A new proof of the main theorem of the paper:
P. Ille, Indecomposable graphs, Discrete Math.
173 (1997) 71-78.
1
Introduction
We consider different structures.
• Given a (finite and nonempty) set S and a positive integer k, a binary
structure [3] is a function B : (S × S) − {(x, x) : x ∈ S} −→ {0, . . . k − 1}.
The set S is the basis of B which is denoted by B, and the integer k is
the rank of B which is denoted by rk(B).
• A digraph D is defined by a (finite and nonempty) vertex set V (D) and an
arc set A(D), where an arc of D is an ordered pair of distinct vertices of
D. To each digraph D corresponds a unique binary structure B defined
by B = V (D), rk(B) = 2 and B −1 ({1}) = A(D).
• A (simple) graph G is defined by a (finite and nonempty) vertex set V (G)
and an edge set E(D), where an edge of G is a pair of vertices of G.
Similarly, to each graph G corresponds a unique binary structure B defined
by B = V (G), rk(B) = 2 and for x 6= y ∈ B, B(x, y) = 1 if {x, y} ∈
E(G). Let G be a graph. With any X ⊆ V (G) associate the induced
subgraph G(X) of G by X defined by V (G(X)) = X and E(G(X)) =
E(G) ∩ {{x, y} : x 6= y ∈ X}. For each vertex x of G, the neighbourhood
{y ∈ V (G) : {x, y} ∈ E(G)} of x in G is denoted by NG (x) and its degree
| NG (x) | is denoted by dG (x). A graph G is multipartite by a partition P
of V (G) if for every X ∈ P , E(G(X)) = ∅. It is bipartite when | P |= 2. A
bipartite graph G by a partition {X, Y } is complete if for any x ∈ X and
y ∈ Y , {x, y} ∈ E(G).
Observation 1 Let G be a bipartite graph by {X, Y } which is connected and
not complete. Assume that for every x ∈ X, dG (x) = 1 or | Y | and that for
every y ∈ Y , dG (y) = 1 or | X |. Then, we have | {x ∈ X : dG (x) =| Y |} |=|
{y ∈ Y : dG (y) =| X |} |= 1.
Proof . Since G is not complete and connected, we have | X |> 1 and | Y |> 1.
Suppose that | {x ∈ X : dG (x) =| Y |} |> 1. For every y ∈ Y , we have dG (y) > 1
1
and hence dG (y) =| X |. Consequently, G would be complete bipartite. It follows
that | {x ∈ X : dG (x) =| Y |} |≤ 1. As | Y |> 1, consider y 6= y 0 ∈ Y . Because
G is connected, there exists a sequence z0 = y, . . . , zn = y 0 of vertices of G
satisfying {zi , zi+1 } ∈ E(G) for i ∈ {0, . . . , n − 1}. Since E(G(Y )) = ∅, we
have n ≥ 2. In particular, we have {z0 , z1 } ∈ E(G) and {z1 , z2 } ∈ E(G).
We obtain that z1 ∈ X and z0 , z2 ∈ NG (z1 ). Consequently, dG (z1 ) =| Y |
and hence | {x ∈ X : dG (x) =| Y |} |= 1. Symmetrically, we show that
| {y ∈ Y : dG (y) =| X |} |= 1.
Given a binary structure B, with each X ⊆ B associate the binary substructure B(X) of B of basis X and of rank rk(B) which is the restriction of B to
(X × X) − {(x, x) : x ∈ X}. For convenience, B(X) is denoted by B − Y if
X = B − Y and by B − x if X = B − {x}.
Given two binary structures B and C, such that rk(B) = rk(C), an isomorphism from B onto C is a bijection f : B −→ C such that for any x, y ∈ B,
with x 6= y, B(x, y) = C(f (x), f (y)). When such an isomorphism exists, B and
C are said to be isomorphic, which is denoted by B ' C.
With a binary structure B associate an equivalence relation ≡ defined on
(B × B) − {(x, x) : x ∈ B} as follows: given (u, v), (x, y) ∈ (B × B) − {(x, x) :
x ∈ B}, (u, v) ≡ (x, y) if the function {u, v} −→ {x, y}, defined by u 7→ x and
v 7→ y, realizes an isomorphism from B({u, v}) onto B({x, y}). Given X ⊆ B
and x ∈ B − X, x ∼ X means that for u, v ∈ X, (x, u) ≡ (x, v). Given disjoint
subsets X and Y of B, X ∼ Y signifies that for x, x0 ∈ X and y, y 0 ∈ Y ,
(x, y) ≡ (x0 , y 0 ).
Let B be a binary structure. A subset I of B is an interval of B if for every
x ∈ B − I, x ∼ I. For instance, ∅, B and {x}, where x ∈ B, are intervals of B,
called trivial intervals. A binary structure is indecomposable if all its intervals
are trivial, otherwise it is decomposable. By definition, a graph (or a digraph)
has the same intervals as its corresponding binary structure. Hence, a graph
(or a digraph) and its corresponding binary structure are both indecomposable
or not. Consequently, we examine the indecomposability in the case of binary
structures.
The purpose of this note is to present a simpler and more concise demonstration of the following result which constitutes the main result of [2].
Theorem 1 ([2]) Given an indecomposable binary structure B, consider a subset X of B sush that | X |≥ 3 and B(X) is indecomposable. If | B − X |≥ 6, then
there exist x 6= y ∈ B − X such that B − {x, y} is indecomposable.
2
Preliminaries
Some results are recalled.
Proposition 1 ([3, 5]) For every indecomposable binary structure B, with |
B |≥ 3, there exists X ⊆ B such that | X |= 3 or 4 and B(X) is indecomposable.
2
Given a binary structure B, let X be a proper subset of B such that | X |≥ 3
and B(X) is indecomposable. We consider the following subsets of B − X:
• Ext(X) is the set of x ∈ B − X such that B(X ∪ {x}) is indecomposable;
• [X] is the set of x ∈ B − X such that X is an interval of B(X ∪ {x});
• given u ∈ X, X(u) is the set of x ∈ B − X such that {u, x} is an interval
of B(X ∪ {x}).
The family {Ext(X), [X]} ∪ {X(u)}u∈X is denoted by p[B,X] .
Remark 1 By definition, we have Ext(X) ∩ [X] = ∅ and Ext(X) ∩ X(u) = ∅
for every u ∈ X. Furthermore, suppose that there is x ∈ [X] ∩ X(u), where
u ∈ X. We obtain that X and {u, x} are intervals of B(X ∪ {x}). Therefore,
[X] − {u, x} = X − {u} is an interval of B(X ∪ {x}) and hence a non trivial
interval of B(X). Lastly, suppose that there is x ∈ X(u) ∩ X(v), where u 6= v ∈
X. We obtain that {u, x} and {v, x} are intervals of B(X ∪ {x}). It follows
that {u, x} ∪ {v, x} = {u, v, x} is an interval of B(X ∪ {x}) and thus {u, v} is
a non trivial interval of B(X).
Consequently, p[B,X] constitutes a partition of B − X.
Remark 2 Let x and y be distinct elements of B − X.
1. If x, y ∈ [X], then X is a non trivial interval of B(X ∪ {x, y}).
2. If there is u ∈ X such that x, y ∈ X(u), then {u, x, y} is a non trivial
interval of B(X ∪ {x, y}).
In both cases, B(X ∪ {x, y}) is decomposable.
Proposition 2 ([1]) Given a binary structure B, consider a proper subset X
of B such that | X |≥ 3 and B(X)is indecomposable.
1. Given x ∈ [X] and y ∈ B − (X ∪ [X]), if B(X ∪ {x, y}) is decomposable,
then X ∪ {y} is an interval of B(X ∪ {x, y}).
2. Given x ∈ X(u), where u ∈ X, and y ∈ B − (X ∪ X(u)), if B(X ∪ {x, y})
is decomposable, then {u, x} is an interval of B(X ∪ {x, y}).
3. Given x 6= y ∈ Ext(X), if B(X ∪ {x, y}) is decomposable, then {x, y} is
an interval of B(X ∪ {x, y}).
Theorem 2 ([1]) Given an indecomposable binary structure B, consider a subset X of B such that | X |≥ 3 and B(X)is indecomposable. If | B − X |≥ 2, then
there exist x 6= y ∈ B − X such that B(X ∪ {x, y}) is indecomposable. More
precisely, we have:
1. If [X] 6= ∅, then there are x ∈ [X] and y ∈ B − (X ∪ [X]) such that
B(X ∪ {x, y}) is indecomposable.
3
2. For every u ∈ X, if X(u) 6= ∅, then there are x ∈ X(u) and y ∈ B − (X ∪
X(u)) such that B(X ∪ {x, y}) is indecomposable.
This theorem leads us to introduce the following graph. Given a binary
structure B, consider a subset X of B such that | X |≥ 3, | B − X |≥ 2 and B(X)
is indecomposable. The graph G[B,X] is defined on V (G[B,X] ) = B − X by:
given x 6= y ∈ B − X, {x, y} ∈ E(G[B,X] ) if B(X ∪ {x, y}) is indecomposable.
Remark 3 By Remark 2, if Ext(X) = ∅, then G[B,X] is multipartite by p[B,X] .
From Proposition 1 and Theorem 2 follows:
Corollary 1 ([1]) For every indecomposable binary structure B, with | B |≥ 3,
there exist x, y ∈ B such that B − {x, y} is indecomposable.
The next result is a direct consequence of Theorem 2.
Corollary 2 Given an indecomposable binary structure B, consider a subset X
of B such that | X |≥ 3, | B − X |≥ 3 and B(X)is indecomposable. Then, the
assertions below are equivalent.
1. For any x 6= y ∈ B − X, B − {x, y} is decomposable.
2. | B − X | is odd and for every proper subset Z of B − X, if | Z | is odd,
then B(X ∪ Z) is decomposable.
3
Proof of Theorem 1
Lemma 1 Given an indecomposable binary structure B, consider a subset X
of B such that | X |≥ 3, | B − X |≥ 4 and B(X)is indecomposable. Assume
that for Z ⊆ B − X, with | Z |= 1 or 3, B(X ∪ Z) is decomposable. Then, both
assertions below hold.
1. Given distinct elements a, a0 and b of B − X, if {a, b}, {a0 , b} ∈ E(G[B,X] ),
then {a, a0 } is an interval of B(X ∪ {a, a0 , b}). In particular, there is
M ∈ p[B,X] such that a, a0 ∈ M .
2. Given distinct elements a, a0 and b of B − X, if there is M ∈ p[B,X]
such that a, a0 ∈ M and if {a, b} ∈ E(G[B,X] ) and {a0 , b} 6∈ E(G[B,X] ),
then either M = [X] and X ∪ {a, b} is an interval of B(X ∪ {a, a0 , b}) or
M = X(u), where u ∈ X, and {u, a0 } is an interval of B(X ∪ {a, a0 , b}).
Proof . In both assertions, set Y = X ∪ {a, b} so that B(Y ) is indecomposable.
By hypothesis, we have Ext(X) = Ext(Y ) = 0.
For the first assertion, we have to verify that a0 ∈ Y (a). Now, if a0 ∈ [Y ],
then a0 ∼ Y and hence a0 ∼ X ∪ {b}, which contradicts {a0 , b} ∈ E(G[B,X] ). If
a0 ∈ Y (u), where u ∈ X ∪ {b}, then we obtain the same contradiction because
{u, a0 } would be an interval of B(X∪{a0 , b}). Consequently, a0 ∈ Y (a). It follows
4
that {a, a0 } is an interval of B(X ∪{a, a0 }) and hence B(X ∪{a}) ' B(X ∪{a0 }).
Therefore, a and a0 belong to the same element of p[B,X] .
In the second assertion, B(X ∪ {a, b}) is indecomposable, B(X ∪ {a0 , b})
is decomposable and B(X ∪ {a, a0 }) is decomposable by Remark 2. It follows
that a0 6∈ Y (a) ∪ Y (b). Indeed, if a0 ∈ Y (a), that is, if {a, a0 } is an interval of
B(X ∪ {a, a0 , b}), then B(X ∪ {a, b}) ' B(X ∪ {a0 , b}), and if a0 ∈ Y (b), then
B(X ∪ {a, b}) ' B(X ∪ {a, a0 }). Therefore, if a0 ∈ Y (u), where u ∈ Y , then
u ∈ X and thus a0 ∈ X(u). Clearly, if a0 ∈ [Y ], then a0 ∈ [X]. We conclude
as follows: if M = [X], then a0 ∈ [Y ] and if M = X(u), where u ∈ X, then
a0 ∈ Y (u).
Proposition 3 Given an indecomposable binary structure B, consider a subset
X of B such that | X |≥ 3, | B − X |≥ 4 and B(X)is indecomposable. Assume
that for Z ⊆ B − X, with | Z |= 1 or 3, B(X ∪ Z) is decomposable. For every
connected component C of G[B,X] , the following assertions are satisfied.
1. | C |> 1.
2. There exist distinct elements M and N of p[B,X] such that G[B,X] (C) is
bipartite by {M ∩ C, N ∩ C}.
3. B(X ∪ C) is indecomposable.
Proof . For the first assertion, denote by W the set of the elements x of B − X
such that dG[B,X] = 0. We have to establish that W = ∅. Since Ext(X) = ∅
and since B is indecomposable, it suffices to prove that B − (W ∩ [X]) is an
interval of B and that for each u ∈ X, {u} ∪ (W ∩ X(u)) is also. To begin,
consider B − (W ∩ [X]). Given x ∈ W ∩ [X], we have only to verify for each
y ∈ (B − X) − (W ∩ [X]) that x ∼ X ∪ {y}. As x ∈ W , B(X ∪ {x, y}) is
decomposable. If y 6∈ [X], then x ∼ X ∪ {y} by Proposition 2.1. If y ∈ [X],
then y 6∈ W and there exists z ∈ B − X such that {y, z} ∈ E(G[B,X] ). Since
x ∈ W , {x, z} 6∈ E(G[B,X] ). It follows from Lemma 1.2 that x ∼ X ∪ {y, z}.
Now, consider {u} ∪ (W ∩ X(u)), where u ∈ X. Given x ∈ W ∩ X(u), we have
only to verify for each y ∈ (B − X) − (W ∩ X(u)) that y ∼ {u, x}. As x ∈ W ,
B(X ∪{x, y}) is decomposable. If y 6∈ X(u), then y ∼ {u, x} by Proposition 2.2.
If y ∈ X(u), then y 6∈ W and we conclude as previously by applying Lemma
1.2.
For the second assertion, it follows from the first that there are a 6= b ∈ C
such that {a, b} ∈ E(G[B,X] ). By Remark 2, there exist distinct elements M
and N of p[B,X] such that a ∈ M and b ∈ N . By Remark 3, it is sufficient to
prove that C ⊆ M ∪ N . Let c be an element of C − {a, b}. Since G[B,X] (C) is
connected, there exists a sequence a0 = a, . . . , ap = c of elements of C such that
p ≥ 1 and {ai , ai+1 } ∈ E(G[B,X] ) for i ∈ {0, . . . , p − 1}. By applying Lemma
1.1 to the edges {a, b} and {a0 , a1 } of G[B,X] , when b 6= a1 , we obtain that
a1 ∈ N . Therefore, a0 ∈ M and a1 ∈ N . So, assume that p ≥ 2. For every
i ∈ {0, . . . , p − 2}, we have {ai , ai+1 }, {ai+1 , ai+2 } ∈ E(G[B,X] ). By Lemma 1.1,
5
ai and ai+2 belong to the same element of p[B,X] . It ensues that c ∈ M if p is
even and c ∈ N if p is odd.
For the last assertion, we demonstrate the following: if I is an interval of
B(X ∪ C) such that | I |≥ 2, then I = X ∪ C. Since B(X) is indecomposable,
we have I ∩ X = ∅ or | I ∩ X |= 1 or I ∩ X = X. To begin, suppose that
I ∩ X = ∅. We will demonstrate that I would be an interval of B as well, which
contradicts its indecomposability. Given an element x of B − X and distinct
elements i and j of I, we have to verify that x ∼ {i, j}. As I is an interval
of B(X ∪ C), we may assume that x 6∈ X ∪ C. For any k 6= k 0 ∈ I, we have
I ∩ (X ∪ {k, k 0 }) = {k, k 0 } is an interval of B(X ∪ {k, k 0 }). It follows that
B(X ∪ {k}) ' B(X ∪ {k 0 }). Therefore, all the elements of I belong to the same
element of p[B,X] . For instance, assume that I ⊆ M . Now, we prove that there
exists α ∈ N ∩ C satisfying for every k ∈ I, {k, α} ∈ E(G[B,X] ). Let k0 be an
element of I. Because G[B,X] (C) is connected, there exists α ∈ C such that
{k0 , α} ∈ E(G[B,X] ). Since G[B,X] (C) is bipartite by {M ∩ C, N ∩ C}, we have
α ∈ N ∩ C and hence α 6∈ I. For any k ∈ I, I ∩ (X ∪ {k0 , k, α}) = {k0 , k} is an
interval of B(X ∪ {k0 , k, α}). It follows that B(X ∪ {k0 , α}) ' B(X ∪ {k, α})
and thus {k, α} ∈ E(G[B,X] ). To continue, we distinguish two cases. We recall
that, since x 6∈ X ∪ C, {x, α} 6∈ E(G[B,X] ) and {x, k} 6∈ E(G[B,X] ) for every
k ∈ I.
• Assume that M = [X]. If x 6∈ [X], then, by Proposition 2.1, k ∼ X ∪ {x}
for every k ∈ I. Given u ∈ X, we obtain that (x, i) ≡ (u, i) and (x, j) ≡
(u, j). But, (u, i) ≡ (u, j) because I is an interval of B(X ∪ C). If x ∈ [X],
then, by Lemma 1.2, x ∼ X ∪ {k, α} for every k ∈ I. In particular, we
have x ∼ {k, α} for every k ∈ I. Consequently, x ∼ {i, j}.
• Assume that M = X(u), where u ∈ X. If x 6∈ X(u), then, by Proposition
2.2, x ∼ {u, k} for every k ∈ I. Therefore, x ∼ {i, j}. If x ∈ X(u), then,
by Lemma 1.2, k ∼ {u, x} for every k ∈ I. In particular, we obtain that
(x, i) ≡ (u, i) and (x, j) ≡ (u, j). However, (u, i) ≡ (u, j) because I is an
interval of B(X ∪ C).
Therefore, we established that either I ∩ X = X or | I ∩ X |= 1. Suppose that
there is u ∈ X such that I ∩ X = {u}. As | X |≥ 2, I − {u} =
6 ∅. Furthermore,
we have I − {u} ⊆ X(u). For example, assume that M = X(u). Consider any
elements a of I − {u} and b of N ∩ C. Since b 6∈ M , we have b 6∈ I. It follows
that I ∩ (X ∪ {a, b}) = {u, a} is an interval of B(X ∪ {a, b}). So, B(X ∪ {a, b}) is
decomposable, that is, {a, b} 6∈ E(G[B,X] ). Consequently, G[B,X] (C) would not
be connected. It ensues that I ∩ X = X. Since X ⊆ I, we have C − I ⊆ [X].
As previously, we verify that if C − I 6= ∅, then G[B,X] (C) is not connected.
Indeed, assume that M = [X] and consider any elements a of C − I and b of
N ∩ C. Since b 6∈ M , we have b ∈ I. It follows that I ∩ (X ∪ {a, b}) = X ∪ {b}
is an interval of B(X ∪ {a, b}). To conclude, we obtain that C − I = ∅ and thus
I = X ∪ C.
Corollary 3 Given an indecomposable binary structure B, consider a subset X
of B such that | X |≥ 3, | B − X |≥ 4 and B(X)is indecomposable. Assume that
6
for any x 6= y ∈ B − X, B − {x, y} is decomposable. Then, G[B,X] is connected,
| p[B,X] |= 2 and G[B,X] is bipartite by p[B,X] but non complete bipartite.
Proof . By Corollary 2, | B − X | is odd and for every proper subset Z of B − X,
if | Z | is odd, then B(X ∪ Z) is decomposable. For a contradiction, suppose that
G[B,X] is not connected. As | B −X | is odd, there exists a connected component
C of G[B,X] (C) such that | C | is odd. Since C is a proper subset of B − X,
B(X ∪ C) is decomposable, which contradicts Proposition 3.3. Consequently,
G[B,X] is connected and it follows from Proposition 3.2 that G[B,X] is bipartite
by p[B,X] . Lastly, if G[B,X] ware complete bipartite, then, by Lemma 1.1, both
elements of p[B,X] are intervals of B, which contradicts its indecomposability.
Proof of Theorem 1 . For a contradiction, suppose that for any x 6= y ∈ B − X,
B − {x, y} is decomposable. By the preceding corollary, G[B,X] is connected,
| p[B,X] |= 2 and G[B,X] is bipartite by p[B,X] but non complete bipartite. Denote
the elements of p[B,X] by M and N .
In order to apply Observation 1 to G[B,X] , we will prove that for every
a ∈ M , dG[B,X] (a) = 1 or | N | and that for every b ∈ N , dG[B,X] (b) = 1
or | M |. Otherwise, suppose for instance that there are an element a of M
and distinct elements b, b0 et b00 of N such that {a, b}, {a, b0 } ∈ E(G[B,X] ) and
{a, b00 } 6∈ E(G[B,X] ). Since G[B,X] is connected, there is a0 ∈ M − {a} such that
{a0 , b00 } ∈ E(G[B,X] ). Set Y = X ∪ {a, b} so that B(Y ) is indecomposable. By
Lemma 1.1, we have b0 ∈ Y (b). Suppose that a0 ∈ Y (b), that is, {a0 , b} is an
interval of B(Y ∪{a0 }) and hence of B(X ∪{a0 , b}). Then, B(X ∪{a0 }) ' B(X ∪
{b}) so that a and b should belong to the same element of p[B,X] . Consequently,
a0 6∈ Y (b). Now, we distinguish two cases.
1. Assume that N = [X]. By Lemma 1.2, we have b00 ∈ [Y ]. But, a0 6∈ [Y ]
because [Y ] ⊆ [X] = N .
2. Assume that N = X(u), where u ∈ X. By Lemma 1.2, we have b00 ∈ Y (u).
But, a0 6∈ Y (u) because Y (u) ⊆ X(u) = N .
In both cases, | p[B,Y ] |≥ 3, which contradicts Corollary 3 applied to B(Y ).
By Observation 1 applied to G[B,X] , there exist a ∈ M and b ∈ N such that
G[B,X] is decomposed as follows:
∗ NG[B,X] (a) = N and NG[B,X] (b) = M ;
∗ for every a0 ∈ M − {a}, NG[B,X] (a0 ) = {b};
∗ for every b0 ∈ N − {b}, NG[B,X] (b0 ) = {a}.
Finally, set Y = X ∪ {a, b} so that B(Y ) is indecomposable. By Lemma 1.1
applied to B(X), we obtain that M −{a} ⊆ Y (a) and N −{b} ⊆ Y (b). It follows
that M −{a} = Y (a), N −{b} = Y (b) and p[B,Y ] = {M −{a}, N −{b}}. Consider
any elements a0 of M − {a}, b0 of N − {b} and suppose that B(Y ∪ {a0 , b0 }) is
decomposable. By Proposition 2.2 applied to B(Y ), {a, a0 } is an interval of
7
B(Y ∪ {a0 , b0 }). Therefore, we should obtain that B(X ∪ {a, b0 }) ' B(X ∪
{a0 , b0 }). But, B(X ∪ {a, b0 }) is indecomposable whereas B(X ∪ {a0 , b0 }) is not.
Consequently, for any a0 ∈ M − {a} and b0 ∈ N − {b}, B(Y ∪ {a0 , b0 }) is
indecomposable. In other words, G[B,Y ] is complete and bipartite by p[B,Y ] ,
which contradicts Corollary 3 applied to B(Y ).
4
Epilogue
To begin, we recall the first important result on the indecomposable binary
structures which improves Corollary 1.
Theorem 3 ([4]) For every indecomposable binary structure B, with | B |≥ 7,
there exist x =
6 y ∈ B such that B − {x, y} is indecomposable.
We will establish this theorem for an indecomposable binary structure B
such that | B |≥ 8. The following is another consequence of Corollary 3.
Corollary 4 Given an indecomposable binary structure B, consider a subset X
of B such that | X |≥ 3, | B − X |≥ 4 and B(X)is indecomposable. Assume that
for any x 6= y ∈ B − X, B − {x, y} is decomposable. Given u ∈ X such that
X(u) ∈ p[B,X] , there exists x ∈ B − X such that B − {u, x} is indecomposable.
Proof . Consider u ∈ X such that X(u) ∈ p[B,X] . By Corollary 3, G[B,X] is
connected, | p[B,X] |= 2 and G[B,X] is bipartite by p[B,X] but non complete
bipartite. Therefore, there are a ∈ X(u) and b, b0 ∈ B − (X ∪ X(u)) such that
{a, b} ∈ E(G[B,X] ) and {a, b0 } 6∈ E(G[B,X] ).
Since a ∈ X(u), that is, since {u, a} is an interval of B(X ∪ {a}), we have
B(X) ' B((X − {u}) ∪ {a}). Thus, B(X 0 ) is indecomposable, where X 0 = (X −
{u}) ∪ {a}. We will prove that | p[B,X 0 ] |≥ 3. Clearly, u ∈ X 0 (a) because {u, a}
is an interval of B(X ∪ {a}) = B(X 0 ∪ {u}). For a contradiction, suppose that
b ∈ X 0 (a) as well. By Remark 2, {a, u, b} would be an interval of B(X 0 ∪{u, b}) =
B(X ∪ {a, b}). Therefore, {a, u, b} ∩ (X ∪ {b}) = {u, b} would be an interval of
B(X ∪ {b}) or, equivalently, b ∈ X(u), which contradicts {a, b} ∈ E(G[B,X] ). It
follows that b 6∈ X 0 (a). Now, we distinguish two cases.
1. Assume that p[B,X] = {X(u), [X]}. By Lemma 1.2 applied to B(X),
b0 ∼ X ∪ {a, b} and hence b0 ∈ [X 0 ]. As b ∈ [X], we have b ∼ X. However,
we do not have b ∼ X ∪ {a} because {a, b} ∈ E(G[B,X] ). Therefore, we do
not have b ∼ {a, w} for w ∈ X − {u}. Consequently, b 6∈ [X 0 ].
2. Assume that p[B,X] = {X(u), X(v)}, where v ∈ X − {u}. It follows from
Lemma 1.2 that {v, b0 } is an interval of B(X∪{a, b, b0 }) = B(X 0 ∪{u, b, b0 }).
In particular, we obtain that b0 ∈ X 0 (v). As {a, b} ∈ E(G[B,X] ), {v, b} is
not an interval of B(X ∪ {a, b}). But, {v, b} is an interval of B(X ∪ {b}).
Necessarily, {v, b} is not an interval of B({v, b, a}) and thus b 6∈ X 0 (v).
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By recalling that u ∈ X 0 (a) and b 6∈ X 0 (a), we obtain that | p[B,X 0 ] |≥ 3
in both cases. It results from Corollary 3 that there exist x0 6= y 0 ∈ B − X 0
such that B − {x0 , y 0 } is indecomposable. Obviously, u ∈ {x0 , y 0 } because we
assume that B − {x, y} is decomposable for any x 6= y ∈ B − X. Consequently,
{x0 , y 0 } = {u, x}, where x ∈ B − X.
An immediate consequence of Corollaries 3 and 4 follows.
Corollary 5 Let B be an indecomposable binary structure. If there is a subset
X of B such that | X |≥ 3, | B − X |≥ 4 and B(X)is indecomposable, then there
exist x 6= y ∈ B such that B − {x, y} is decomposable.
Proof . Assume that B − {x, y} is decomposable for any x =
6 y ∈ B − X. By
Corollary 3, | p[B,X] |= 2 and Ext(X) = ∅ by Corollary 2. Necessarily, p[B,X]
contains at least an element of type X(u), where u ∈ X.
It suffices to apply Proposition 1 and Corollary 5 to obtain Theorem 3 for
an indecomposable binary structure B such that | B |≥ 8.
References
[1] A. Ehrenfeucht, G. Rozenberg, Primitivity is hereditary for 2-structures,
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[2] P. Ille, Indecomposable graphs, Discrete Math. 173 (1997) 71–78.
[3] P. Ille, La décomposition intervallaire des structures binaires, La Gazette
des Mathématiciens 104 (2005), 39–58.
[4] J.H. Schmerl, W.T. Trotter, Critically indecomposable partially ordered
sets, graphs, tournaments and other binary relational structures, Discrete
Math. 113 (1993) 191–205.
[5] D.P. Sumner, Graphs indecomposable with respect to the X-join, Discrete
Math. 6 (1973) 281–298.
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