∪ Z will be called an xextension of S. In an analogous way, given S ∈ D∗e (G)[Py 6= ∅], we define an extension of Py (S) and a y-extension of S. Observe that if e is a local minimum edge and dG (y) ≥ 2, then for every z ∈ NG (x), dG (z) ≥ dG (y) ≥ 2 and therefore every S ∈ D∗e (G)[Px 6= ∅] has an x-extension and every x-extension of every S ∈ D∗e (G)[Px 6= ∅] is an element of D∗e (G)[Px = ∅]. In the case that dG (x) ≥ 2 there are analogous implications. Lemma 6 Let G be a graph and e = xy ∈ E(G). If xy is a local minimum edge with min{dG (x), dG (y)} ≥ 2 then i) Every S ∈ D∗e (G)[Px 6= ∅] has at least 2dG (y)−1 − 1 different x-extensions. ii) Every S ∈ D∗e (G)[Py 6= ∅] has at least 2dG (x)−1 − 1 different y-extensions. Proof. Let S ∈ D∗e (G)[Px 6= ∅]. The lemma will be proved by induction over |Px (S)|. If Px (S) = {w} it follows that (NG (w) \ {x}) ∩ V (S) = ∅, and for every non-empty Z ⊆ NG (w) \ {x}, the graph induced by V (S) ∪ Z is a x-extension of S. Since e is local minimum, dG (w) ≥ dG (y) and the result follows. Suppose that |Px (S)| ≥ 2 and let M =

S

NG (w) \ {x}. Clearly M ∩ V (S) = ∅.

w∈Px (S)

If for every z ∈ M , Px (S) ⊆ NG (z), then for every non-empty subset Z ⊆ M , the graph induced by V (S) ∪ Z is an x-extension of S, which implies that S has 2|M | − 1 different x-extensions, and since e is local minimum, |M | ≥ dG (y) − 1 and the result follows. If for some z ∈ M , Px (S) \ NG (z) 6= ∅ let S 0 = S \ (Px (S) ∩ NG (z)). Clearly S 0 ∈ D∗e (G)[Px 6= ∅] with |Px (S 0 )| < |Px (S)|. Given an x-extension V (S 0 ) ∪ Z 0 of S 0 , since NG (z) ∩ Px (S 0 ) = ∅ 5

and z 6∈ V (S) , it follows that z 6∈ V (S 0 ) ∪ Z 0 and it is not hard to see that V (S 0 ) ∪ Z 0 ∪ {z} is an x-extension of S. So, for every x-extension of S 0 there is an x-extension of S, and since by induction hypothesis S 0 has at least 2dG (y)−1 − 1 x-extensions, i) follows. In an analogous way we prove ii).

Proof of Theorem 4. Let G be a graph and e = xy ∈ E(G) be a local minimum edge. If min{dG (x), dG (y)} = 1 the result follows from [1, Theorem 2] . So let suppose that dG (y) ≥ dG (x) ≥ 2 and let k = |T (e)|. We claim that |D∗e (G)[Px 6= ∅, Py 6= ∅]| ≤ (

2dG (x)−(k+1) − 1 )|D∗e (G)[Px = ∅, d(x) ≥ 2]|. 2dG (y)−1 − 1

(5)

Indeed. Let S ∈ D∗e (G)[Px 6= ∅, Py 6= ∅]. Clearly any x-extension S 0 of S has no x-pendant edges, and since x and y have pendant edges in S and e is not an isolated edge, it follows that min{dS (y), dS (x)} ≥ 2 and so dS 0 (x) ≥ 2. Therefore S 0 ∈ D∗e (G)[Px = ∅, d(x) ≥ 2]. Let B = {X, Y } be a bipartite graph where X = D∗e (G)[Px 6= ∅, Py 6= ∅], Y = D∗e (G)[Px = ∅, d(x) ≥ 2] and, for S ∈ X and S 0 ∈ Y , SS 0 ∈ E(B) if and only if S 0 is an x-extension of S. From Lemma 6 it follows that each element of X has degree at least 2dG (y)−1 − 1 and therefore |X|(2dG (y)−1 − 1) ≤ |E(B)|. Let us suppose there is S 0 ∈ Y with degree greater than 2dG (x)−(k+1) − 1. Since for every S ∈ X, dS (y) ≥ 2 we have that y 6∈ Px (S), and so for every S ∈ X, ∅ = 6 Px (S) ⊆ NG (x) \ {T (e) ∪ {y}}. This implies that S 0 is an x-extension of a pair of different S0 , S1 ∈ X such that Px (S0 ) = Px (S1 ). By definition V (S 0 ) = V (S0 ) ∪ Z0 = V (S1 ) ∪ Z1 where Z0 ∩ V (S0 ) = Z1 ∩ V (S1 ) = ∅. Observe that if there is u ∈ V (S1 ) such that u ∈ Z0 ∩ V (S1 ), then u ∈ Z0 which implies, by definition, that there is w ∈ Px (S0 ) = Px (S1 ) such that wu ∈ E(G) and so w 6∈ Px (S1 ) which is a contradiction. Therefore Z0 ∩ V (S1 ) = ∅ and Z1 ∩ V (S0 ) = ∅. Thus, since x ∈ V (S0 ) ∩ V (S1 ), Z0 = Z1 which implies that V (S0 ) = V (S1 ) and so S0 = S1 which is a contradiction. Therefore each S 0 ∈ Y has degree at most 2dG (x)−(k+1) − 1 which implies that |E(B)| ≤ |Y |(2dG (x)−(k+1) − 1) and the claim follows. Now we claim that |D∗e (G)[Px = ∅, Py 6= ∅, d(x) ≥ 2| ≤ (

2dG (y)−(k+1) − 1 )|D∗e (G)[Px = Py = ∅]| 2dG (x)−1 − 1

(6)

Indeed. Given S ∈ D∗e (G)[Px = ∅, Py 6= ∅, d(x) ≥ 2] it is not hard to see that any y-extension of S is an element of D∗e (G)[Px = Py = ∅]. 6

Let B = {X, Y } be a bipartite graph where X = D∗e (G)[Px = ∅, Py 6= ∅, d(x) ≥ 2], Y = D∗e (G)[Px = Py = ∅] and, for S ∈ X and S 0 ∈ Y , SS 0 ∈ E(B) if and only if S 0 is an y-extension of S. From Lemma 6 it follows that each element of X has degree at least 2dG (x)−1 − 1 and so |X|(2dG (x)−1 − 1) ≤ |E(B)|. Let us suppose there is S 0 ∈ Y with degree greater than 2dG (y)−(k+1) − 1. Since for every S ∈ X, dS (x) ≥ 2, we see that x 6∈ Py (S) and therefore for every S ∈ X, ∅ 6= Py (S) ⊆ NG (y)\{T (e)∪{x}}, which implies that S 0 is an y-extension of a pair of different S0 , S1 ∈ X such that Py (S0 ) = Py (S1 ). As in the previous claim, we see that this is not possible, so |E(B)| ≤ |Y |(2dG (y)−(k+1) − 1) and the claim follows. Clearly D∗e (G)[Px = ∅, d(x) ≥ 2] = D∗e (G)[Px = ∅, Py 6= ∅, d(x) ≥ 2] ∪ D∗e (G)[Px = Py = ∅, d(x) ≥ 2]. On the other hand, given S ∈ D∗e (G)[Px = Py = ∅], since e is not an isolated edge, min{dS (x), dS (y)} ≥ 2 which implies that D∗e (G)[Px = Py = ∅, d(x) ≥ 2] = D∗e (G)[Px = Py = ∅]. Thus |D∗e (G)[Px = ∅, d(x) ≥ 2]| = |D∗e (G)[Px = ∅, Py 6= ∅, d(x) ≥ 2]| + |D∗e (G)[Px = Py = ∅]| and therefore, from (5) we see that |D∗e (G)[Px 6= ∅, Py 6= ∅]| ≤ ( (

2dG (x)−(k+1) − 1 )|D∗e (G)[Px = ∅, Py 6= ∅, d(x) ≥ 2]|+ 2dG (y)−1 − 1

2dG (x)−(k+1) − 1 )|D∗e (G)[Px = Py = ∅]|) 2dG (y)−1 − 1

and from (6) it follows that |D∗e (G)[Px 6= ∅, Py 6= ∅]| ≤ ( (

2dG (x)−(k+1) − 1 2dG (y)−(k+1) − 1 )( d (x)−1 )|D∗e (G)[Px = Py = ∅]|)+ 2dG (y)−1 − 1 2G −1

2dG (x)−(k+1) − 1 )|D∗e (G)[Px = Py = ∅]|) = 2dG (y)−1 − 1

|D∗e (G)[Px = Py = ∅]((

2dG (x)−(k+1) − 1 2dG (y)−(k+1) − 1 2dG (x)−(k+1) − 1 )( ) + ). 2dG (y)−1 − 1 2dG (x)−1 − 1 2dG (y)−1 − 1 7

Thus, to end the proof, by Lemma 5, we only need to show that 2dG (x)−(k+1) − 1 2dG (y)−(k+1) − 1 2dG (x)−(k+1) − 1 ( d (y)−1 )( d (x)−1 )+ ≤ 2G −1 2G −1 2dG (y)−1 − 1

(

2

if |T (e)| = k = 0;

1

if |T (e)| = k ≥ 1.

If k = 0, (

2dG (x)−(k+1) − 1 2dG (y)−(k+1) − 1 2dG (x)−(k+1) − 1 2dG (x)−1 − 1 )( ) + = (1 + ) 2dG (y)−1 − 1 2dG (x)−1 − 1 2dG (y)−1 − 1 2dG (y)−1 − 1

and since dG (x) ≤ dG (y) the result follows. If k ≥ 1 first observe that min{

2dG (x)−(k+1) − 1 2dG (y)−(k+1) − 1 1 , }≤ d (x)−1 d (y)−1 G G 2 2 −1 2 −1

and therefore (

2dG (x)−(k+1) − 1 2dG (y)−(k+1) − 1 1 2dG (x)−(k+1) − 1 2dG (y)−(k+1) − 1 )( ) = ( )( d (y)−1 )≤ . d (y)−1 d (x)−1 d (x)−1 G G G G 4 2 −1 2 −1 2 −1 2 −1

Finally, since dG (x) ≤ dG (y), 2dG (x)−(k+1) − 1 1 ≤ d (y)−1 G 2 2 −1 and thus we see that (

2dG (x)−(k+1) − 1 2dG (y)−(k+1) − 1 2dG (x)−(k+1) − 1 1 1 )( ) + ≤ + 4 2 2dG (y)−1 − 1 2dG (x)−1 − 1 2dG (y)−1 − 1

and the theorem follows.

References [1] M. H. El-Zahar, A Graph-Theoretical Version of the Union-Closed Sets Conjecture, Journal of Graph Theory , 26, (1997), 155 – 163. [2] I. Rival (Ed.), Graphs and order, Reidel, Dordrecht-Boston, (1985), p. 25. [3] B. Poone, Union-closed families, J. Combin. Theory A, 59 (1992), 253–268.

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