Infinite Gammoids

Finite strict gammoids, introduced in the early 1970's, are matroids defined via finite digraphs equipped with some set of sinks: a set of vertices is independent if it admits a linkage to these sinks. In particular, an independent set is maximal (i.e. a base) precisely if it is linkable onto the sinks. In the infinite setting, this characterization of the maximal independent sets need not hold. We identify a type of substructure as the unique obstruction. This allows us to prove that the sets linkable onto the sinks form the bases of a (possibly non-finitary) matroid if and only if this substructure does not occur. Infinite matroid theory has seen vigorous development (e.g. [1], [5] and [7]) since Bruhn et al. [6] in 2010 gave five equivalent sets of axioms for infinite matroids in response to a problem proposed by Rado [19] (see also Higgs [13] and Oxley [15]). Our contribution to the development focusses on the class of gammoids. In this first paper, the main object of investigation is the bases of infinite strict gammoids. (The second one considers other aspects including duality and minors [2].) The concept of gammoids originated from the transversal matroids of Edmonds and Fulkerson [11]. A transversal matroid can be defined by taking as its independent sets the subsets of a fixed vertex class of a bipartite graph matchable to the other vertex class. Perfect [17] introduced the class of gammoids by replacing matchings in bipartite graphs with disjoint directed paths in digraphs. Later, Mason [14] started the study of a subclass of gammoids known as strict gammoids.


Introduction
Infinite matroid theory has seen vigorous development (e.g.[1], [5] and [7]) since Bruhn et al. [6] in 2010 gave five equivalent sets of axioms for infinite matroids in response to a problem proposed by Rado [19] (see also Higgs [13] and Oxley [15]).Our contribution to the development focusses on the class of gammoids.In this first paper, the main object of investigation is the bases of infinite strict gammoids.(The second one considers other aspects including duality and minors [2].) The concept of gammoids originated from the transversal matroids of Edmonds and Fulkerson [11].A transversal matroid can be defined by taking as its independent sets the subsets of a fixed vertex class of a bipartite graph matchable to the other vertex class.Perfect [17] introduced the class of gammoids by replacing matchings in bipartite graphs with disjoint directed paths in digraphs.Later, Mason [14] started the study of a subclass of gammoids known as strict gammoids.
To be precise, let a dimaze (short for di rected maze) be a digraph with a fixed subset of the vertices of out-degree 0, called exits.A dimaze contains another dimaze, if, in addition to digraph containment, the exits of the former include those of the latter.In the context of digraphs, any path or ray (i.e.infinite path) is forward oriented.A set of vertices of (the digraph of) the dimaze is independent if it is linkable to the exits by a collection of disjoint directed paths.The set of all linkable sets is the linkability system of the dimaze.When the linkability system is the set of independent sets of a matroid on the vertex set of the dimaze, we say that the dimaze defines a matroid.Any matroid arising in this way is called a strict gammoid.
Mason [14] proved that every finite dimaze defines a strict gammoid.When a dimaze is infinite, Perfect [17] gave sufficient conditions for when some subset of the linkability system gives rise to a matroid.Any such matroid is finitary, in the sense that a set is independent as soon as all its finite subsets are.Since finitary matroids were the only ones known at that time, infinite dimazes whose linkability systems are non-finitary 1 were not considered to define matroids.
With infinite matroids canonically axiomatized in a way that allows for non-finitary ones, a natural question is whether every infinite dimaze now defines a matroid.In general, the answer is still negative, as the linkability system may fail to satisfy one of the infinite matroid axioms (IM), which asks for the existence of certain maximal independent sets.Observe that in any finite dimaze, a set is linkable onto the exits if and only if it is maximally independent.However, in an infinite dimaze, sets which are linkable onto the exits need not be maximally independent.It turns out that if they all are, we have a matroid.
So the question arises: In which dimazes is every set that is linkable onto the exits maximally independent?Investigation of this question leads us to the following example of a dimaze.An alternating ray is a digraph obtained from an undirected ray, i.e. a 1-way infinite path, by directing the edges in such a way that the first vertex and infinitely many others have out-degree 0. An alternating comb is a dimaze constructed by linking all the vertices of out-degree 0 of an alternating ray onto a set of exits by (possibly trivial) disjoint directed paths which meet the ray exactly at their initial vertices (see for example Figure 3a).It will be easy to see that, in an alternating comb, the set of vertices of outdegree 2 is linkable either onto the exits or to a proper subset thereof; hence, the set is not maximally independent and yet linkable onto the exits.
By proving that alternating combs form the unique obstruction to the characterization of maximal independent sets as sets linkable onto the exits, we are able to establish the following.
Theorem.Given a dimaze, the vertex sets linkable onto the exits form the bases of a matroid if and only if the dimaze contains no alternating comb.The independent sets of this matroid are precisely the linkable sets of vertices.
The non-trivial direction implies that a dimaze that fails to define a matroid contains an alternating comb.While a dimaze containing an alternating comb may still define a matroid, the set of bases is a proper subset of the sets linkable onto the exits and can be difficult to describe.
We collect definitions and give examples of infinite dimazes which do not define a matroid in Section 1.In Section 2, we prove that the independence augmentation axiom as well as a variant (Lemma 2.8) hold in general.(These are applied in [8] to characterize nearly finitary linkability systems.)After rephrasing a proof of the linkage theorem of Pym [18], we prove the main result (Theorem 2.5).In Section 3, we consider the question whether every strict gammoid can be defined by a dimaze without any alternating comb.We answer this negatively via the intermediate step of showing that any tree, when viewed as a bipartite graph, defines a transversal matroid, a statement which is of independent interest.

Preliminaries
In this section, we present relevant definitions.For notions not found here, we refer to [6] and [16] for matroid theory, and [9] for graph theory.
Given a set E and a family of subsets I ⊆ 2 E , let I max denote the maximal elements of I with respect to set inclusion.For a set I ⊆ E and x ∈ E, we also write I + x and I − x for I ∪ {x} and I \ {x} respectively.The set E is the ground set and the elements in I are the independent sets of M .Equivalently, matroids can be defined with base axioms.A collection of subsets B of E is the set of bases of a matroid if and only if the following three axioms hold: (BM) The set I of all subsets of elements in B satisfies (IM).
the electronic journal of combinatorics 22(1) (2015), #P1.53 The ground sets we consider will be sets of vertices of digraphs or bipartite graphs.We usually identify a matroid with its set of independent sets.
Connectivity in finite matroids stems from graph connectivity and is a well established part of the theory.In the infinite setting, Bruhn and Wollan [7] gave the following rankfree definition of connectivity that is compatible with the finite one.For an integer k 0, a k-separation of a matroid is a partition of E into X and Y such that both |X|, |Y | k and for any bases B X , B Y of M \ Y and M \ X respectively, the number of elements to be deleted from B X ∪ B Y to get a base of M is less than k.A matroid is k-connected if there are no l-separations for any l < k.We will show that there are highly connected strict gammoids that are in a sense far from being finitary.
Next, suppose we are given a digraph D, and a subset B 0 of the vertices of outdegree 0. 2 The pair (D, B 0 ) is called a dimaze and B 0 the set of exits.A dimaze (D, B 0 ) contains a dimaze (D , B 0 ), if D is a subdigraph of D and B 0 ⊆ B 0 .A linkage P is a set of disjoint directed paths such that the terminal vertex of each path is in B 0 .Let Ini(P) and Ter(P) be respectively the set of initial vertices and that of terminal vertices of paths in P. A set I ⊆ V (D) is linkable or independent if there is a linkage P from I, i.e.Ini(P) = I.Suppose further that Ter(P) = B 0 , then I is linkable onto B 0 .The collection of linkable sets is called the linkability system.Note that, by adding trivial paths if required: Any linkable set in (D, B 0 ) can be extended to one linkable onto B 0 . ( Definition 1.2.Let (D, B 0 ) be a dimaze.The pair of V (D) and the set of linkable subsets is denoted by M L (D, B 0 ).A strict gammoid is a matroid isomorphic to M L (D, B 0 ) for some (D, B 0 ).A gammoid is a matroid restriction of a strict gammoid.Given a gammoid M , (D, B 0 ) is called a presentation of M if M = M L (D, B 0 ) X for some X ⊆ V (D).We say that the dimaze (D, B 0 ) defines a matroid if M L (D, B 0 ) is a matroid. 3iven vertex subsets A and B of a digraph D, an A-B separator S is a set of vertices such that there are no paths from A to B avoiding S. A separator is on a linkage P if it consists of exactly one vertex on each path in P. The celebrated Aharoni-Berger-Menger theorem [3] states that there exist a linkage from a subset of A to B and an A-B separator on this linkage.
Mason [14] (see also [17]) showed that M L (D, B 0 ) is a matroid for any finite dimaze (D, B 0 ).However, this is not the case for infinite dimazes.For example, let D be a complete bipartite graph between an uncountable set X and a countably infinite set B 0 with all the edges directed towards B 0 .Then I ⊆ X is independent if and only if I is countable.So there is no maximal independent set in X, hence M L (D, B 0 ) does not satisfy the axiom (IM).
Example 1.3.Here is a counterexample whose digraph is locally finite.Let D be the digraph obtained by directing upwards or leftwards the edges of the subgraph of the grid Z × Z induced by {(x, y) : y > 0 and y x 0} and let B 0 := {(0, y) : y > 0}, see Figure 1.Then I := {(x, x) : x > 0} is linkable onto a set J ⊆ B 0 if and only if J is infinite.Therefore, I ∪ (B 0 \ J) is independent if and only if J is infinite.Hence, I does not extend to a maximal independent set in X := I ∪ B 0 .The following dimazes play an important role in our investigation.An undirected ray is a graph with an infinite vertex set {x i : i 1} and the edge set {x i x i+1 : i 1}.We orient the edges of an undirected ray in different ways to construct three dimazes: 1. R A by orienting (x i+1 , x i ) and (x i+1 , x i+2 ) for each odd i 1 and the set of exits is empty; 2. R I by orienting (x i+1 , x i ) for each i 1 and x 1 is the only exit; 3. R O by orienting (x i , x i+1 ) for each i 1 and the set of exits is empty.
A subdivision of R A , R I and R O is called alternating ray, incoming ray and (outgoing) ray, respectively.
Let Y = {y i : i 1} be a set disjoint from X.We extend the above types of rays to combs by adding edges (and their terminal vertices) to the corresponding digraphs and declaring the resulting sinks to be the exits: 1. C A by adding no edges to R A ; 2. C I by adding the edges (x i , y i ) to R I for each i 2; the electronic journal of combinatorics 22(1) (2015), #P1.53

C O by adding the edges (x
Any subdivision of C A , C I and C O is called alternating comb, incoming comb and outgoing comb, respectively.The subdivided ray in any comb is called the spine and the paths to the exits are the spikes. A dimaze (D, B 0 ) is called H-free for a set H of dimazes if it does not have a subdimaze isomorphic to a subdivision of an element in H.A (strict) gammoid is called H-free if it admits an H-free presentation.In general, an H-free gammoid may admit a presentation that is not H-free (see Figure 3 for H = {C A }).
For notations about paths, we follow [9, p. 7]; and for a path P , we also write P for V (P ).

Dimazes and matroid axioms
The aim of this section is to give a sufficient condition for a dimaze (D, B 0 ) to define a matroid.As (I1) and (I2) hold for M L (D, B 0 ), we need only consider (I3) and (IM).

Linkability system and proof of (I3)
We prove that (I3) holds in any M L (D, B 0 ) using a classical result due to Grünwald [12], which can be formulated as follows (see also [9,Lemmas 3.3.2 and 3.3.3]).Let (D, B 0 ) be a dimaze and P a linkage.A (finite) P-alternating walk is a sequence W = w 0 e 0 w 1 e 1 . . .e n−1 v n of vertices w i and distinct edges e i of D, such that every e i ∈ W is incident with w i and w i+1 , and the following properties hold for each 0 i < n: (W1) e i = (w i+1 , w i ) if and only if e i ∈ E(P); Lemma 2.1.Let (D, B 0 ) be a dimaze, P a linkage, and Ini(P) ⊆ X ⊆ V .
(i) If there is a P-alternating walk from X \ Ini(P) to B 0 \ Ter(P), then there is a linkage P with Ini(P) Ini(P ) ⊆ X and Ter(P) Ter(P ) ⊆ B 0 .
(ii) If there is not any P-alternating walk from X \ Ini(P) to B 0 \ Ter(P), then there is an X-B 0 separator on P.
Then we have a linkage Q from B and another P from I. We may assume P misses some v 0 ∈ B 0 .
If there is a P-alternating walk from (B ∪ I) \ Ini(P) to B 0 \ Ter(P), then we can extend I in B \ I by Lemma 2.1(i).
On the other hand, if no such walk exists, we draw a contradiction to the maximality of B. In this case, by Lemma 2.1(ii), there is a (B ∪ I)-B 0 separator S on P. For every v ∈ B, let Q v be the path in Q starting from v. Let s v be the first vertex of S that Q v meets and P v the path in P containing s v .Let us prove that Suppose v and v are distinct vertices in B such that Q v s v P v and Q v s v P v meet each other.As P and Q are linkages, without loss of generality, we may assume Q v s v meets s v P v at some s / ∈ S. Then Q v sP v is a path from B to B 0 avoiding the separator.This contradiction shows that Q is indeed a linkage from B to B 0 .As Q does not cover v 0 , B + v 0 is independent which contradicts the maximality of B.

Linkage theorem and (IM)
For any dimaze (D, B 0 ), now that we know M L (D, B 0 ) satisfies (I3), it remains to investigate (IM).If D is finite, then the following holds: A set is maximally independent if and only if it is linkable onto the exits.
( †) When D is infinite, ( †) need not hold; for instance, the dimaze in Example 1.3, which does not even define a matroid.Using the Aharoni-Berger-Menger theorem [3] and the linkage theorem [18] (see also [10]), we prove that when ( †) holds, M L (D, B 0 ) is a matroid.Now the natural question is: in which dimazes is every set, that is linkable onto the exits, a maximal independent set?Consider the alternating comb given in Figure 3a.Using the notation there, the set X := {x i : i 1} can be linked onto B 0 by the linkage {(x i , y i−1 ) : i 1} or to B 0 − x 0 by the linkage {(x i , y i ) : i 1}.Hence, X is a nonmaximal independent set that is linkable onto B 0 .More generally, if a dimaze (D, B 0 ) contains an alternating comb C, then the vertices of out-degree 2 on C together with B 0 − C is a non-maximal set linkable onto B 0 .So an answer to the above question must exclude dimazes containing an alternating comb.We will prove that dimazes without any alternating comb are precisely the answer.
One might think that the following proof strategy should work: If the characterization of maximal independent sets does not hold, then there are two linkages, a blue one from a set and a red one from a proper superset, both covering the exits.To construct an alternating comb, one starts with finding an alternating ray.For that, a first attempt is to "alternate" between the red and blue linkages, i.e. to repeat the following: go forward along the red linkage, change to the blue one at some common vertex, and then go backwards on the blue linkage, and change again to the red one.It is not the case that this construction always gives rise to an alternating ray (because vertices might be visited twice).But supposing that we do get an alternating ray, a natural way to extend it to an alternating comb is to use the terminal segments of one fixed linkage.However, this alternating ray can have two distinct vertices of in-degree 2 which lie on the same path of the fixed linkage.
Appropriate choices to alternate between the linkages will be provided by the proof of the linkage theorem of Pym [18].So we outline the proof, rephrased for our purpose.
Linkage Theorem.Let D be a digraph and two linkages be given: the "red" one, P = {P x : x ∈ X P }, from X P onto Y P and the "blue" one, Proof outline.We construct a sequence of linkages converging to a linkage with the desired properties.For each integer i 0, we will specify a vertex on each path in P. For each x ∈ X P , let f 0 x := x.Let Q 0 := Q.For each i > 0 and each x ∈ X P , let f i x be the last vertex v on f i−1 x P x such that (f i−1 For y ∈ Y Q , let t i y be the first vertex v ∈ Q y such that the terminal segment vQ y does not contain any f i x .Let and It can be shown that Q i is a linkage.Moreover, for any x ∈ X P , {f i x } i 0 eventually settles at a vertex f ∞ x as i → ∞; similarly for any y ∈ Y Q , {t i y } i 1 settles at some t ∞ y .Then Q ∞ , defined as the union of the following three sets, is a linkage satisfying the requirements.
We can now prove the following.
Lemma 2.3.Given a dimaze (D, B 0 ), suppose that every independent set linkable onto the exits is maximal, then the dimaze defines a matroid.
Proof.Since (I1) and (I2) are obviously true for M L (D, B 0 ), and that (I3) holds by Proposition 2.2, to prove the theorem, it remains to check that (IM) holds.Let I be independent and a set X ⊆ V such that I ⊆ X be given.Suppose there is a "red" linkage from I to B 0 .Apply the Aharoni-Berger-Menger theorem on X and B 0 to get a "blue" linkage Q from B ⊆ X to B 0 and an X-B 0 separator S on the blue linkage.Consider the subdigraph of D induced by those vertices separated from B 0 by S. Let H be obtained from this digraph by deleting the edges with initial vertex in S. Since every linkage from H to B 0 goes through S, a subset of V (H) is linkable in (D, B 0 ) if and only if it is linkable in (H, S).Use the linkage theorem to find a linkage Let Y ⊇ X ∞ be any independent set in M L (H, S).By applying the linkage theorem on a linkage from Y to S and Q ∞ in (H, S), we may assume that Y is linkable onto S by a linkage Q .Concatenating Q with segments of paths in Q starting from S and adding trivial paths from the electronic journal of combinatorics 22(1) (2015), #P1.53 Applying the above statement on X ∞ shows that X ∞ ∪ (B 0 \ V (Q)) is also maximal in M L (D, B 0 ).It follows that Y cannot be a proper superset of X ∞ .Hence, X ∞ is maximal in M L (H, S), and so also in M L (D, B 0 ) ∩ 2 X .This completes the proof that M L (D, B 0 ) is a matroid.
Next we show that containing an alternating comb is the only reason that the characterization ( †) fails.
Proof.The forward direction follows trivially from (1).
For the backward direction, let I be a non-maximal subset that is linkable onto B 0 , by a "blue" linkage Q.Since I is not maximal, there is x 0 / ∈ I such that I + x 0 is linkable to B 0 as well, by a "red" linkage P. Construct an alternating comb inductively as follows: Running the proof of the linkage theorem on P and Q, we get a linkage Q ∞ from I + x 0 onto B 0 consisting of only B ∞ , as Y P ⊆ Y Q and X Q ⊆ X P .So each path in Q ∞ consists of a red initial and a blue terminal segment.
Start the construction with x 0 .For k 1, if x k−1 is defined, let Q k be the blue path containing p k−1 := f ∞ x k−1 .We will prove that p k−1 / ∈ I so that we can define q k to be the last vertex on Q k pk−1 that is on a path in Q ∞ .Since the blue segments of Q ∞ are disjoint, q k lies on a red path P x k .We continue the construction with x k .

Claim 1. p k−1 /
∈ I and hence, the blue segment q k Q k p k−1 is non-trivial.The red segment q k P x k p k is also non-trivial.
Proof.We prove the claim by induction.Clearly, p 0 / ∈ I, so the claim holds for k = 1.For k 2, assume that p k−2 / ∈ I.We argue that q k−1 = p k−1 .Suppose they are equal for a contradiction.Then the path P Hence, we have p k−1 = q k−1 .This shows that the red segment q k−1 P x k−1 p k−1 is non-trivial, and so p k−1 / ∈ I.
We now show that p 1 Q 1 ∪ ∞ k=2 q j Q j p j−1 ∪ q k P x k p k Q k+1 is an alternating comb.Claim 2. x j = x k for any distinct j and k.
Proof.For l 0, let i l be the least integer such that x k is on the segment P x k q k , and Hence, x j = x k for any j = k.Since q k P x k p k Q k+1 is a segment of the path on x k in the linkage Q ∞ , it is disjoint from qj Q j pj−1 by the definition of q j .Moreover, by Claim 2, all the segments of the form q k P x k p k Q k+1 are disjoint, and so are those of the form q j Q j p j−1 .Hence, we have an alternating comb.This contradiction completes the proof.
We have all the ingredients to prove the main result.
Theorem 2.5.Given a dimaze, the vertex sets linkable onto the exits form the bases of a matroid if and only if the dimaze contains no alternating comb.The independent sets of this matroid are precisely the linkable sets of vertices.
Proof.For the first statement, the backward direction follows from Lemma 2.3 and Lemma 2.4.To see the forward direction, suppose there is an alternating comb C. Let B 1 be the union of the vertices of out-degree 2 on C with B 0 − C. Then B 1 is linkable onto B 0 , and so is B 1 + v for any v ∈ B 0 ∩ C.But B 1 and B 1 + v violate the base axiom (B2).The second statement follows from the first and (1).
Corollary 2.6.Any dimaze which does not define a matroid contains an alternating comb.
We remark that even after forbidding alternating combs (or any ray at all), there are dimazes defining interesting strict gammoids.The existence of wild matroids, i.e. matroids containing a circuit which has an infinite intersection with a cocircuit, was first demonstrated in [5].It turns out that strict gammoids are a rich source of wild matroids.
Lemma 2.7.Suppose that M L (D, B 0 ) is a strict gammoid such that there is a circuit containing infinitely many vertices linkable to a fixed exit b in B 0 .Then M L (D, B 0 ) is a wild matroid.
Proof.The fundamental cocircuit of b with respect to B 0 , consisting of all the vertices linkable to b, intersects the given circuit at infinitely many vertices.
free dimaze, and {v i , b 1 : i 1} is an infinite circuit satisfying the lemma.

Nearly finitary linkability system
Although forbidding alternating combs ensures that we get a strict gammoid, not every strict gammoid arises this way.It turns out that when a dimaze gives rise to a nearly finitary ( [4]) linkability system, the dimaze defines a matroid regardless of whether it contains an alternating comb or not.We will show this using the proof of the linkage theorem.
Lemma 2.8.Let (D, B 0 ) be a dimaze.Then M L (D, B 0 ) satisfies the following: ( * ) For all independent sets I and J with J \ I = ∅, for every v ∈ I \ J there exists u ∈ J \ I such that J + v − u is independent.
Proof.We may assume that I \ J = {v}.Let Q = (Q y ) y∈Y Q be a "blue" linkage from J onto some Y Q ⊆ B 0 and P a "red" one from I. The linkage theorem yields a linkage Q ∞ , which we will show to witness the independence of a desired set.We use the notations introduced in its proof.For each y ∈ Y Q , let t 0 y be the initial vertex of Q y .
For i > 0 it is not hard to derive the following facts from the definitions of Q i , f i x and t i y : x Claim.For i 0, either Q i = Q ∞ or there is some x i such that: no such x otherwise.Proof.With x 0 := v, the claim clearly holds for i = 0. Given i > 0, to prove the claim, we may assume that Note that by (2) only x i−1 can be a vertex such that So by (4), we have x i−1 ∈ Ini(Q i ).Case (i): Suppose that there exists x ∈ I − x i−1 such that f i x i−1 and f i x are on the same path Q y i .By D i−1 , t 0 y i ∈ I and x is unique.Then D i holds for x i := x.In particular, by (3), J \ I ⊆ Ini(A i ).
We now prove U i .As Case (ii): Suppose that there does not exist any x ∈ I − x i−1 such that f i x is on the path Q y i containing f i x i−1 , if such a path exists.In this case, t 0 y i ∈ J \ I.By D i−1 , (2) and (4), we have If for some integer i > 0, case (ii) holds, then by (3), only u := t 0 y i ∈ J \ I can fail to be in Ini(A ∞ ).Otherwise, case (i) holds for each integer i 0, so that J \ I is a subset of Ini(A i ) and hence a subset of Ini(Q ∞ ).In either situation, since I = X P ⊆ Ini(Q ∞ ), we conclude that there is some u ∈ J \ I such that J + v − u is independent.
Let (E, I) be a set system.The finitarisation I fin of I consists of sets which have all their finite subsets in I. (E, I) is called nearly finitary if for any maximal element B ∈ I fin there is an Theorem 2.9.Let (D, B 0 ) be a dimaze.If M L (D, B 0 ) is nearly finitary, then it is a matroid.
The theorem shows that dimazes which contain an alternating comb may also define matroids.
Example 2.10.We construct a dimaze (D, B 0 ) which defines a nearly finitary linkability system, by identifying the corresponding exits of n > 1 copies of C O (see Figure 2).Note that (D, B 0 ) contains an alternating comb; and M L (D, B 0 ) is not finitary (a vertex not in B 0 together with all reachable vertices in B 0 form an infinite circuit4 ).We check that M L (D, B 0 ) is nearly finitary.Suppose B is a maximal element in M L (D, B 0 ) fin .Let I be the set obtained from B by deleting the last vertex, if exists, of B on each ray in D − B 0 ; and T := B \ I. Fix an enumeration i 1 , i 2 , . . .for I such that D contains a ray starting in i k+1 that avoids I k := {i 1 , . . ., i k } for each k 0 with i k+1 / ∈ B 0 .For any integer k 1, let T k consist of exactly one vertex on each ray in D − B 0 (that hits I k ): the first one in B after the last vertex of I k .Note that there are only finitely many linkages from I k to B 0 avoiding T k .In fact, there is at least one: the restriction to I k of a linkage of the finite subset I k ∪ T k of B. Applying the infinity lemma ([9, Proposition 8.2.1]), with the kth set consisting of the finite non-empty collection of linkages from I k to B 0 avoiding T k , we obtain a linkage from I to B 0 .Hence, I ∈ M L (D, B 0 ).As B is arbitrary and |T | n, we conclude that M L (D, B 0 ) is nearly finitary.
On the other hand, Theorem 2.9 does not imply Theorem 2.5: there are non-nearly finitary C A -free strict gammoids.In fact they can be highly connected and have nonnearly finitary duals.Lemma 2.11.Given a matroid M and a base B, if E \ B contains infinitely many elements which are not in any finite circuit, then M is not nearly finitary.
Example 2.12.For any integer k 2, there is a k-connected strict gammoid M = M L (D, B 0 ) such that the underlying graph of D is rayless, and neither M nor its dual is nearly finitary.
Consider a rooted tree T of depth 3 where each internal vertex has infinitely many children, and each edge is directed towards L 0 ∪ L 2 where L i is the set of vertices at distance i from the root for 0 i 3. Let D be a digraph with V = V (T ) ∪ X ∪ Y , where each of X and Y is an extra set of k vertices; and As no vertex in L 1 lies in a finite circuit, applying Lemma 2.11 with the base B 0 shows that M is not nearly finitary.Similarly, as no vertex in L 2 lies in a finite cocircuit, the same lemma with V \ B 0 shows that M * is not nearly finitary.
For any l < k, it is not difficult to see that in any bipartition of V into sets P, Q of size at least l, there is a linkage from P 1 ⊆ P \ B 0 to Q ∩ B 0 and from Q 1 ⊆ Q \ B 0 to P ∩ B 0 of size at least l.It follows that P 1 ∪ (P ∩ B 0 ) ∪ Q 1 ∪ (Q ∩ B 0 ) contains at least l vertices more than B 0 .Hence, (P, Q) is not an l-separation.So M is k-connected.
So far we have seen that if a dimaze (D, B 0 ) is C A -free or that M L (D, B 0 ) is nearly finitary, then M L (D, B 0 ) is a matroid.However, there are examples of strict gammoids that lie in neither of the two classes.All our examples of dimazes that do not define a matroid share another feature other than possessing an alternating comb: there is an independent set I that cannot be extended to a maximal in I ∪ B 0 .In view of this, we propose the following.Conjecture 2.13.Suppose that for all I ∈ M L (D, B 0 ) and B ⊆ B 0 , there is a maximal independent set in I ∪ B extending I. Then (IM) holds for M L (D, B 0 ).

Dimazes with alternating combs
We have seen in Section 2 that forbidding alternating combs in a dimaze guarantees that it defines a strict gammoid.However, the alternating comb in Figure 3 defines a matroid.On the other hand, this strict gammoid is isomorphic to the one defined by the incoming comb via the isomorphism given in the figure.So one might hope that every strict gammoid is C A -free.In general, this is not the case and the aim of this section is to construct a counterexample.

Finite circuits, cocircuits and alternating combs
We first give a necessary condition of any strict gammoid defined by a C A -free dimaze.
Lemma 3.1.If a dimaze (D, B 0 ) is C A -free, then M L (D, B 0 ) contains a finite circuit or a finite cocircuit.
Proof.Suppose the lemma does not hold.Then every finite subset of V is independent and coindependent, and B 0 is infinite.We construct a sequence (R k : k 1) of finite subdigraphs of D that gives rise to an alternating comb for a contradiction.
Let v 1 / ∈ B 0 and R 1 a path from v 1 to B 0 .For k 1, we claim that there is a path . Indeed, since any finite set containing v k is independent, there is a path from v k avoiding any given finite set disjoint from v k .Hence, there is a set and Ter(P) = B 0 , there is a path P ∈ P that is disjoint from R k and ends in V (F) ∩ B 0 .As the terminal vertex of P is in V (F), but its initial vertex is not, there is an edge (v k+1 , w k ) of P such that w k is on some path P k ∈ F, but v k+1 is not (on P k ).Then the vertices v k+1 , w k and the path P k satisfy the requirements of the claim.By induction, the claim holds for all k 1. Let ) is an alternating comb in (D, B 0 ).This contradiction completes the proof.
A matroid is infinitely connected if it does not have any k-separation for any integer k.The only infinitely connected finite matroids are uniform matroids of rank about half of the size of the ground set (see [16,Chapter 8]) and they are strict gammoids.It seems natural to look for an infinitely connected infinite matroid among strict gammoids, but the previous lemma gives us a partial negative result because the bipartition of any finite circuit of size k against the rest is a k-separation.It remains open whether there is an infinitely connected infinite gammoids.

Trees and transversal matroids
To give a strict gammoid not definable by a dimaze without alternating comb, we need only construct a strict gammoid without any finite circuit or cocircuit.A particular example is furnished by turning a transversal matroid defined on a tree to a strict gammoid.We prove a more general result that any tree gives rise to a transversal matroid.The definitions are recalled here.
Given a bipartite graph G, fix an ordered bipartition (V, W ) of V (G); this induces an ordered bipartition of any subgraph of G.A subset of V is independent if it is matchable to W .Let M T (G) be the pair of V and the collection of independent sets.It is clear

Denote the following statement by A(α):
There is a pairwise disjoint collection P α := {P v : v ∈ I ∩ C α \ S α } of m α -alternating paths such that each P v starts from v ∈ I ∩ C α \ S α with a downward edge and ends at the first vertex r v in I α \ I.
Start the inductive construction with m 0 , which is the set of upward edges that is contained in every matching of I.It is not hard to see that C 0 ∩ I = ∅, so that A(0) holds trivially.
If β is a successor ordinal, let By A(β − 1), the paths in P β−1 are disjoint.So m β−1 ∆E(P β−1 ) is a matching.Using the definition of S β−1 , we see that m β is indeed a matching.Observe also that If β is a limit ordinal, define m β by e ∈ m β ⇐⇒ ∃β < β such that e ∈ m α ∀α with β α < β.
As m α is a matching for every ordinal α < β, we see that m β is a matching in this case, too.
Suppose that a vertex u ∈ (V ∩ I) ∪ W is matched to different vertices by m α and m α for some α, α β.Then there exists some ordinal α + 1 between α and α such that u is matched by an upward m α -edge and by a downward m α +1 -edge.Hence, the change of the matching edges is unique.This implies that for any α, α with α α β, by ( 5) and ( 6), we have Moreover, for an upward m β -edge vw with v ∈ V , we have v ∈ I 0 or ∃α < β such that v ∈ C α and w / ∈ W α .
As there is no infinite decreasing sequence of ordinals, we have an m β -alternating path P v = v 0 w 0 v 1 w 1 • • • that stops at the first vertex r v ∈ V \ I.
The disjointness of the P v 's follows from that every vertex has a unique upward neighbour and, as we just saw, that vP v cannot contain any vertex v ∈ C β .So A(β) holds.
We can now go on with the construction.
Let γ |V | be the least ordinal such that C γ = ∅. 6Let C := α<γ C β and U := V \ (I 0 ∪ C); so V is partitioned into I 0 , C and U .As C γ = ∅, every vertex in U can be matched downwards to a vertex that is not in W γ .These edges together with m γ form a matching m B of B := U ∪ I γ , which we claim to be a witness for (IM).By (8), Suppose B is not maximally independent for a contradiction.Then there is an m Balternating path P = v 0 w 0 v 1 w 1 • • • such that v 0 ∈ V \ B that is either infinite or ends with some w n ∈ W \ V (m B ).We show that neither occurs.Claim 1. P is finite.
Proof.Suppose P is infinite.Since v 0 / ∈ B, P has a subray R = w i P such that w i v i+1 is an upward m B -edge.So w j v j+1 ∈ m B for any j i.As vertices in U are matched downwards, R ∩ U = ∅.As m B ∆E(R) is a matching of B ⊇ I in which every vertex in R ∩ V is matched downwards, R ∩ I 0 = ∅ too.So for any j i, there exists a unique β j such that v j ∈ C β j .
Choose k i such that β k is minimal.But with a similar argument used to prove A(β), we have β k > β k+1 .Hence P cannot be infinite.
Claim 2. P does not end in W \ V (m B ).
Proof.Suppose that P ends with w n ∈ W \ V (m B ). Certainly, v n can be matched downwards (either to w n−1 or w n ) in a matching of B ⊇ I. Hence, v n / ∈ I 0 .It is easy to check that for v ∈ C α , N (v) ⊆ W α+1 .Hence, as w n ∈ W \ W γ , v n / ∈ C. Hence, v n ∈ U .It follows that for each 0 < i n, v i is matched downwards and so does not lie in I 0 .As v 0 / ∈ B, v 0 ∈ C. It follows that w 0 ∈ W γ and v 1 ∈ C. Repeating the argument, we see that v n ∈ C, which is a contradiction.
We conclude that B is maximal.So (IM) holds and M T (G) is a matroid.Corollary 3.5.Let (D, B 0 ) be a dimaze such that the underlying graph of D is a tree and B 0 is a vertex class of a bipartition of D with edges directed towards B 0 .Then M L (D, B 0 ) is a matroid.

Definition 1 . 1 .
[6] A matroid M is a pair (E, I) where E is a set and I ⊆ 2 E which satisfies the following: (I1) ∅ ∈ I. (I2) If I ⊆ I and I ∈ I, then I ∈ I. (I3) For all I ∈ I \ I max and I ∈ I max , there is an x ∈ I \ I such that I + x ∈ I. (IM) Whenever I ∈ I and I ⊆ X ⊆ E, the set {I ∈ I : I ⊆ I ⊆ X} has a maximal element.

Figure 1 :
Figure 1: A locally finite dimaze which does not define a matroid

Figure 2 :
Figure 2: A dimaze that defines a nearly finitary linkability system

Figure 3 :
Figure 3: An alternating comb and an incoming comb defining isomorphic strict gammoids