Proof. Partition $\mathbb {N}$ by residues mod r – that is, $\mathbb {N} = A_0 \cup \ldots \cup A_{r-1}$ , where $A_i=\{n \in \mathbb {N}: n \equiv i \pmod r\}$ . We define an r-coloring as follows: if $m \in A_i$ and $n>m$ , color the edge $mn$ with color i. Note that if $n\not \equiv i\ (\bmod \ r)$ , then n has at most $\lceil (n-1)/r\rceil $ neighbors of color i.

(i) Let T be an infinite D-ary tree and suppose we have a copy of T of color i. For all $n\in \mathbb {N}$ , let $V^{\prime }_n$ be the set of vertices in $V(T)\cap [n]$ which are not congruent to $i \pmod r$ . Since every vertex $m\in V^{\prime }_n$ can only have successors (of color i) in $A_i \cap [n-1]$ , we must have $D \cdot |V_n^{\prime }| \leq \lceil (n-1)/r\rceil $ . So we have

$$\begin{align*}\frac{|V(T)\cap [n]|}{n}\leq \frac{\lceil\frac{n}{r}\rceil+|V_n^{\prime}|}{n}\leq \frac{\lceil\frac{n}{r}\rceil+\frac{1}{D}\lceil\frac{n-1}{r}\rceil}{n} \xrightarrow{n \to \infty} \frac{1}{r}(1+\frac{1}{D}),\end{align*}$$

and thus, $\operatorname {\mathrm {\overline {Rd}}}_r(T)\leq \frac {1}{r}(1+\frac {1}{D})$ .

(ii) Let $0<d_1< d_2< \dots $ be an increasing sequence of integers. Let T be a tree in which every vertex on level i has degree $d_i$ . We can repeat the argument from case (i), except now we have $|V_n^{\prime }|/n \to 0$ as $n \to \infty $ and thus $\frac {|V(T)\cap [n]|}{n}\leq \frac {\lceil \frac {n}{r}\rceil +|V_n^{\prime }|}{n} \xrightarrow {n \to \infty } \frac {1}{r}$ .

Proof. Let $\epsilon>0$ be given, let c be the number of components of G, and let k be an integer with $k>c/\epsilon $ . Partition $\mathbb {N}$ by residues mod k – that is, $\mathbb {N} = A_0 \cup \ldots \cup A_{k-1}$ , where $A_i=\{n \in \mathbb {N}: n \equiv i\ (\bmod \ k)\}$ . For all $0\leq i\leq k-1$ , color all edges inside $A_i$ with green, and for all $0\leq i<j\leq k-1$ , color the edges between $A_i$ and $A_j$ with the half graph coloring where the vertices in $A_i$ have cofinite red degree to $A_j$ and the vertices in $A_j$ have cofinite blue degree to $A_i$ . Note that we have only used three colors, but this can be considered as an r-coloring for all $r\geq 3$ .

Note that for all $0\leq i\leq k-1$ , every vertex in $A_i$ has finite red degree to $A_0\cup \dots \cup A_i$ . If there is a red copy of G, then let $0\leq i\leq k-1$ be maximum such that $V(G)\cap A_i$ is infinite. But this is a contradiction because every vertex in $V(G)\cap A_i$ has finite red degree. Similarly, for all $0\leq i\leq k-1$ , every vertex in $A_i$ has finite blue degree to $A_i\cup \dots \cup A_{k-1}$ . If there is a blue copy of G, then let $0\leq i\leq k-1$ be minimum such that $V(G)\cap A_i$ is infinite. But this is a contradiction because every vertex in $V(G)\cap A_i$ has finite blue degree. Therefore, every monochromatic copy of G is green and thus has upper density at most $c/k<\epsilon $ .

Proof. Assume first that $\chi (G) < \infty $ and partition $\mathbb {N}$ by residues mod $\chi (G)-1$ . Color all edges inside the sets red and all edges between the sets blue. There is no blue copy of G, so every copy of G lies entirely inside one of the sets, all of which have density $\frac {1}{\chi (G)-1}$ .

If $\chi (G) = \infty $ , this construction shows that $\operatorname {\mathrm {\overline {Rd}}}(G) \leq 1/(k-1)$ for every $k \geq 2$ , and therefore, $\operatorname {\mathrm {\overline {Rd}}}(G) = 0$ .

Proof. (i) Since $\mathcal {R}$ contains an infinite clique, we have $\chi (\mathcal {R}) = \infty $ , and thus, the result follows from Example 2.4.

(ii) Let G be a graph on vertex set $\mathbb {N}$ where $[\frac {n(n+1)}{2}, \frac {(n+1)(n+2)}{2}]$ induces a clique for all $n\in \mathbb {N}$ . G is locally finite, connected, and contains a clique of order n for all $n\in \mathbb {N}$ . So $\chi (G)=\infty $ , and thus, the result follows from Example 2.4.

Proof. Let A and B be the parts of $K_{\mathbb {N},\mathbb {N}}$ and partition both of them into r parts $A_1, \ldots , A_r$ and $B_1, \ldots , B_r$ , each of density $1/(2r)$ . For all $i,j \in [r]$ , color every edge between $A_i$ and $B_j$ by $(i-j) \mod r$ . It is easy to see that every part is incident to exactly one other part of each color, and therefore, every monochromatic path can cover at most two parts, finishing the proof.

Fact 2.7. G is weakly expanding if

1. (i) G has finite independence number, or

2. (ii) G has bounded maximum degree and finitely many isolated vertices, or

3. (iii) G is a tree with bounded leaf degree, or

4. (iv) for all $n\in \mathbb {N}$ , G has finitely many vertices of degree n.

Proof. Suppose G is weakly expanding and let f be the function guaranteed by the definition.

Let $a_n$ be an increasing sequence of natural numbers such that $a_0=1$ and for all $k\geq 1$ ,

(2.1) $$ \begin{align} a_k> k( a_{k-1}+f(a_{k-1})). \end{align} $$

For all $u,v\in \mathbb {N}$ with $u<v$ , color the edge $uv$ red if $u\in [a_{2n-1},a_{2n})$ and blue if $u\in [a_{2n},a_{2n+1})$ for some $n\in \mathbb {N}$ .

Suppose there is a, say, blue copy of G in this 2-coloring with vertex set U. We must have that $A_n:=U\cap [a_{2n-1},a_{2n})$ induces an independent set for all $n\in \mathbb {N}$ , and because of the coloring, we have $N_B(A_n)\subseteq [0, a_{2n-1})$ , and thus, $|N_B(A_n)|\leq a_{2n-1}$ . Thus, by the definition of weakly expanding, we have $|A_n|<f(a_{2n-1})$ . We conclude that for all $n\in \mathbb {N}$ ,

$$ \begin{align*} |U\cap [0, a_{2n})|=|U\cap [0,a_{2n-1})|+|A_n|<a_{2n-1}+f(a_{2n-1})\stackrel{2.1}{<}\frac{1}{2n}a_{2n},\end{align*} $$

and thus, $\operatorname {\mathrm {\underline {d}}}(U)=0$ .

Proof. Let $a_n$ be an increasing sequence of natural numbers and let $A_i=[a_i, a_{i+1})$ for all $i\in \mathbb {N}$ . For all $u\in A_i$ and $v\in A_j$ with $u<v$ , color the edge $uv$ red if j is odd and blue if j is even. Let $A^0$ be the union of all even indexed intervals and let $A^1$ be the union of all odd indexed intervals. We note that every vertex in $A^0$ has finite blue degree to $A^1$ and every vertex in $A^1$ has finite red degree to $A^0$ .

Let D be a finite dominating set in G and suppose there is a monochromatic, say, blue copy of G with vertex set V. Since D is finite, there exists an index t such that $D\subseteq A_1\cup A_2\cup \dots \cup A_t$ . Now for all i such that $2i+1>t$ , there are no blue edges from $A_{2i+1}$ to D contradicting the fact that D is a dominating set. So G has finite intersection with say $A^1$ , and thus, if $a_n$ is increasing fast enough, G has lower density 0.

Proof. Let $\{X, Y\}$ be a partition of $\mathbb {N}$ into two sets of lower density 0 (for instance, as we did in Example 2.8 and Example 2.9). Color all edges inside X or inside Y with blue, and color all edges between X and Y red. Note that since G is connected, any blue copy of G is completely contained in X or Y and thus has lower density 0.

If $\chi (G)\geq 3$ , then there is no red copy of G, and we are done. If G is bipartite and one of the parts is finite, then G intersects either X or Y in only finitely many vertices, and thus, any red copy of G will have lower density 0.

Proof. Take the Rado coloring of $K_{\mathbb {N}}$ .

If K has at least two infinite parts, or infinitely many vertices in finite parts, then K contains a spanning copy of $K_{\mathbb {N}, \mathbb {N}}$ ; let $(A,B)$ be such a spanning copy of $K_{\mathbb {N}, \mathbb {N}}$ . Let $a_1,a_2,\ldots $ be the elements of A. Then B is contained in the neighborhood of $a_1,\ldots ,a_n$ , and hence has density $\le 2^{-n}$ , for each n, by (2.2). Hence, B must have density $0$ . The same goes for A.

Now suppose K has exactly one infinite part and exactly n vertices in finite parts. Then by (2.2), we have $\operatorname {\mathrm {\overline {Rd}}}(K)\leq \frac {1}{2^{n}}$ since the infinite part is the intersection of the neighborhoods of the n vertices in finite parts.

To see $\operatorname {\mathrm {\overline {Rd}}}(K)\geq \frac {1}{2^{2n-1}}$ , we are given an arbitrary 2-coloring of $K_{\mathbb {N}}$ , and we choose an arbitrary vertex v. Either $\operatorname {\mathrm {\overline {d}}}(N_R(v))\geq 1/2$ or $\operatorname {\mathrm {\overline {d}}}(N_B(v))\geq 1/2$ , and we choose the color with largest upper density. We repeat this process inside the chosen neighborhood for $2n-1$ steps, at which point we have n vertices whose common neighborhood in color, say, red has upper density at least $\frac {1}{2^{2n-1}}$ and the infinite part of K can be embedded in such a way that it spans this set.

Proof. Consider the Rado coloring of $K_{\mathbb {N}}$ . Suppose now that V is the vertex set of a monochromatic copy of G, say with color i. Then for each N, we have

$$\begin{align*}V \subseteq^* \bigcap_{n=1}^N \bigcup_{v\in F_n} N_i(v). \end{align*}$$

Note that

$$\begin{align*}d\left(\bigcap_{n=1}^N \bigcup_{v\in F_n} N_i(v)\right) = \prod_{n=1}^N (1 - 2^{-|F_n|}). \end{align*}$$

Hence,

$$\begin{align*}\operatorname{\mathrm{\overline{d}}}(V)\le \prod_{n=1}^\infty (1 - 2^{-|F_n|}). \end{align*}$$

It is well known that an infinite product $\prod _{n=1}^\infty \alpha _n$ , with $\alpha _n\in (0,1)$ , converges to $0$ if and only if

$$\begin{align*}\sum_{n=1}^\infty \log(\alpha_n) = -\infty. \end{align*}$$

In our case, we have $|F_n| \le \log _2(n)$ for all sufficiently large n, so

$$\begin{align*}\log(1 - 2^{-|F_n|}) \le \log\left(1 - \frac{1}{n}\right) \leq -1/n. \end{align*}$$

By the limit comparison test and the divergence of the harmonic series, we have $d(V) = 0$ .

Fact 8.2. Let T be a tree of unbounded radius. Either for all $v\in V(T)$ , there is an infinite path in T starting with v or there exists $v_0\in V(T)$ such that T contains an increasing star having $v_0$ as the center.

Proof. Let T be a tree, let $v\in V(T)$ , and suppose there is no infinite path in T starting with v (since T is connected, this implies that there is no infinite path in T at all). Since T has unbounded radius, we can do the following: let $Q_1$ be a path from v to a leaf $u_1$ , which has some length $k_1$ . Now there must exist a path $Q_2$ of length $k_2>k_1$ from v to a leaf $u_2$ , and so on. This process gives an infinite set of leaves U and an increasing sequence $k_1, k_2, \dots $ such that there is a path from v to $u_i$ of length $k_i$ . Now we apply the Star-Comb lemma [Reference Diestel11, Lemma 8.2] to the set U. Since T has no infinite path, there must exist a subdivision of an infinite star with center $v_0$ such that all the leaves, call them $U^{\prime }$ , are in U. We claim that for all k, there exists a path from $v_0$ to $U^{\prime }$ which has length greater than k, which would prove the lemma. If not, then there exists k such that every path from $v_0$ to $U^{\prime }$ has length at most k. However, since there is a path from v to $v_0$ , this would imply that there exists a $k^{\prime }$ such that every path from v to $U^{\prime }$ has length at most $k^{\prime }$ . But this contradicts the fact that the lengths of the paths from v to $U^{\prime }$ form an increasing sequence.

Proof of Lemma 8.1.

First, suppose that a spanning copy of T can be found in every infinitely connected graph H. It is known that there exist infinitely connected graphs with arbitrarily high girth (see [Reference Diestel11, Chapter 8, Exercise 7]); for instance, let $H_0$ be a cycle of length k, and for all $i\geq 1$ , let $H_i$ be the graph obtained by adding a vertex $x_i$ and internally disjoint paths of length $\lceil k/2\rceil $ from $x_i$ to every vertex in $H_{i-1}$ . Then let $H=\cup _{i\geq 0}H_i$ . So H is infinitely connected and has girth k. This proves that T is acyclic (i.e., T is a forest) because otherwise there would be an infinitely connected graph in which every cycle is longer than the shortest cycle in T.

Let H be the infinite blow-up of a one-way infinite path (i.e., replace each vertex with an infinite independent set and each edge with a complete bipartite graph). Clearly, H is infinitely connected. If T is a spanning subgraph of H, then T has a component of unbounded radius or T has infinitely many components.

Thus, T must be a forest with a component of unbounded radius or infinitely many components.

Next suppose T is a forest with a component of unbounded radius or infinitely many components. If T has infinitely many components $T_1, T_2, \dots $ , we may select for all $i\geq 1$ , $t_i\in V(T_i)$ and add the edge $t_it_{i+1}$ for all $i\geq 1$ to get a tree with unbounded radius which contains T as a spanning subgraph. If T has finitely many components $T_1, \dots , T_k$ , at least one of which has unbounded radius, we may for all $i\in [k-1]$ add an edge from $t_{i}\in V(T_i)$ to $t_{i+1}\in V(T_{i+1})$ to get a tree with unbounded radius which contains T as a spanning subgraph. Thus, it suffices to prove the result when T is a tree with unbounded radius.

By Fact 8.2, there exists a vertex $t_0$ such that either there is an infinite path starting with $t_0$ (in which case we say T is of Type 1), or an increasing star having $t_0$ as the center (in which case we say T is of Type 2). Now starting with $t_0$ , fix an enumeration of $V(T) = \{t_0, t_1, t_2 \ldots \}$ such that for all $i\geq 1$ , $T[\{t_0, \dots , t_i\}]$ is connected (in fact, for all $i\geq 1$ , $t_i$ has exactly one neighbor in $\{t_0, \dots , t_{i-1}\}$ ). Also fix an enumeration of $V(H) = \{v_0, v_1, v_2, \ldots \}$ . We will build an embedding f of T into H recursively, in finite pieces, at each stage ensuring that we add the first vertices of and into the domain and range of f respectively.

Initially, let $f(t_0)=v_0$ (we think of $t_0$ as being the root of the tree and $v_0$ as the embedding of the root in H) and let $t_{last}:=t_0$ and $v_{last}:=v_0$ . We now show that Algorithm 1 gives the desired embedding.

Note that if T is of Type 2, then $t_{last}=t_0$ and $v_{last}=v_0$ throughout the process.

To see that f is a well-defined surjective embedding of T into H, first note that we can always follow lines 4 and 12 of Algorithm 1 since H is infinitely connected and in particular every vertex has infinite degree. Line 5 is always possible since there is either an infinite path starting at $v_0$ or an increasing star having $v_0$ as the center. Line 11 is always possible by the enumeration of $V(T)$ . So f is well defined.

We alternate between embedding the vertex t of smallest index from T which has not yet been embedded into an available vertex from H in such that way that the parent $t^{\prime }$ of t has already been embedded and $f(t)$ is adjacent to $f(t^{\prime })$ , and embedding a path $t_0t_1\dots t_\ell $ to a vertex such that $f(t_0)f(t_1)\dots f(t_\ell )$ is a path in H and $f(t_\ell )$ is the vertex of smallest index from $V(H)$ which has yet to be mapped to. So f will be a surjective embedding of T.

Proof. Let $t_1 \in V(T)$ be a vertex of infinite degree and let $v_1=v$ from the statement of the theorem (again we think of $t_1$ as being the root of the tree and $v_1$ as the embedding of the root in H). We will build an embedding f of T into H recursively, in finite pieces, at each stage adding one more child of every previously embedded $t \in T$ (unless all children have been embedded already). The embedding strategy is very similar to that in the proof of Lemma 8.1. Initially, let $f(t_1) = v_1$ . We will use the following Algorithm 2.

First, note that we can always follow line 4 of Algorithm 1 since every vertex in H has infinite degree. Let $f :V(T) \to V(H)$ be the function produced by Algorithm 2. We need to prove that f is well defined, an embedding of T and that $N_H(v) \subseteq \operatorname {\mathrm {dom}} f$ .

Since we always embed the smallest not yet embedded neighbor of every previously embedded $t \in V(T)$ in line 4, every other vertex will be embedded eventually as well. Therefore, f is well defined. Furthermore, by construction of f, it defines a proper embedding (whenever a new vertex $t \in T$ is embedded, its parent $t^{\prime }$ is already embedded, and we make sure that $f(t)$ is adjacent to $f(t^{\prime })$ ). Finally, note that we are infinitely often in line 4 when $t = t_1$ since $N_T(t_1)$ is infinite. Since we always choose the smallest available vertex in , it follows that $ N_H(v) \subseteq \operatorname {\mathrm {ran}} f$ .

Proof of Theorem 1.13.

It clearly suffices to prove the result for trees, so let T be an infinite tree and suppose the edges of $K_{\mathbb {N}}$ are colored with two colors. If T does not have an infinite path, it must have at least one vertex of infinite degree (by König’s infinity lemma [Reference König20]), and therefore, the theorem follows immediately from Lemmas 8.5 and 8.3. So suppose T has an infinite path. By Lemma 8.4, there is an infinite set A with $\operatorname {\mathrm {\overline {d}}}(A) \geq 1/2$ and a color c, so that the induced subgraph on A is infinitely connected in c. By Lemma 8.1, there is a monochromatic copy of T spanning A, and we are done.

Proof of Lemma 8.4.

Fix a $2$ -coloring of $K_{\mathbb {N}}$ . We define a sequence of sets $R_\alpha , S_\alpha $ , for all ordinals $\alpha $ , as follows. Let $S_0 = \mathbb {N}$ . For each $\alpha $ , let $R_\alpha $ be the set of vertices in $S_\alpha $ whose blue neighborhood has finite intersection with $S_\alpha $ , and let . If $\lambda $ is a limit ordinal, then we let $S_\lambda $ be the intersection of the sets $S_\alpha $ , for $\alpha < \lambda $ .

Note that the sets $R_\alpha $ are pairwise disjoint, and hence, there is some countable ordinal $\gamma $ such that $R_\alpha = \emptyset $ for all $\alpha \ge \gamma $ . Let $\gamma ^*$ be the minimal ordinal such that $R_{\gamma ^*}$ is finite; it follows then that $R_\beta = \emptyset $ for all $\beta> \gamma ^*$ . Set

(Note that $\gamma ^*$ may be $0$ , in which case $R = \emptyset $ .)

Suppose that R is infinite. Then $\gamma ^*> 0$ and $R_\alpha $ is infinite for all $\alpha < \gamma ^*$ . Let $u,v\in R$ with $u\in R_\alpha $ and $v\in R_\beta $ for some $\alpha \le \beta < \gamma ^*$ . It follows that the red neighborhoods of both u and v are cofinite in $R_\beta $ . Since $R_\beta $ is infinite, this implies that there is a red path of length $2$ connecting u and v, even after removing a finite set of vertices. Hence, R is infinitely connected in red.

Set $S = S_{\gamma ^*+1}$ . Then , so $R\cup S$ is cofinite. Moreover, since $R_{\gamma ^*+1} = \emptyset $ , it follows that for every $v\in S$ , the blue neighborhood of v has infinite intersection with S. Now suppose that S is not infinitely connected in blue. Then there is a finite set $F\subseteq S$ and a partition such that X and Y are both nonempty, and every edge between X and Y is red. Note that X and Y must both be infinite, since if $x_0\in X$ and $y_0\in Y$ , then $X\cup F$ and $Y\cup F$ must contain the blue neighborhoods of $x_0$ and $y_0$ (both of which are infinite), respectively. But then the red graph restricted to is infinitely connected.

Proof of Lemma 8.5.

Fix a $2$ -coloring of $K_{\mathbb {N}}$ . Similarly as in the proof of Lemma 8.4, we will construct sets $R_\alpha , B_\alpha , S_\alpha $ for all ordinals $\alpha $ . Let $S_0 = \mathbb {N}$ . For each $\alpha $ , let $R_\alpha $ be the set of vertices in $S_\alpha $ whose blue neighborhood has finite intersection with $S_\alpha $ , let $B_{\alpha }$ be the set of vertices in $S_{\alpha }$ whose red neighborhood has finite intersection with $S_{\alpha }$ , and let . If $\lambda $ is a limit ordinal, then we let $S_\lambda $ be the intersection of the sets $S_\alpha $ , for $\alpha < \lambda $ . As in the proof of Lemma 8.4, the following properties hold.

1. (i) There is a unique ordinal $\gamma ^{*}$ such that $R_\alpha \cup B_\alpha $ is infinite for all $\alpha < \gamma ^*$ , finite for $\alpha = \gamma ^*$ and empty for all $\alpha> \gamma ^*$ . We denote $R = \bigcup _{\alpha <\gamma ^*} R_\alpha $ and $B = \bigcup _{\alpha <\gamma ^*} B_\alpha $ .

2. (ii) $S_\alpha = S_{\alpha ^{\prime }}$ for all ordinals $\alpha , \alpha ^{\prime }> \gamma ^*$ . We denote $S = S_{\gamma ^* + 1}$ .

3. (iii) $R, B, S$ are pairwise disjoint, and $R \cup B \cup S$ is cofinite.

4. (iv) If $v \in R_\gamma $ for some ordinal $\gamma $ , then v has finitely many blue neighbors in $S \cup \bigcup _{\alpha \geq \gamma } R_\alpha $ .

5. (v) If $v \in B_\gamma $ for some ordinal $\gamma $ , then v has finitely many red neighbors in $S \cup \bigcup _{\alpha \geq \gamma } B_\alpha $ .

6. (vi) Every $v \in S$ has infinitely many neighbors of both colors in S.

If $R \cup B$ is empty, then let $A = S$ and choose an arbitrary vertex $v \in S$ . Since A is cofinite in $\mathbb {N}$ , either the blue or the red neighborhood of v in A has upper density at least $1/2$ . Since every vertex in A has infinite degree in both colors, we are done.