Discrete Mathematics 306 (2006) 2790–2797www.elsevier.com/locate/disc

Note

Elements belonging to triangles in 3-connected matroids

Manoel Lemos1

Departamento de Matemática, Universidade Federal de Pernambuco, Cidade Universitária, Recife, Pernambuco 50740-540, Brazil

Received 22 February 2005; received in revised form 8 May 2006; accepted 24 May 2006Available online 20 July 2006

Abstract

An element e of a 3-connected matroid M is said to be superfluous provided M/e is 3-connected. In this paper, we show that a3-connected matroid M with exactly k superfluous elements has at least

max

{5|E(M)| + 30 − 15k

9,

5r∗(M) + 30 − 10k

7

}

elements covered by triangles. For each k, an infinite family of matroids that attain this bound is constructed.© 2006 Elsevier B.V. All rights reserved.

Keywords: Matroid; 3-Connected matroid; Minimally 3-connected matroid; Triad; Triangle

1. Introduction

A minimally 3-connected matroid is a 3-connected matroid without a single-element deletion that is 3-connected.Oxley [7] proved that a minimally 3-connected matroid has many triads. (This is a natural extension to matroids ofthe graph results due to Halin [1] and Mader [6].) So the set of triads of a minimally 3-connected matroid must covera large number of its elements. Leo [5] conjectured a lower bound for this number. Leo’s Conjecture was proved byLemos [3], namely:

Theorem 1.1. Let N be a minimally 3-connected matroid with at least eight elements. Then the number of elementswhich meet a triad of M is at least

5|E(N)| + 30

9.

Leo [5] constructed an infinite family of matroids that attains this bound. The similar question for 2-connectedmatroids also has a solution: for a minimally 2-connected matroid M, Reid and Wu [9] gave a sharp lower bound onthe number of elements meeting some 2-element cocircuit in terms of the total number of elements of M.

1 The author was partially supported by CNPq (Grant no. 302195/02-5) and ProNEx/CNPq (Grant no. 664107/97-4).E-mail address: manoel@dmat.ufpe.br.

0012-365X/$ - see front matter © 2006 Elsevier B.V. All rights reserved.doi:10.1016/j.disc.2006.05.025

M. Lemos / Discrete Mathematics 306 (2006) 2790 –2797 2791

A 3-connected matroid M is said to be cominimally 3-connected provided M∗ is minimally 3-connected. If S(M) ={e ∈ E(M) : M/e is 3-connected}, then M is cominimally 3-connected if and only if S(M)=∅. In this paper we provethe following result:

Theorem 1.2. If M is a 3-connected matroid such that |E(M)|�8, then there are at least

5|E(M)| + 30 − 15|S(M)|9

elements of M meeting some triangle of M.

If N is a minimally 3-connected matroid, then M = N∗ is a cominimally 3-connected matroid and so |S(N∗)| = 0.Therefore, Theorem 1.1 is a consequence of Theorem 2.1. In the last section, for each fixed value of |S(M)|, weconstruct an infinite family of matroids that attain the bound on Theorem 2.1. Thus this result is sharp.

Lemos [3] also has proved the dual of the next result:

Theorem 1.3. Let M be a 3-connected matroid with at least five elements. Then the number of elements which meet atriangle of M is at least

|E(M)| + 10 − 3|S(M)|3

·

Note that Theorem 1.3 is almost a consequence of Theorem 1.2. In this paper, we also prove another bound for thenumber of elements belonging to some triangle of a 3-connected matroid. This bound depends on the corank of thematroid instead of its number of elements.

Theorem 1.4. If M is a 3-connected matroid such that |E(M)|�8 and r(M)�3, then there are at least

5r∗(M) + 30 − 10|S(M)|7

elements of M meeting some triangle of M.

This theorem is also sharp. We prove simultaneously Theorems 1.2 and 1.4 in the next section. An important toolused in the proof of these theorems is a decomposition for 3-connected matroids proposed in Section 3 of Lemos [3].This decomposition has also been used to give new proofs of the main results of [2–4].

2. Proof of the main result

For a 3-connected matroid M, T (M) denotes the set of elements of M covered by some triangle of M. A line L of Mis said to be large provided |L|�3. Now, we prove the next result that implies both Theorems 1.2 and 1.4.

Theorem 2.1. If M is a 3-connected matroid such that |E(M)|�8 and r(M)�3, then

|T (M)|� max

{5|E(M)| + 30 − 15|S(M)|

9,

5r∗(M) + 30 − 10|S(M)|7

}.

Proof. Suppose this result is not true and choose a counter-example M such that (r(M), |S(M)|, |T (M)|) is minimumin the lexicographic order. First, we show that

W = E(M) − [T (M) ∪ S(M)] �= ∅. (2.1)

2792 M. Lemos / Discrete Mathematics 306 (2006) 2790 –2797

Assume W = ∅. In particular, |E(M)| = |T (M)| + |S(M)|. As M is a counter-example to this theorem, it follows thatat least one of the inequalities holds:

9|T (M)| < 5|E(M)| + 30 − 15|S(M)|, (2.2)

7|T (M)| < 5r∗(M) + 30 − 10|S(M)|. (2.3)

If (2.2) occurs, then 4|T (M)| + 10|S(M)| < 30 and so 4|E(M)| < 30; a contradiction to the hypothesis. Hence (2.3)holds. Thus 7|E(M)| + 3|S(M)| − 5r∗(M) < 30 and so 2|E(M)| + 3|S(M)| + 5r(M) < 30. As |E(M)|�8, it followsthat r(M)�2; a contradiction to the hypothesis. Therefore (2.1) follows.

Now, we establish the following auxiliary lemma:

Lemma 2.1. For each e ∈ W , there is a 3-separating set Xe of M (i.e., {Xe, E(M) − Xe} is a 3-separation of M) suchthat:

(X1) e ∈ Le, where Le = clM(Xe) − Xe.(X2) |Le|�2.(X3) rM(Xe) = 3.(X4) Le = clM(E(M) − Xe) ∩ clM(Xe).(X5) There are 3-point lines L1e and L2e of M such that Xe ∪ [Le ∩ T (M)] = L1e ∪ L2e and L1e ∩ L2e = {xe}, for

some xe ∈ Xe ∪ [Le ∩ T (M)].

Proof. Let {X, Y } be a 2-separation of M/e such that min{rM/e(X), rM/e(Y )}�2. By Lemos [3, Lemma 3.3], thereare 3-connected matroids MX and MY such that:

(i) For Z ∈ {X, Y }, clM(Z) ⊆ E(MZ), MZ|clM(Z) = M|clM(Z) and Z is a spanning set of MZ .(ii) L = E(MX) ∩ E(MY ) is a 3-point line of both MX and MY such that clM(X) ∩ clM(Y ) ⊆ L and E(MX) ∪

E(MY ) = E(M) ∪ L.(iii) The only large line of MX or MY that contains an element of L − E(M) is L.

(Using the same terminology set in Section 3 of Lemos [3], for Z ∈ {X, Y }, let MZ be the factor of M with respect toZ having LZ as special line. Note that e ∈ LZ . As e does not belong to a triangle of M, it follows that |LZ ∩ E(M)|�2and so |LZ| = 3. Moreover, we can take LX = LY = L because LX ∩ E(M) = LY ∩ E(M).) Moreover,

{S(MX), S(MY )} is a partition of S(M). (2.4)

To prove the existence of Xe we consider two cases.Case 1: [L ∩ E(M)] ∩ T (M) �= ∅.If f ∈ [L ∩ E(M)] ∩ T (M), then L ∩ E(M) = {e, f }, since e does not belong to a large line of M. As {E(MX) −

L, E(MY ) − L, L ∩ E(M)} is a partition of E(M), it follows that

|E(M)| = |E(MX)| + |E(MY )| − 4. (2.5)

For Z ∈ {X, Y }, an element of Z − L belongs to a triangle of MZ if and only if it belongs to a triangle of M. As fbelongs to a triangle of M, it follows that

|T (M)| = |T (MX)| + |T (MY )| − 5. (2.6)

Observe that

r(M) = rM(X) + rM(Y ) − 2 = r(MX) + r(MY ) − 2�4 (2.7)

and so

r∗(M) = r∗(MX) + r∗(MY ) − 2. (2.8)

M. Lemos / Discrete Mathematics 306 (2006) 2790 –2797 2793

Assume the result holds for both MX and MY . Hence

|T (MZ)|� max

{5|E(MZ)| + 30 − 15|S(MZ)|

9,

5r∗(MZ) + 30 − 10|S(MZ)|7

},

for Z ∈ {X, Y }. Adding these inequalities, we obtain

|T (MX)| + |T (MY )|� 19 [5(|E(MX)| + |E(MY )|) + 60 − 15(|S(MX)| + |S(MY )|)],

|T (MX)| + |T (MY )|� 17 [5(r∗(MX) + r∗(MY )) + 60 − 10(|S(MX)| + |S(MY )|)].

By (2.4)–(2.6) and (2.8), we obtain

|T (M)| + 5� 19 [5(|E(M)| + 4) + 60 − 15|S(M)|],

|T (M)| + 5� 17 [5(r∗(M) + 2) + 60 − 10|S(M)|].

These inequalities can be reordered as

|T (M)|� max

{5|E(M)| + 35 − 15|S(M)|

9,

5r∗(M) + 35 − 10|S(M)|7

};

a contradiction. Hence the result does not hold for either MX or MY , say MY . By the choice of M, |E(MY )|�7 since3�rM(Y ) = r(MY ) < r(M).

As r(MY )�3 and MY is a 3-connected matroid having a large line, it follows that |E(MY )| ∈ {6, 7} and so|Y − L| ∈ {3, 4}. Now we show that

r(MY ) = 3. (2.9)

If r(MY ) > 3, then r(MY ) = 4 and Y − L is a 4-point line of M∗Y . Thus M∗

Y \g = [MY /g]∗ is 3-connected, for eachg ∈ Y − L, and so S(MY ) = Y − L; a contradiction since the result does not hold for MY . Therefore (2.9) follows.Hence E(MY ) = T (MY ) ∪ S(MY ) and so |E(MY )| = |T (MY )| + |S(MY )|. As MY is a counter-example to this resultand, by (2.9), r∗(MY ) = |E(MY )| − 3, it follows that

|T (MY )| < max

{5|E(MY )| + 30 − 15|S(MY )|

9,

5|E(MY )| + 15 − 10|S(MY )|7

}.

Therefore

0 < max

{30 − 4|E(MY )| − 6|S(MY )|

9,

15 − 2|E(MY )| − 3|S(MY )|7

}.

But |E(MY )|�6 and so

0 < max

{6(1 − |S(MY )|)

9,

3(1 − |S(MY )|)7

}.

Thus |S(MY )| = 0. Hence each element of MY belongs to some large line of MY . As MY is 3-connected, it follows thatMY has at least three large lines. Let L1 and L2 be large lines of MY different from L. Observe that L∩(L1 ∪L2) ⊆ {f }.Therefore f ∈ L1 ∪ L2, |L1 ∩ L2| = 1 and |(L1 ∪ L2) − L| = 4. (Remember that |E(MY )|�7.) So (X3) and (X5)hold, when we take Xe = Y − L. Note that Le = L ∩ E(M) = {e, f } and so (X1), (X2) and (X4) also follow.

Case 2: [L ∩ E(M)] ∩ T (M) = ∅.For Z ∈ {X, Y }, let HZ be a matroid (see its geometric representation in the next figure) such that:

(i) L = E(HX) ∩ E(HY );(ii) E(HX) − L, E(HY ) − L and E(M) are pairwise disjoint;

(iii) |E(HX)| = |E(HY )| = 8;(iv) r(HX) = r(HY ) = 3; and

2794 M. Lemos / Discrete Mathematics 306 (2006) 2790 –2797

(v) for Z ∈ {X, Y }, HZ has only three large lines L, L1Z and L2Z each having three elements and L1Z ∪ L2Z =E(HZ) − L.

L1Z L2Z

L c

As MZ is 3-connected, it follows that KZ , the generalized parallel connection across L of MZ and HZ , is 3-connected.Choose c ∈ L − E(M). (It exists because e ∈ L and e does not belong to a large line of M.) Now, we prove thatNZ =KZ\c is 3-connected. If U is a 2-separating set for NZ , then |U | > 2, otherwise T ∗ =U ∪c is a triad of KZ whichis contrary to orthogonality (|T ∗ ∩L|�2 and T ∗ ∩LiZ �= ∅, for some i ∈ {1, 2}, and so |T ∗ ∩LiZ|�2). Hence U − d

is a 2-separating set for KZ\c/d , for d ∈ L1Z ∩ L2Z . But the only 2-separating sets of KZ\c/d are L1Z − d, L2Z − d

and their complements. We have a contradiction because either L1Z or L2Z is not a 2-separating set of NZ . Hence NZ

is 3-connected. By Lemma 3.3 of Lemos [3], we have that S(NZ) = S(MZ), for both Z ∈ {X, Y }, and so, by (2.4),

{S(NX), S(NY )} is a partition of S(M). (2.10)

Observe that for Z ∈ {X, Y }r(NZ) = r(MZ) + 1. (2.11)

As {E(NX) − E(HX), E(NY ) − E(HY ), L ∩ E(M)} is a partition of E(M), it follows that

|E(M)| = |E(NX)| + |E(NY )| + |L ∩ E(M)| − 14. (2.12)

By (2.7), (2.11) and (2.12), we have that

r∗(M) = r∗(NX) + r∗(NY ) + |L ∩ E(M)| − 10. (2.13)

For Z ∈ {X, Y }, an element of clM(Z) = Z ∪ [L ∩ E(M)] belongs to a triangle of NZ if and only if it belongs to atriangle of M. Hence

|T (M)| = |T (NX)| + |T (NY )| − 10. (2.14)

Assume the result holds for both NX and NY . Thus

|T (NZ)|� max

{5|E(NZ)| + 30 − 15|S(NZ)|

9,

5r∗(NZ) + 30 − 10|S(NZ)|7

},

for Z ∈ {X, Y }. Adding these inequalities, we obtain

|T (NX)| + |T (NY )|� 19 [5(|E(NX)| + |E(NY )|) + 60 − 15(|S(NX)| + |S(NY )|)],

|T (NX)| + |T (NY )|� 17 [5(r∗(NX) + r∗(NY )) + 60 − 10(|S(NX)| + |S(NY )|)].

By (2.10), (2.12), (2.13) and (2.14), we obtain

|T (M)| + 10� 19 [5(|E(M)| + 14 − |L ∩ E(M)|) + 60 − 15|S(M)|],

|T (M)| + 10� 17 [5(r∗(M) + 10 − |L ∩ E(M)|) + 60 − 10|S(M)|].

M. Lemos / Discrete Mathematics 306 (2006) 2790 –2797 2795

These inequalities can be reordered as

|T (M)|� 19 [5|E(M)| + 30 − 15|S(M)| + 10 − 5|L ∩ E(M)|],

|T (M)|� 17 [5r∗(M) + 30 − 10|S(M)| + 10 − 5|L ∩ E(M)|];

a contradiction because |L ∩ E(M)|�2. Hence the result does not hold for either NX or NY , say NX.By the choice of M, r(NX)�r(M). By (2.7) and (2.11), r(NX) = r(M) + 3 − r(MY ). As r(MY ) = r(Y )�3, it

follows that r(MY ) = 3 and so r(M) = r(NX). Again, by the choice of M and (2.10), we conclude that S(NX) =S(M) and S(NY ) = ∅. Hence S(MY ) = ∅ and MY is cominimally 3-connected. In particular, T (MY ) = E(MY ). As(r(M), |S(M)|) = (r(NX), |S(NX)|), it follows by the choice of M that |T (M)|� |T (NX)|. Thus |Y − L|�5 becauseY − L ⊆ T (M). (Remember that MY is a rank-3 cominimally 3-connected matroid.) As Y − L is the union of largelines of MY and none of these lines has a common point with L, it follows that Y −L contains just two large lines, eachwith exactly three points. Moreover, the intersection of those two large lines is a point. We can take Xe to be equal toY − L and, in this case, (X1) to (X5) are also satisfied. �

Now we continue with the proof of Theorem 2.1 by proving that

Xe = Xf or Xe ∩ Xf = ∅ when {e, f } is a 2-subset of W . (2.15)

Moreover, when the first case happens Le = Lf = {e, f }. Suppose that (2.15) does not hold for some e and f. Inparticular,

Xe ∩ Xf �= ∅. (2.16)

As E(M) − Xe does not span any element of Xe, by (X1) and (X4), and Xf ∪ [Lf ∩ T (M)] = L1f ∪ L2f , by (X5), itfollows that

|Lif − Xe| �= 2 for i ∈ {1, 2}. (2.17)

Next, we prove that

(Xf ∪ Lf ) ∩ (Xe ∪ Le) does not contain a basis of Xf ∪ Lf . (2.18)

If (Xf ∪ Lf ) ∩ (Xe ∪ Le) contains a basis of Xf ∪ Lf , then Xe ∪ Le spans Xf ∪ Lf . Hence Xf ∪ Lf ⊆ Xe ∪ Le. By(X1) and (X3), Xe ∪ Le = Xf ∪ Lf because both are rank-3 closed sets of M. Hence Xe = Xf and Le = Lf = {e, f },by (X1), (X2) and (X5); a contradiction. Thus, (Xf ∪ Lf ) ∩ (Xe ∪ Le) does not contain a basis of Xf ∪ Lf and(2.18) holds. Now, we show that xf /∈ Xe ∩ Xf . If xf ∈ Xe ∩ Xf , then, by (2.17), (Lif − xf ) ∩ Xe �= ∅, for eachi ∈ {1, 2}. Thus Xe contains a basis for Xf ∪ Lf ; a contradiction to (2.18). Hence xf /∈ Xe ∩ Xf . Therefore, by (X5)and (2.16), there is i ∈ {1, 2} such that (Lif − xf ) ∩ (Xe ∩ Xf ) �= ∅, say i = 1. By (2.17), L1f − xf ⊆ Xe and soxf ∈ Le. In particular, Le = {e, xf } and L1f ⊆ Xe ∪ Le. Hence L1f = Lje, for some j ∈ {1, 2}, say j = 1. Similarly,Lf ={f, xe} and xe ∈ L1f =L1e. Moreover, (Xe ∪xf )∩ (Xf ∪xe)=L1e =L1f . We set X′

f =Xf , when xe =xf , andX′

f = (Xf −xf )∪xe, when xe �= xf . As Xe and X′f are 3-separating sets of M such that Xe ∩X′

f =L1e −xf , it followsby submodularity that Xe ∪ X′

f is a 3-separating set of M. Observe that e, f and xf are elements of M not belongingto Xe ∪ X′

f . As {e, f, xf } ⊆ clM(Xe ∪ X′f ), it follows that {e, f, xf } is contained in a line of M; a contradiction. Thus

(2.15) holds.By (2.15), there are elements e1, . . . , en belonging to W such that Le1 − T (M), Le2 − T (M), . . . , Len − T (M)

partition W. Hence

|W | = n + r , (2.19)

where r is the number of indices i ∈ {1, . . . , n} such that |Lei− T (M)| = 2. As Xe1 , . . . , Xen are pairwise disjoint, by

(2.15), it follows that

|T (M)|� |Xe1 | + |Xe2 | + · · · + |Xen | = 4n + r

2796 M. Lemos / Discrete Mathematics 306 (2006) 2790 –2797

and so

|T (M)| = 4n + r + �, (2.20)

for some non-negative integer �. (Note that the elements belonging to T (M) ∩ [Le1 ∪ Le2 ∪ · · · ∪ Len ] are not countedin 4n + r—they must be counted in �.) As

|T (M)| < max

{5|E(M)| + 30 − 15|S(M)|

9,

5r∗(M) + 30 − 10|S(M)|7

},

it follows that

11n − r + 4� + 10|S(M)| < 30 or 3n − 3r + 2� + 5|S(M)| + 5r(M) < 30. (2.21)

Set Z = E(M) − [Xe1 ∪ Xe2 ∪ · · · ∪ Xen ]. Let B be a basis of M|Z. For i ∈ {1, 2, . . . , n}, choose fi ∈ Xei. Now,

we show that

B ∪ {f1, f2, . . . , fn} is an independent set of M . (2.22)

(It is possible to prove that B ∪ {f1, f2, . . . , fn} is a basis of M, but we do not need this information.) IfB ∪ {f1, f2, . . . , fn} is dependent in M, then there is a circuit C of M such that C ⊆ B ∪ {f1, f2, . . . , fn}. As Bis independent in M, it follows that C ∩ {f1, . . . , fn} �= ∅, say f1 ∈ C. Hence C − f1 ⊆ E(M) − Xe1 spans f1; acontradiction to (X4). So C does not exist and (2.22) holds. By (2.22),

r(Z) + n�r(M). (2.23)

If n�2, then r(Z)�3 and so r(M)�5. (If r(Z) = 2, then Z = {e1, e2} = Le1 = Le2 ; a contradiction.)From (2.21), we have two distinct cases to consider:Case 1: 3n − 3r + 2� + 5|S(M)| + 5r(M) < 30.Hence |S(M)| + r(M)�5 since r �n. By (2.7), r(M)�4 and so r(M) ∈ {4, 5}. Now, we show that

r(M) = 5. (2.24)

If r(M) = 4, then n = 1, by the comment made after inequality (2.23). Moreover, if {X, Y } is a 2-separation of M/e1such that min{rM/e1(X), rM/e1(Y )}�2 and Xe1 ⊆ X, then Y − Le1 ⊆ T (M) ∪ S(M) because n = 1. As |S(M)|�1,it follows that clM(Y ) contains a large line of M and so ��3. (Remember that, when we calculate the bound (2.20) for|T (M)|, we have not counted, when it exists, the element belonging to Le1 ∩ T (M) which may belong to this largeline.) Hence |S(M)| = 0 and so ��5; a contradiction. Thus (2.24) follows.

By (2.24), |S(M)| = 0 and so 3n − 3r + 2� < 5. In particular, M is cominimally 3-connected. By (2.23) and (2.24),n�2. If n = 1, then, by the dual of Theorem 1 of [2], M has at least two triangles not contained in Xe1 ∪ Le1 and so��4; a contradiction. Therefore, n = 2 and so

2� − 3r � − 1. (2.25)

In particular, r ∈ {1, 2}. Observe that E(M) �= clM(Xe1) ∪ clM(Xe2), otherwise, by (X3) and (2.24), {clM(Xe1),

clM(Xe2) − clM(Xe1)} is a 2-separation of M. As M is cominimally 3-connected and W ⊆ Le1 ∪ Le2 , it follows thatthere is a large line L of M such that L is not contained in clM(Xei

), for both i ∈ {1, 2}. Hence ��3. By (2.25), weobtain that 3r �7; a contradiction and the result follows in this case.

Case 2: 11n − r + 4� + 10|S(M)| < 30.Hence |S(M)| + n�2 because r �n. In particular, |S(M)|�1. To conclude the proof, we divide it in two subcases.

If n = 1, then

� + |S(M)| = |E(M)| − |Xe1 ∪ [Le1 − T (M)]|� |E(M)| − 7.

But |E(M)| − |clM(Xe1)|�3 and so E(M)−[Xe1 ∪ e1] contains a large line L of M because |S(M)|�1. In particular,��3. Hence |S(M)| = 0 and M is cominimally 3-connected. In this case, E(M) − [Xe1 ∪ e1] contains another largeline of M and so ��5; a contradiction. Hence n = 2 and so |S(M)| = 0. In particular, E(M) = T (M) ∪ W and soE(M) − (Le1 ∪ Le2) ⊆ T (M). If c ∈ E(M) − [Xe1 ∪ Xe2 ∪ Le1 ∪ Le2 ], then there is a large line L of M such that

M. Lemos / Discrete Mathematics 306 (2006) 2790 –2797 2797

c ∈ L. By (X1) and (X4), L ∩ Xei= ∅ or |L − Xei

|�1, for i ∈ {1, 2}. If L ∩ Xe1 = L ∩ Xe2 = ∅, then ��3; acontradiction. Thus |L − Xei

| = 2 and c ∈ Lei, for some i ∈ {1, 2}; a contradiction. Thus c does not exist. Hence

E(M) = Xe1 ∪ Xe2 ∪ Le1 ∪ Le2 ; a contradiction because {Xe1 ∪ Le1 , [Xe2 ∪ Le2 ] − Le1} is a 2-separation of M, by(X3) and the comment made after (2.23). The result also follows in this case. �

3. An extremal example

Let k and t be integers such that k�0 and t �2. In this section, we construct a 3-connected matroid Mk,t such that

|T (Mk,t )| = 5t ,

|S(Mk,t )| = k,

|E(Mk,t )| = 9t + 3k − 6,

r∗(Mk,t ) = 7t + 2k − 6.

Observe that

|T (Mk,t )| = 5|E(Mk,t )| + 30 − 15|S(Mk,t )|9

= 5r∗(Mk,t ) + 30 − 10|S(Mk,t )|7

.

Therefore, Mk,t attains both lower bounds given by Theorem 2.1, for all integers k and t such that k�0 and t �2. Whenk = 0, we obtain the matroids constructed by Leo [5].

Let A1, A2, . . . , At+k−1, B1, B2, . . . , Bt−2, C1, C2, . . . , Ck, D1, D2, . . . , Dt be pairwise disjoint sets such that|Ai | = 2, for every i ∈ {1, . . . , t + k − 1}, |Bi | = 2, for every i ∈ {1, . . . , t − 2}, |Ci | = 1, for every i ∈ {1, . . . , k}and |Di | = 5, for every i ∈ {1, . . . , t}. For i ∈ {1, . . . , t}, let Li and L′

i be 3-subsets of Di such that Di = Li ∪ L′i .

Consider the following matroids:

(i) Let M0 be a rank-3 simple matroid over A1 ∪ D1 having just two large lines, namely L1 and L′1.

(ii) For i ∈ {1, ..., t − 2}, let Mi be a matroid over Ai ∪ Bi ∪ Ai+1 isomorphic to U3,6.(iii) For i ∈ {1, ..., t − 2}, let Ni be a rank-3 simple matroid over Bi ∪ Di+1 having just two large lines, namely Li+1

and L′i+1.

(iv) For i ∈ {t − 1, ..., t + k − 2}, let Mi be a matroid over Ai ∪ Ci−t+2 ∪ Ai+1 isomorphic to U3,5.(v) Let Mt+k−1 be a rank-3 simple matroid over At+k−1 ∪ Dt having just two large lines, namely Lt and L′

t .

Let Mk,l be the generalized parallel connection of M0, M1, ..., Mt+k−1, N1, ..., Nt−2. Note that L1, L′1, L2, L

′2, ...,

Lt , L′t are the large lines of Mk,t and S(Mk,t ) = C1 ∪ C2 ∪ · · · ∪ Ck .

References

[1] R. Halin, Untersuchugen über minimale n-fach zusammenhängende graphen, Math. Ann. 182 (1969) 175–188.[2] M. Lemos, On 3-connected matroids, Discrete Math. 73 (1989) 273–283.[3] M. Lemos, Elements belonging to triads in 3-connected matroids, Discrete Math. 285 (2004) 167–181.[4] J.W. Leo, Triads and triangles in 3-connected matroids, Discrete Math. 194 (1999) 173–193.[5] J.W. Leo, private communication.[6] W. Mader, Ecken vom Grad n in minimalen n-fach zusammenhängenden Graphen, Arch. Math. (Basel) 23 (1972) 219–224.[7] J.G. Oxley, On matroid connectivity, Quart. J. Math. Oxford Ser. 32 (2) (1981) 193–208.[9] T.J. Reid, H. Wu, On elements in small cocircuits in minimally k-connected graphs and matroids, Discrete Math. 243 (2002) 273–282.

Further reading

[8] J.G. Oxley, Matroid theory, Oxford University Press, New York, 1982.