Density of monochromatic infinite paths
Abstract.
For any subset , we define its upper density to be . We prove that every -edge-colouring of the complete graph on contains a monochromatic infinite path, whose vertex set has upper density at least . This improves on results of Erdős and Galvin, and of DeBiasio and McKenney.
1. Introduction
A -edge-colouring of a graph is an assignment of 2 colours, red and blue, to each edge of . We say that is monochromatic if all the edges of are coloured with the same colour. Given an arbitrary -edge-colouring of , what is the size of the largest monochromatic path contained as a subgraph? This was answered by Gerencsér and Gyárfás [GerencserGyarfas1967], who proved that every -edge-coloured contains a monochromatic path of length at least . This result is sharp.
Now consider the infinite complete graph on the vertex set . For any subset , the upper density of is defined as
Given a subgraph of , we define the upper density of to be that of . Trying to generalise the results known in the finite case, it is natural to ask what are the densest paths which can be found in any -edge-coloured . This problem was considered first by Erdős and Galvin [ErdosGalvin1993]. Other variants of this problem have been studied as well. For example, it is possible to consider other monochromatic subgraphs rather than paths, edge-colourings with more than two colours, use different notions of density or consider monochromatic sub-digraphs of infinite edge-coloured digraphs, etc. Results along these lines have been obtained by Erdős and Galvin [ErdosGalvin1991, ErdosGalvin1993], DeBiasio and McKenney [DeBiasioMcKenney2016] and Bürger, DeBiasio, Guggiari and Pitz [Guggiari2017].
We focus on the case of monochromatic paths in -edge-coloured complete graphs. By a classical result of Ramsey Theory, any -edge-colouring of contains a monochromatic infinite complete graph, and therefore, also a monochromatic infinite path . However, this argument alone cannot guarantee a monochromatic path with positive upper density, as it was shown by Erdős [Erdos1964] that there exist -edge-colourings of the infinite complete graph where every infinite monochromatic complete subgraph has upper density zero. Rado [Rado1978] showed that in every -edge-coloured there are monochromatic paths, of distinct colours, which partition the vertex set. This immediately implies that every -edge-coloured contains an infinite monochromatic path with .
Erdős and Galvin [ErdosGalvin1993] proved that for every -edge-colouring of there exists a monochromatic path with and showed an example of a -edge-colouring of such that every monochromatic path satisfies . DeBiasio and McKenney [DeBiasioMcKenney2016] improved the lower bound and showed that for every -edge-colouring of , there exists a monochromatic path with . In this paper, we improve the lower bound on .
Theorem 1.1.
Every -edge-colouring of contains a monochromatic path with .
In Section 2 we state our main lemma (Lemma 2.1) and use it to deduce Theorem 1.1. In Section 3 we collect some useful tools that will be used during the proof of Lemma 2.1, which is done in Section 4.
1.1. Notation
Given a graph , we write and for its vertex and edge set, respectively; and . Given , we write for the subgraph of induced by . If are disjoint, we write for the bipartite graph with classes and consisting precisely of those edges in with one endpoint in and the other in .
Let be a -edge-coloured graph. Throughout the paper, we assume its colours to be red and blue. For a vertex and a subset , we write the red neighbourhood of in for the set , that is, the set of vertices in connected to with red edges. We define analogously for blue. For all , we also define whenever is finite, otherwise.
For every , let and . For every set and we write for .
We write to mean that for all there exists such that for all the following statements hold. Hierarchies with more constants are defined in a similar way and are to be read from right to left.
2. Monochromatic path-forests
Our proof follows the strategies of Erdős and Galvin [ErdosGalvin1993] and of DeBiasio and McKenney [DeBiasioMcKenney2016], where they reduce the problem of finding monochromatic paths to the problem of finding collections of monochromatic disjoint paths satisfying certain conditions, which are then joined together to form an infinite path.
Consider a -edge-coloured . We say a vertex is red (or blue) if has infinitely many red (or blue, respectively) neighbours in . Note that it is possible for a vertex to be both red and blue. A -edge-colouring of is restricted if there is no vertex that is both red and blue. We write and for the set of red and blue vertices of , respectively.
A path-forest is a collection of vertex-disjoint paths. Let be a -edge-coloured graph. A path-forest of is said to be red if every edge of is red and all endpoints of every path in are red. We further assume that, for every path in , its vertices alternate between red and blue. Note that a red path-forest may contain isolated red vertices. A blue path-forest is defined similarly.
Our main lemma states that given a restricted -edge-coloured , there exists a monochromatic path-forest and an arbitrary long interval such that has size which is linear in .
Lemma 2.1.
Let and . For every restricted -edge-coloured , there exists an integer and red and blue path-forests and , respectively, such that
We defer the proof of Lemma 2.1 to Section 4. Note that we can always add any vertex which is both red and blue to a monochromatic path-forest, as an isolated vertex. Thus Lemma 2.1 implies the following corollary, which is valid for arbitrary -edge-colourings.
Corollary 2.2.
Let and . For every -edge-coloured , there exists an integer and red and blue path-forests and , respectively, such that
We use it now to deduce Theorem 1.1. The proof is based on the proofs of [ErdosGalvin1993, Theorem 3.5] and [DeBiasioMcKenney2016, Theorem 1.6].
Proof of Theorem 1.1.
Consider an arbitrary -edge-colouring of . Suppose that there exist two red vertices and a finite subset of such that does not contain a red path between and . For , let be the set of vertices reachable from using red paths in . Let . Then and are infinite; and are pairwise disjoint and there are no red edges between any for distinct . Thus there is an infinite blue path on the vertex set . Since is finite, , so we are done. An analogous argument is true if red is swapped with blue. Hence, we might assume that
| (2.1) |
For all , let . If the vertex is red, set to be the red path with the vertex and to be empty. Otherwise, set to be empty and . Set . Suppose that, for some , we have already found an integer and red and blue paths and , respectively, such that the endpoints of are red, the endpoints of are blue; and
| (2.2) |
We construct , and as follows. Let and . Considering the induced subgraph of on , by Corollary 2.2, there exists a monochromatic path-forest and such that . Let . By the choice of , note that
Suppose is red (if not, interchange the colours in what follows). Let . Apply (2.1) repeatedly to join the endpoints of the paths in and obtain a red path containing and with red vertices as endpoints.
By construction, we have and (2.2) holds for all . Without loss of generality, we may assume that for infinitely many values of . Let . Therefore, is a monochromatic path and . ∎
3. Preliminaries
In this section, we consider two ways of extending a path forest.
Proposition 3.1.
Let be a graph. Let be a path-forest and let be the set of vertices with degree at most one in . Let be such that . Then there exist such that is a path-forest.
Proof.
Since , there exist at least two neighbours of in , which are not endpoints of the same path in . ∎
Proposition 3.2.
Let be a graph and a path-forest. Let and . Suppose that
-
(i)
, and
-
(ii)
for every , .
Then there exists a path-forest ; every path in has both endpoints in ; is a path-forest and .
Proof.
Without loss of generality, we may assume that for all . We proceed by induction on . It is trivial if (by setting to be empty). So we may assume that . Note that by (i). Pick be such that and are not connected in . By (ii) and , there exists . Set and . It is easy to check that also satisfy the corresponding (i) and (ii). Therefore, by our induction hypothesis, the proposition holds. ∎
The next lemma is a useful statement about difference inequalities. We include its proof for completeness.
Lemma 3.3.
Let , be given and let be a strictly increasing sequence of non-negative integers. Suppose there exists such that for every ,
Then .
Proof.
Suppose . Choose sufficiently small such that and let and . Since is a strictly increasing sequence of non-negative integers, there exists such that
Then, for , , which implies, for every ,
| (3.1) |
Consider the function given by . It is immediate that is continuous. Since , it follows that for all .
4. Proof of Lemma 2.1
4.1. The path-forests algorithm
To satisfy the conditions stated in Lemma 2.1, we consider an algorithm that will build path-forests considering one extra vertex at a time, in increasing order.
Our algorithm is based on the following simple idea. Suppose that is a red vertex and we have constructed red and blue path-forests and , respectively. We can add to without any difficulty, forming a new red path-forest. We would like to add to the blue path-forest as well. However, we will add to the blue path-forest using only forward edges or only backward edges. Namely, when we say “add to using forward edges” (or backward edges) we mean to add the blue edges to for some blue vertices (or , respectively). We remark that the red (or blue) path-forest will contain all the red (or blue) vertices that have been considered so far, but it might be possible that some vertices are never included in the path-forest of the opposite colour.
Here we give an outline of Algorithm 4.1. There is a positive even integer which will be chosen before running the algorithm. The algorithm will consider each in order to decide whether to add it to the path-forest of the opposite colour by using forward or backward edges, with a preference toward forward edges. In fact, the algorithm will add a vertex using forward edges straight away, if possible, but will only add vertices using backward edges in batches. Roughly speaking, will be an (ordered) set of red vertices such that is joined to almost all blue vertices with red edges. Once is large enough, we will set aside a subset of “of size ”, which will be the ‘smaller’ endpoints of the backward edges. We continue the algorithm and collect a set of blue vertices, which could not be included in the red path-forest by using red forward edges. Once has vertices, we then add most of the vertices of into the red path-forest using red backward edges between and .
During the course of the algorithm, we will also construct a function , which will help us to define the sets at any given step. The role of is the following: a red vertex will be part of only when , similarly with the blue vertices. Imprecisely speaking, for a red vertex we would like to be “the last” of the blue vertices connected to via forward blue edges (this makes sense since the colouring is restricted); if no such blue vertices exist we just define . If is chosen like this, then when the algorithm reaches step , the red vertex will now be connected to “most” of the upcoming blue vertices using only red edges, which makes suitable to belong in .
Before presenting the algorithm, we will need the following notation. Suppose that after round number , we have constructed red and blue path-forests and , respectively. Given an ordered vertex set and , define
We view to be the number of additional degree that we can (theoretically) add to while keeping being a path-forest. Suppose an even is given and . If , then we define in the following way: let be minimal such that and then select to be minimal with respect to inclusion such that ; and let . Note that, by choice, for all . Note as well that . (Referring to the outline above, we will set .)
We make the following crucial definition. For all and , we define
We are now ready to describe the algorithm. We will verify that this algorithm is well-defined in Lemma 4.2.
Algorithm 4.1.
Fix an even . Given any restricted -edge-colouring of , we now construct monochromatic path-forests as follows. Initially, let be empty for all . Now suppose that we are at round number , and we have already constructed monochromatic path-forests , an ordered vertex subset , vertex subsets for and a function .
We now construct as follows by considering the vertex . Suppose (and if , interchange the roles of and in what follows). Our algorithm works in four steps.
-
Step 1:
Adding to the red path-forest.
Set . -
Step 2:
Updating available and waiting blue vertices.
Let be obtained from by adding the vertices with at the end of the ordering and . If and , then set ; otherwise set . -
Step 3:
Classifying .
We now classify into one of four types, which will use to determine whether (and how) can be added to the blue path-forest . Let . That is, is the blue neighbourhood of , that theoretically we can use to attach to using blue forward edges without creating a vertex of degree . If , then we set to be the smallest such that . We say that is-
•
of type if ;
-
•
of type if and ;
-
•
of type if , and ;
-
•
of type if , and .
-
•
-
Step 4:
Trying to add to the blue path-forest.
Depending on the type of , we have three different cases.-
Step 4a:
is of type .
We add to using forward edges. By Proposition 3.1 (with playing the roles of ) there exist such that is a blue path-forest. Further choose and such that is maximised (which is well-defined as and the colouring is restricted, so is finite). Define and for all . Set , , and . -
Step 4b:
is of type or .
In this case, we will not add to at all. Define and for all . Set . Let be obtained from by adding to the end of the ordering. If and , set ; otherwise set . Finally, set . -
Step 4c:
is of type .
In this case, we will try to add to using backwards edges if has reached the correct size. Define , and as in Step 44b.If , then set and and finish this step. Otherwise, we have . By Proposition 3.2 (with playing the roles of ), we obtain a blue path-forest such that is a blue path-forest which covers all but at most vertices of . Let . Adding the new blue edges to form means we need to redefine accordingly, as follows: if , then redefine ; otherwise redefine . Finally, define .
-
Step 4a:
4.2. Correctness and analysis of the algorithm
First we show that Algorithm 4.1 is well-defined. For , define (and ) to be the set of vertices (and , respectively) of type , as in Step 3 of Algorithm 4.1. Similarly, define for .
Lemma 4.2.
Let be even. Then Algorithm 4.1 is well defined.
Proof.
Suppose that has a restricted -edge-colouring. We prove by induction on that , , , , , , , , given by Algorithm 4.1 satisfy the following properties (and similar statements hold if we interchange and ):
-
(i)
for all and for all ;
-
(ii)
if and , then ;
-
(iii)
, and ;
-
(iv)
;
-
(v)
if , then ;
-
(vi)
;
-
(vii)
if and , then ;
-
(viii)
if , then and for all , .
Note that these properties imply the lemma. By our construction, (i)–(vii) hold.
To see (viii), let to be the smallest such that . Consider any . Clearly by (i) and (iv). So the first assertion of (viii) holds. Let , which is defined at round number . For all , we have . Hence . If is not of type , then . If is of type , then would contradict the maximality of in Step 44a. Hence we have for all . ∎
Recall that for every and , . In the next two lemmas, we collect some useful information from the algorithm.
Lemma 4.3.
Let be even. Suppose that has a restricted -edge-colouring. Let , , , , , , , , be as defined by Algorithm 4.1. Then the following holds for all (and similar statements hold if we interchange and ):
-
(i)
;
-
(ii)
are nested;
-
(iii)
if there exists such that for all , then ;
-
(iv)
if with , then and ;
-
(v)
if with , then ;
-
(vi)
if , then ;
-
(vii)
;
-
(viii)
for ;
-
(ix)
if for some , then .
Proof.
Now we prove (vii). By our construction, we have . Partition into (with possibly empty) such that, for all , , and . In other words, is a partition of into sets of ‘consecutive’ vertices. Consider any . Let . Since , Step 44c implies that we have , and . Moreover, all but at most vertices of are added to (at round number ). Therefore,
Hence (vii) holds.
To see (viii), note that is a decreasing sequence in and it decreases if and only if we join some vertices of to some vertices in with red edges to form the red path-forest. Each such vertex of reduces by at most .
Lemma 4.4.
Let be even. Suppose that has a restricted -edge-colouring. For all , let be as defined by Algorithm 4.1. Then there exist for all and such that (where similar statements hold if we interchange and ):
-
(i)
, for every ;
-
(ii)
;
-
(iii)
if for some , then ;
-
(iv)
.
Proof.
Let . Let . Note that (here we view as an unordered set). Hence
| (4.1) |
as for all . Note that for . Hence
implying (i).
Let . Since is a red path-forest, is a bipartite graph with vertex classes and . If with , then by Lemma 4.3(v). If with , then we must have and so . Hence if with , then . Therefore,
| (4.2) |
On the other hand, since ,
4.3. Evolutions of and
To prove Lemma 2.1, we will consider the path-forests , for every , as constructed by Algorithm 4.1. If, given and , for some we have , then we are done. Therefore, assuming this is not the case, we will deduce information about the evolution of the parameters and whenever increases, which we will use to finish the proof. (It also suffices to use Lemmas 4.3 and 4.4 instead of appealing to Algorithm 4.1.)
First, we show that if then there exists such that (or we are already done). That is, almost all vertices have degree in the red path-forest at round number .
Lemma 4.5.
Let be even. Suppose that has a restricted -edge-colouring. Let be as defined by Algorithm 4.1. Suppose . Then there exists such that or .
Proof.
Lemma 4.6.
Let be even and . Suppose that has a restricted -edge-colouring. Let , , , , , , , , be as defined by Algorithm 4.1. Suppose that . Then there exists such that or .
Proof.
Let . Suppose the contrary, that is, for all we have
| (4.4) |
Note that Lemma 4.3(iii) and (vi) imply that
| (4.5) |
for all .
Given , define to be the minimum such that , which exists by Lemma 4.5 and . Analogously, define . Define and . This defines sequences such that, for all ,
For convenience, let and for every , let . For every and , let
Lemma 4.3(vii) and (4.4) imply that
and a similar inequality also holds for . In summary, we have for ,
| (4.6) |
Consider any . Write . Lemma 4.3(viii) implies that
and a similar inequality holds for . Hence by combining both inequalities and using Lemma 4.4(iv), we have
where the last inequality follows from . Hence, for all ,
| (4.7) |
Claim 4.7.
For all , .
Now we are ready to prove Lemma 2.1.
Proof of Lemma 2.1.
Let . Choose such that is even, and
| (4.8) |
Let , , , , , , , , be as defined by Algorithm 4.1. Lemma 4.3(i) implies that for all ,
| (4.9) |
Lemma 4.3(vii) together with (4.9) imply that for all (and a similar bound is true replacing by ):
| (4.10) |
We might suppose that for all we have
| (4.11) |
or else we are done. Together with Lemma 4.4(iv) and (4.8),
| (4.12) |
Without loss of generality, we may assume that . Define to be the minimum such that , which exists by Lemma 4.6 and (4.11). Note that by (4.12). Similarly, define to be the minimum such that . Now define to be the minimum such that . Note that exists by Lemma 4.5 and (4.11), and that .
Claim 4.8.
There exist
| (4.14) |
such that
Proof of the claim.
Since , we have . Let and be given by Claim 4.8. Let
Thus, and . Let and . Since is the least real root of the polynomial and , it follows that .
Now we use the previous bounds to get
where the last line follows from (4.8), (4.13) and . Rearranging, we get , and recalling that we have
| (4.17) |
A similar argument (by estimating ) shows that
| (4.18) |
Next, we would like to estimate and . By the choice of , Lemma 4.4(iv) and (4.11),
where the last inequality follows from (4.8). Together with Lemma 4.3(viii) and the choice of we get
| (4.19) |
Using Claim 4.8, we get
where the last inequality follows from (4.8), (4.13) and . Rearranging, we get . Recalling that , we get
| (4.20) |
A similar argument (by estimating ) shows that
| (4.21) |
Remark
After the submission of this paper, we learned that Corsten, DeBiasio, Lamaison and Lang [CDLL18] have obtained an improved version of Theorem 1.1.
Acknowledgements
We thank an anonymous referee for their helpful suggestions.
References
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