Proof. Consider the colouring of $E(K_{\mathbb{N}})$ in which the edge $uv$ is red iff $a-1$ divides $v-u$ . The graph formed by the blue edges has chromatic number $a-1$ and thus does not contain $H$ as a subgraph. Every monochromatic copy $H'$ of $H$ in this colouring is red.

The red graph consists of $a-1$ cliques $C_1, \dots, C_{a-1}$ , each with $\bar d(C_i)=1/(a-1)$ . The $b$ components that concentrate $V(H')$ must be each contained in a clique $C_i$ . Because modifying finitely many elements does not affect the density of a set, we have $\bar d(H')\leq \bar d(C_1\cup \dots \cup C_b)=b/(a-1)$ . We conclude that $\rho (H)\leq b/(a-1)$ .

Proof of Theorem 1.6. Denote $\lambda =\liminf \limits _{n\rightarrow \infty }\frac{\mu (H,n)}{n}$ . We assume that $\lambda \gt 0$ , as otherwise the statement becomes the trivial inequality $\rho (H)\leq 1$ .

Let $\epsilon \gt 0$ . Let $g$ be a 1-Lipschitz function such that the upper limit in (1) is less than $h(\gamma )+\epsilon$ , for $\gamma =\frac{\lambda -1}{\lambda +1}$ . Take an infinite set of vertices $v_1, v_2, \dots$ and arrange them from left to right in this order. Colour these vertices red and blue, in such a way that, for every $n\in{\mathbb{N}}$ , among the $n$ leftmost vertices there are exactly $\lfloor (n+g(n))/2\rfloor$ red vertices (this is possible because $g$ is 1-Lipschitz). Form a two-coloured complete graph by giving each edge the colour of its leftmost endpoint.

There must be infinitely many vertices of each colour. Indeed, if there are finitely many red vertices then the non-decreasing function $\frac{x+g(x)}{2}$ is bounded, while if there are finitely many blue vertices the non-decreasing function $\frac{x-g(x)}{2}$ is bounded. Note that if $x\pm g(x)$ is bounded, then $\gamma x\pm g(x)$ has an upper bound, as $\gamma \leq 1$ . In the first case $\Gamma ^+_\gamma (g,n)=+\infty$ for every $n$ large enough, while in the second case $\Gamma ^-_\gamma (g,n)=+\infty$ .

Let the red vertices be $r_1, r_2, r_3, \dots$ and the blue vertices be $b_1, b_2, b_3, \dots$ , according to the left-to-right order. Let $\alpha _i$ be the smallest value such that $r_{\alpha _i}$ has at most $\lambda (\alpha _i-i)$ blue vertices to its left, and $\beta _i$ the smallest value such that $b_{\beta _i}$ has at most $\lambda (\beta _i-i)$ red vertices to its left. The following discussion will not only prove the existence of $\alpha _i$ and $\beta _i$ , but also give a bound on them.

For a fixed value of $i$ , let $w=\frac{2}{1+\lambda }(\lambda i+2\lambda +2)$ . Let $z^+=\Gamma ^+_\gamma (g,w)$ and $z^-=\Gamma ^-_\gamma (g,w)$ . By continuity of $g$ and definition of $\Gamma ^+_\gamma$ and $\Gamma ^-_\gamma$ , we have

\begin{equation*}g(z^+)=w-\gamma z^+,\,\,\,\,\,g(z^-)=\gamma z^--w.\end{equation*}

One can check that the following identity holds by substution of $g(z^-)$ :

\begin{equation*}\frac {z^-+g(z^-)}{2}+2=\lambda \left (\frac {z^--g(z^-)}{2}-i-2\right ).\end{equation*}

Among the $\lfloor z^-\rfloor$ leftmost vertices there are $\left \lfloor \frac{\lfloor z^-\rfloor +g(\lfloor z^-\rfloor )}{2}\right \rfloor$ red vertices and $\lfloor z^-\rfloor -\left \lfloor \frac{\lfloor z^-\rfloor +g(\lfloor z^-\rfloor )}{2}\right \rfloor$ blue vertices. Observe that

\begin{align*} \left \lfloor \frac{\lfloor z^-\rfloor +g(\lfloor z^-\rfloor )}{2}\right \rfloor & \leq \frac{\lfloor z^-\rfloor +g(\lfloor z^-\rfloor )}{2} \leq \frac{z^-+g(z^-)}{2} \lt \lambda \left (\frac{z^--g(z^-)}{2}-i-2\right )\\[5pt] & \leq \lambda \left (\lfloor z^-\rfloor -\left \lfloor \frac{\lfloor z^-\rfloor +g(\lfloor z^-\rfloor )}{2}\right \rfloor -i\right ). \end{align*}

If the last blue vertex among those $\lfloor z^-\rfloor$ is $b_{\tau }$ , then the number of red vertices to its left is less than $\lambda (\tau -i)$ , meaning that $\beta _{i}\leq \tau$ , and in particular $\beta _i$ exists. Hence

Analogously, one has the identity

\begin{equation*}\frac {z^+-g(z^+)}{2}+2=\lambda \left (\frac {z^++g(z^+)}{2}-i-2\right )\end{equation*}

and the inequality

\begin{equation*}\lfloor z^+\rfloor -\left \lfloor \frac {\lfloor z^+\rfloor +g(\lfloor z^+\rfloor )}{2}\right \rfloor \leq \lambda \left (\left \lfloor \frac {\lfloor z^+\rfloor +g(\lfloor z^+\rfloor )}{2}\right \rfloor -i\right ).\end{equation*}

Hence we find

\begin{equation*}\alpha _i\leq \left \lfloor \frac {\lfloor z^+\rfloor +g(\lfloor z^+\rfloor )}{2}\right \rfloor \leq \frac { z^++g(z^+)}{2}=\frac {1-\gamma }{2}z^++\frac {w}{2}.\end{equation*}

Adding the two values together, for $i$ large enough we have

(3) \begin{equation} \alpha _i+\beta _i\leq \frac{1-\gamma }{2}(z^++z^-)+w+2\leq \left (\frac{1-\gamma }{2}h(\gamma )+\epsilon +1\right )\frac{2\lambda }{1+\lambda }i+o(i). \end{equation}

Let $\phi\;:\;\mathbb{N}\rightarrow \{v_1, v_2, \dots \}$ be an arbitrary bijection satisfying $\phi ([\alpha _j+\beta _j])=\{r_1, r_2, \dots, r_{\alpha _j}, b_1, b_2, \dots, b_{\beta _j}\}$ for every $j$ . The function $\phi$ defines a colouring of $E(K_{\mathbb{N}})$ , where the colour of the edge $ij$ is the colour of the edge $\phi (i)\phi (j)$ .

Let $R$ and $B$ be the sets of positive integers $i$ whose image $\phi (i)$ is red or blue, respectively. Let $H'\subseteq K_{\mathbb{N}}$ be a monochromatic copy of $H$ in this colouring. Suppose that $H'$ is red. Let $n$ be a positive integer, and let $B_n=V(H')\cap [n]\cap B$ . Because the vertices of $B_n$ form a monochromatic blue clique in our colouring of $E(K_{\mathbb{N}})$ , the set $B_n$ must be independent in $H'$ .

Let $j$ be the minimum value such that $\phi (B\cap [n])\subseteq \{b_1, b_2, \dots, b_{\beta _j}\}$ . We claim first that there are at least $(1-O(\epsilon ))j$ vertices in $[n]$ which do not belong to $H'$ . Indeed, let $B'_{\!\!n}=V(H')\cap \phi ^{-1}(\{b_1, b_2, \dots, b_{\beta _{j-1}}\})\subseteq B_n$ . From the construction of the colouring, the vertices that are connected to a vertex of $\{b_1, b_2, \dots, b_{\beta _{j-1}}\}$ through a red edge are precisely the red vertices to the left of $b_{\beta _{j-1}}$ , of which there are at most $\lambda (\beta _{j-1}-(j-1))$ . This means that $\mu (H,|B'_{\!\!n}|)\leq \lambda (\beta _{j-1}-(j-1))$ . For $j$ large enough, this implies $|B'_{\!\!n}|\leq (1+o(1))(\beta _{j-1}-(j-1))$ , and since the vertices in $\phi ^{-1}(\{b_1, b_2, \dots, b_{\beta _{j-1}}\})$ which are not in $B'_{\!\!n}$ are not in $V(H')$ ,

\begin{equation*}|[n]\setminus V(H')|\geq \beta _{j-1}-|B'_{\!\!n}|\geq (1+o(1))(j-1)-o(1)\beta _{j-1}.\end{equation*}

Since $\beta _j\leq \alpha _j+\beta _j=O(j)$ by (3), the right hand side in the inequality above is $(1-o(1))j$ .

Observe next that we cannot have $\beta _j=\beta _{j+1}$ . This is because $b_{\beta _j-1}$ , which is to the leftFootnote ³ of $b_{\beta _j}$ , has more than $\lambda ((\beta _j-1)-j)=\lambda (\beta _j-(j+1))$ red vertices to its left. We thus have, by minimality of $j$ , that $b_{\beta _{j+1}}\notin \phi ([n])$ , and by construction of $\phi$ we have $\phi ([n])\subset \{r_1,r_2,\dots, r_{\alpha _{j+1}}, b_1, b_2, \dots, b_{\beta _{j+1}}\}$ . This leads to the desired bound:

\begin{equation*}\frac {|V(H')\cap [n]|}{n}\leq 1-\frac {(1-o(1))j}{n}\leq 1-\frac {(1-o(1))j}{\alpha _{j+1}+\beta _{j+1}}\leq 1-\frac {1-o(1)}{\left (\frac {1-\gamma }{2}h(\gamma )+\epsilon +1\right )\frac {2\lambda }{1+\lambda }}\end{equation*}

which for $\epsilon$ small enough and $n$ large enough can take values arbitrarily close to $f(\lambda )$ .

The case in which $H'$ is monochromatic blue is analogous. Indeed, besides the direction of the rounding, it is equivalent to taking the function $-g(x)$ instead of $g(x)$ .

Proof of Theorem 1.41. Let $\chi\;:\;E(K_{\mathbb{N}})\rightarrow \{R,B\}$ be an edge-colouring. Let $\mathcal{F}$ be an inclusion-maximal family of pairwise disjoint monochromatic infinite cliques in $\chi$ . Then ${\mathbb{N}}\setminus V(\mathcal{F})$ is finite, because otherwise by Ramsey’s theorem there would be an infinite monochromatic clique in $\chi$ restricted to ${\mathbb{N}}\setminus V(\mathcal{F})$ , contradicting the maximality of $\mathcal{F}$ . Let $\mathcal{F}_R$ and $\mathcal{F}_B$ be the families of red and blue cliques in $\mathcal{F}$ . Since $\bar d(V(\mathcal{F}_R)\cup V(\mathcal{F}_B))=1$ , we have $\max \{\bar d(V(\mathcal{F}_R)),\bar d(V(\mathcal{F}_B))\}\geq 1/2$ . Wlog assume $\bar d(\mathcal{F}_R))\geq 1/2$ . We can suppose that $\mathcal{F}_R$ contains infinitely many cliques, because otherwise we can take one clique $K\in \mathcal{F}_R$ and divide it into infinitely many infinite cliques. Let $K_1, K_2, \dots$ , be the cliques in $\mathcal{F}_R$ . We can partition the vertex set of $H$ into infinitely many parts $S_1, S_2, \dots$ , each of which is made up of infinitely many components of $H$ . Now take any $\Phi\;:\;V(H)\rightarrow V(K_{\mathbb{N}})$ which is a bijection from each $S_i$ to each $K_i$ . The image of $H$ is a monochromatic graph $H'$ and $\bar d(V(H'))=\bar d(\mathcal{F}_R)\geq 1/2$ .

Proof of Theorem 3.1. Let $\chi\;:\;E(K_{\mathbb{N}})$ be given. Apply Lemma 3.3 to this edge-colouring to obtain an $a$ -good colouring where at most $a-1$ shades of each colour are non-empty. Assign the colour red to the vertices in $X$ . Apply Lemma 3.4 to obtain $C$ and $W$ . Remove from $W$ every component which uses a vertex of $X$ (this does not affect $\bar d(W)$ because it only removes finitely many vertices). By Lemma 3.5, we can find a monochromatic $H'\subseteq K_{\mathbb{N}}$ with $\bar d(H')\geq b/(a-1)\bar d(W)\geq b/(a-1)f(s/r)$ .

Proof of Lemma 3.3. For each vertex $v$ , we will denote by $c(v)$ and $s(v)$ the colour and the shade that we assign to it, respectively. The colour assigned to a vertex might change while the algorithm is running, but the shade of each vertex is final once assigned and it will match the colour that the vertex has at that time.

At some points, the shade assigning algorithm will call the basic colouring algorithm to colour an infinite set $V=\{v_1, v_2, \dots \}$ of vertices. We will first describe this algorithm.

Basic colouring algorithm: First, the colour $c(v_1)$ is assigned, satisfying that $N_{c(v_1)}(v_1)\cap V$ is infinite. Once the colours of $v_1, \dots, v_{n-1}$ have been assigned, assuming by induction that $ (\cap _{i=1}^{n-1}N_{c(v_i)}(v_i) )\cap V$ is infinite, the colour $c(v_n)$ is chosen so that $ (\cap _{i=1}^{n}N_{c(v_i)}(v_i) )\cap V$ is infinite.

The colouring produced satisfies that $ (\cap _{i=1}^{n}N_{c(v_i)}(v_i) )\cap V$ is infinite for every $n$ . We say that a colour $C$ is dominant in this colouring if, for every $n$ , $ (\cap _{i=1}^{n}N_{c(v_i)}(v_i) )\cap V$ contains infinitely many vertices $v$ with $c(v)=C$ . Observe that at least one of the colours is dominant.

Now we define the shade assigning algorithm:

1. 1. For every $v\in{\mathbb{N}}$ , start with $c(v)$ and $s(v)$ unassigned.

2. 2. If finitely many vertices $v$ remain with $s(v)$ unassigned, assign $s(v)=X$ and END.

3. 3. Let $V$ be the set of vertices without a shade. Colour $V$ with the basic colouring algorithm. Choose a colour $C$ that is dominant. Let $i$ be the minimum value such that $C_i$ is empty. For every $v\in V$ with $c(v)=C$ , set $s(v)=C_i$ .

4. 4. If $i=a-1$ , set $s(v)=\bar C_a$ for every $v\in V$ with $c(v)=\bar C$ and END. If $i\neq a-1$ , return to Step 2.

The algorithm runs the loop $2$ – $4$ at most $2a-3$ times before ending. Whenever a set $C_i$ with $i\leq a-1$ is defined, the colour $C$ is dominant in the corresponding colouring, meaning that in particular $ (\cap _{v\in S}N_C(v) )\cap C_i$ is infinite for every finite non-empty $S\subseteq C_i$ , as it is a superset of the colour $C$ vertices of $ (\cap _{i=1}^nN_{c(v_i)}(v_i) )\cap V$ for $n$ large enough. For the same reason, for any finite subset $S$ of vertices whose shade is not assigned when $C_i$ is defined, we have that $ (\cap _{v\in S}N_{\bar C}(v) )\cap C_i$ is infinite. If $C_a$ is defined at some point in the algorithm (namely at the end), then $\bar C_1, \bar C_2, \dots, \bar C_{a-1}, C_a$ are defined in this order. This proves that the colouring that we obtained is $a$ -good.

To conclude the proof of Lemma 3.3, simply observe that $X$ is nonempty only if the algorithm terminates at Step 2, the set $(R_a\cup B_{a-1})$ is nonempty only if the algorithm terminates at Step 4 with $C=B$ and $(B_a\cup R_{a-1})$ is nonempty only if the algorithm terminates at Step 4 with $C=R$ .

Proof. Take an orientation of every edge in $G$ from $X$ to $Y$ and give it an infinite capacity. Connect every vertex in $X$ to a source $\sigma$ through an edge with capacity $r$ , and every vertex in $y$ to a sink $\tau$ through an edge with capacity $s$ . Let $D$ be the maximum flow in this network. $D$ is the maximum value for which a function $h$ as in the statement exists (by the integrality theorem, there exists a maximum flow in which the flow of every edge is an integer). $D$ is also the minimum value for which a cut $(C_1, C_2)$ with $\sigma \in C_1$ and $\tau \in C_2$ exists. Observe that $(C_1, C_2)$ is a cut with finite capacity iff $(C_2\cap X)\cup (C_1\cap Y)$ is a vertex cover of $G$ , in which case the capacity of the cut is $r|C_2\cap X|+s|C_1\cap Y|$ . Our lemma follows from the Ford–Fulkerson theorem.

Proof. Let $\lambda =s/r$ . Our constants will follow the hierarchy

\begin{equation*}\eta, N^{-1}\ll \gamma \ll \kappa \ll \xi \ll \epsilon, \lambda .\end{equation*}

That is, after $\epsilon$ and $\lambda$ are given we pick $\xi$ small enough, after fixing $\xi$ we pick $\kappa$ small enough, and so on.

For every red vertex $v$ , we define its blue degree $d_B(v)$ as the number of blue vertices $w$ such that $vw$ is blue. Let $v_1, v_2, \dots, v_{|R|}$ be the set of red vertices, sorted from smallest to largest blue degree, and let $d_i=d_B(v_i)$ . Define additionally $d_0=0$ and $d_k=d_{|R|}$ for $k\gt |R|$ . Let $g\;:\;[0,+\infty )\rightarrow [0,+\infty )$ be the function that satisfies $g(k)=d_k$ for every integer $k$ and which is linear between every pair of consecutive integers.

By Lemma 3.9 there exists $\tau \in [\gamma n, \kappa n ]$ for which $\frac{\ell ^+_\lambda (g,\tau )+\ell ^-_\lambda (g,\tau )}{\tau }\geq \frac{f(\lambda )}{1-f(\lambda )}-\xi$ . Adding 1 on each side of the expression, $\frac{\ell ^+_\lambda (g,\tau )+\ell ^-_\lambda (g,\tau )+\tau }{\tau }\geq \frac{1}{1-f(\lambda )}-\xi$ . Let $t=\left (\frac{1}{1-f(\lambda )}-\xi \right )\tau$ . Then, since $t\leq \ell ^+_\lambda (g,\tau )+\ell ^-_\lambda (g,\tau )+\tau$ , we have either $|R\cap [t]|\lt \ell ^-_\lambda (g,\tau )$ or $|B\cap [t]|\leq \ell ^+_\lambda (g,\tau )+\tau$ . We consider both cases, in the former we will have $C=B$ and in the latter (mostly) $C=R$ :

Case 1: $|R\cap [t]|\lt \ell ^-(g,\tau )$ . Let $R'=R\cap [t]$ . Let $G'$ be the graph of blue edges in $G$ between $R'$ and $B$ . Let $h$ , $Z$ and $D$ be as in Lemma 3.7 applied to $G'$ , with $X=B$ and $Y=R'$ . Suppose that $D\leq s(|R'|-\tau )$ . Every vertex $v\in R'\setminus Z$ must have all its blue neighbours in $B\cap Z$ , and so $d_B(v)\leq |B\cap Z|$ . Therefore

\begin{equation*}d_{|R'|- |Z\cap R'|}\leq |Z\cap B|=\frac {D-s|Z\cap R'|}{r}\leq \frac sr(|R'|-|Z\cap R'|-\tau ).\end{equation*}

Setting $x=|R'|-|Z\cap R'|$ , this expression rearranges to $x-\frac{g(x)}\lambda \geq \tau$ , so by definition of $\ell ^-_\lambda$ this means that $x\geq \ell ^-_\lambda (g,\tau )$ . But this is a contradiction, because $x\leq |R'|\lt \ell ^-_\lambda (g,\tau )$ . This means that we have $D\gt s(|R'|-\tau )$ , and

\begin{equation*}\frac {|B\cap [t]|}{t}+\frac {D}{st}\geq \frac {t-|R'|}{t}+\frac {s(|R'|-\tau )}{st}=1-\frac \tau t=1-\frac {1}{\frac {1}{1-f(\lambda )}-\xi }\geq f(\lambda )-\epsilon .\end{equation*}

Case 2: $|B\cap [t]|\leq \ell ^+(g, \tau )+\tau$ . Let $B'=B\cap [t]$ . Let $G'$ be the graph of red edges between $R$ and $B'$ . Let $h$ , $Z$ and $D$ be as in Lemma 3.7 applied to $G'$ , with $X=R$ and $Y=B'$ . Suppose that $D\lt s(|B'|-\tau -\eta n-\frac 1\lambda )$ . Every edge between $R\setminus Z$ and $B'\setminus Z$ is blue. Every vertex $v$ has at most $\eta n$ vertices to which it is not connected, and so $d_B(v)\geq |B'\setminus Z|-\eta n$ for all $v\in R\setminus Z$ .

If $R\setminus Z$ is empty, then $r|R|\leq D\leq s|B'|\leq st\leq s\frac{\tau }{1-f(\lambda )}\leq s\frac{\kappa }{1-f(\lambda )}n$ . This leads to $|R|\leq \frac{\kappa s}{r(1-f(\lambda ))}n\leq (1-f(\lambda ))n$ and $|B|\geq f(\lambda )n$ . In this case we can take $t'=n$ , $h=0$ and $C=B$ for Lemma 3.10. Thus we can assume that $R\setminus Z$ is not empty, and so $d_{|R\cap Z|+1}\geq |B'\setminus Z|-\eta n$ .

\begin{align*} d_{|R\cap Z|+1} &\geq |B'|-|B'\cap Z|-\eta n\geq |B'|-\frac{D-r|R\cap Z|}{s}-\eta n\\[5pt] &=\frac{s|B'|-D}{s}+\frac 1\lambda |R\cap Z|-\eta n\geq \tau +\eta n+\frac 1\lambda +\frac 1\lambda |R\cap Z|-\eta n\\[5pt] &\geq \tau +\frac 1\lambda (|R\cap Z|+1). \end{align*}

Setting $x=\frac 1\lambda (|R\cap Z|+1)$ , this expression rearranges to $g(\lambda x)-x\geq \tau$ , so by definition of $\ell ^+_\lambda$ this means that $x\geq \ell ^+_\lambda (g,\tau )$ . On the other hand, $x=\frac{|R\cap Z|+1}{\lambda }\leq \frac{D}{s}+\frac 1\lambda \lt |B'|-\tau -\eta n-\frac 1\lambda +\frac 1\lambda \lt |B'|-\tau \leq \ell ^+_\lambda (g,\tau )$ , which is a contradiction. This means that we have $D\geq s(|B'|-\tau -\eta n-\frac 1\lambda )$ , and

\begin{equation*}\frac {|R\cap [t]|}{t}+\frac {D}{st}\geq \frac {t-|B'|}{t}+\frac {s(|B'|-\tau -\eta n-\frac 1\lambda )}{st}\geq 1-\frac \tau t-\frac \eta \gamma -\frac 1{\lambda \gamma N}\geq f(\lambda )-\epsilon .\end{equation*}

Proof of Lemma 3.4. Let $\lambda =s/r$ . We first claim that, for every $\epsilon \gt 0$ , there exists $\gamma (\epsilon )\gt 0$ and $N(\epsilon )$ such that, for every $n\gt N$ , there exist $t\in [\gamma n,n]$ , a colour $C$ and a subgraph $\mathcal{F}\subseteq K_{\mathbb{N}}$ contained in $[n]$ in which every component is either an isolated vertex of colour $C$ or a $K_{r,s}^C$ using only a shade of each colour, with

\begin{equation*}\frac {|V(\mathcal {F})\cap [t]|}{t}\geq f(\lambda )-\epsilon .\end{equation*}

Let $a$ be the total number of shades (from both colours). Our constants will follow the hierarchy

\begin{equation*}N^{-1}\ll M^{-1}\ll \rho \ll \delta \ll \zeta \ll \gamma, \eta \ll \epsilon,r^{-1},s^{-1},a^{-1}.\end{equation*}

Let $G$ be the restriction of our colouring to $[n]$ . Take a partition of $[n]$ into $\ell =a\lceil \rho ^{-1}\rceil$ parts $\{Z_1, \dots, Z_\ell \}$ , such that each $Z_i$ is contained in one shade, and $\max Z_i-\min Z_i\lt \rho n$ . Applying Lemma 3.6 to $G$ with $d=2\delta$ , we find a subgraph $H\subseteq G$ and a partition $[n]=\{V_0, V_1, \dots, V_m\}$ , with $\ell \leq m\leq M$ , as in the statement of Lemma 3.6, replacing $\epsilon$ with $\delta$ .

We suppose that the labelling of the parts is such that $\min V_1\lt \min V_2\lt \cdots \lt \min V_m$ . We define an auxilliary graph $H'$ as follows: the vertex set is $[m]$ . The colour of every vertex $i$ is the same as the colour of each of its vertices in $G$ . Between any two vertices $ij$ , we draw an edge if the bipartite graph $V_iV_j$ is nonempty in $H$ , and we colour it in the most dense colour in $V_iV_j$ .

Let $y=|V_1|=\dots =|V_m|$ . Then $\frac{(1-\delta )n}{m}\leq y\leq \frac{n}{m}$ . The minimum degree in $H'$ is at least $(1-\eta )m$ . Indeed, given $i$ and $v\in V_i$ , we have $d_{H'}(i)\geq \frac{d_H(v)-\delta n}{y}\geq \frac{d_{G}(v)-4\delta n}{y}\geq (1-\frac{4\delta }{1-\delta })m\gt (1-\eta )m$ .

Apply Lemma 3.10 to $H'$ , with parameters $\epsilon/2, r,s$ to obtain a colour $C$ , a function $h\;:\;E(H')\rightarrow{\mathbb{N}}$ and a value $\tau \in [\gamma m,m]$ as in the statement of Lemma 3.10, replacing $t$ with $\tau$ . Subdivide each $V_i$ with colour $C$ into $r$ parts $V_{i,1},\dots, V_{i,r}$ , each of size at least $\lfloor y/r\rfloor$ and each $V_i$ with colour $\bar C$ into $s$ parts $V_{i,1},\dots, V_{i,s}$ , each of size at least $\lfloor y/s\rfloor$ . Construct a matching $\mathcal{M}$ of pairs $(V_{i,k}, V_{j,k'})$ , where for any fixed values of $i$ and $j$ , the number of pairs $(V_{i,k},V_{j,k'})$ in $\mathcal{M}$ is $h(ij)$ .

Within each pair $(V_{i,k}, V_{j,k'})$ , where $V_i$ has colour $C$ and $V_j$ has colour $\bar C$ , find a maximum family $\mathcal{F}_{i,k,j,k'}$ of disjoint copies of $K_{r,s}^C$ . Since $N\gg M,\delta ^{-1},r,s$ , and therefore $\delta y\gg r,s$ , then $\min \{|V_{i,k}\setminus V(\mathcal{F}_{i,k,j,k'})|,|V_{j,k'}\setminus V(\mathcal{F}_{i,k,j,k'})|\}\lt \delta y$ . That is because otherwise the bipartite graph between $V_{i,k}\setminus V(\mathcal{F}_{i,k,j,k'})$ and $V_{j,k'}\setminus V(\mathcal{F}_{i,k,j,k'})$ would have density at least $ \delta$ in the edges of colour $C$ , and for $\delta y$ large enough this implies the existence of a copy of $K_{r,s}^C$ , which would contradict the maximality of $\mathcal{F}_{i,k,j,k'}$ .

Let $\mathcal{F}$ be the union of all families $\mathcal{F}_{i,k,j,k'}$ . Let $t=\min V_\tau$ . We will now bound $\frac{|(V(\mathcal{F})\cup C)\cap [t]|}{t}$ . If $v\geq t+\rho n$ , and $v\in V_i$ with $i\neq 0$ , then $\min V_i\gt \max V_i-\rho n\geq v-\rho n\geq t=\min V_\tau$ , and thus $i\gt \tau$ . This means that $|(\cup _{i=1}^\tau V_i)\setminus [t]|\leq \rho n$ , and $t\geq \tau y-\rho n\geq \frac{(1-\delta )\tau n}{m}-\rho n$ . On the other hand, if $v\leq t$ then either $v\in V_0$ or $v\in V_i$ with $\min V_i\leq v\leq t=\min V_\tau$ , and thus $i\leq \tau$ . This implies that $t\leq \sum _{i=0}^\tau |V_i|\leq \delta n+\tau y\leq \delta n+\frac{\tau n}m$ .

Every $V_i$ with colour $C$ and $i\in [\tau ]$ will trivially be contained in $(V(\mathcal F)\cup C)\cap (\cup _{i=1}^\tau V_i)$ . For any $V_i$ with colour $\bar C$ and $i\in [\tau ]$ , there are $\sum _{e\ni i}h(e)$ parts $V_{i,k}$ which are paired up with a different part $V_{j,k'}$ . We either have $|V_{i,k}\setminus V(\mathcal{F})|\leq \delta y$ or $|V_{j,k'}\setminus V(\mathcal{F})|\leq \delta y$ . In the first case, $|V_{i,k}\cap V(\mathcal{F})|\geq \lfloor y/s\rfloor -\delta y\geq (1/s-1/y-\delta )y$ . In the second case, $|V_{j,k'}\cap V(\mathcal{F})|\geq \lfloor y/r\rfloor -\delta y\geq (1/r-1/y-\delta )y$ . But $\mathcal{F}$ is a family of copies of $K_{r,s}$ , so $|V_{i,k}\cap V(\mathcal{F})|=\frac rs|V_{j,k'}\cap V(\mathcal{F})|\geq (1/s-\lambda ^{-1}(1/y+\delta ))y$ . In either case we have $|V_{i,k}\cap V(\mathcal{F})|\geq (1-\zeta )y/s$ .

Putting our bounds together:

\begin{align*} \frac{|(V(\mathcal{F})\cup C_{G})\cap [t]|}{t} & \geq \frac{|(V(\mathcal{F})\cup C_{G})\cap (\cup _{i=1}^\tau V_i)|-\rho n}{t}\\[5pt] & \geq \frac{y|C_{H'}\cap [\tau ]|}{t}+\left (1-\zeta \right )\frac ys\frac{\sum _{v\in (\bar C_{H'}\cap [\tau ])}\sum _{e\ni v}h(e)}{t}-\frac{\rho n}{t}\\[5pt] & \geq \left (1-\zeta \right )\frac{\tau y}{t}\left (\frac{|C_{H'}\cap [\tau ]|}{\tau }+\frac{\sum _{v\in (\bar C_{H'}\cap [\tau ])}\sum _{e\ni v}h(e)}{s\tau }\right )-\frac{\rho n}{t}\\[5pt] & \geq \left (1-\zeta \right )\frac{\tau y}{t}\left (f(\lambda )-\frac \epsilon 2\right )-\frac{\rho }{\frac{\tau (1-\delta )}{m}-\rho }\\[5pt] & \geq \left (1-\zeta \right )\frac{\tau y}{\delta n+\tau y}\left (f(\lambda )-\frac \epsilon 2\right )-\frac{\rho }{\gamma (1-\delta )-\rho }\\[5pt] & \geq \left (1-\zeta \right )\frac{1}{1+\delta \frac{n}{my}\frac m\tau }\left (f(\lambda )-\frac \epsilon 2\right )-\frac \epsilon 4\\[5pt] & \geq \left (1-\zeta \right )\frac{1}{1+\frac{\delta }{(1-\delta )\gamma }}\left (f(\lambda )-\frac \epsilon 2\right )-\frac \epsilon 4\\[5pt] & \geq f(\lambda )-\epsilon . \end{align*}

To conclude the proof of our initial claim, notice that $t\geq \left (\frac{(1-\delta ) \tau } m-\rho \right )n\geq ((1-\delta )\gamma -\rho ) n\geq \gamma 'n$ for a constant $\gamma '\gt 0$ .

We are now ready to construct $W$ . Take a sequence $f(s/r)\gt \epsilon _1\gt \epsilon _2\gt \cdots \gt 0$ with $\epsilon _i\rightarrow 0$ . Start by applying the claim with $\epsilon =\epsilon _1$ and $n_1=N(\epsilon )$ to obtain a subgraph $\mathcal{F}_1$ with colour $C_1$ with density at least $f(s/r)-\epsilon _1$ in $[t_1]$ . Now proceed by induction and set $n_i=\max \{N(\epsilon _i/2), 2n_{i-1}(r+s)/(\epsilon _i\gamma (\epsilon _i/2))\}$ . Applying the claim with $\epsilon =\epsilon _i/2$ we find a subgraph $\mathcal{F}_i'$ with colour $C_i$ contained in $[n_i]$ and with density at least $f(s/r)-\epsilon _i/2$ in $[t_i]$ , for some $t_i\in [\gamma (\epsilon )n_i,n_i]$ . Remove from $\mathcal{F}_i'$ all components that intersect $[n_{i-1}]$ (this represents at most $n_{i-1}(r+s)$ vertices) to obtain $\mathcal{F}_i$ . Then $\mathcal{F}_i$ is disjoint from all previous $\mathcal{F}_j$ , and by the choice of $n_i$ , it still has density at least $f(s/r)-\epsilon _i$ in $[t_i]$ .

Select a colour $C$ such that $C_i=C$ for infinitely many $i$ . Let $W=\cup _{C_i=C}\mathcal{F}_i$ . Then by construction $\bar d(W)\geq f(s/r)$ , since the $t_i$ tend to infinity, and the components of $W$ are isolated vertices of colour $C$ or $K_{r,s}^C$ . This concludes the proof of Lemma 3.4.

Proof of Lemma 3.5. Without loss of generality, assume that $C$ is red, let $S_j$ denote the vertices in $W$ of shade $R_j$ , plus the blue vertices contained in a copy of $K_{r,s}^R$ in $W$ in which the red side has shade $R_j$ . Removing from $W$ the finite sets $S_j$ does not affect its density, so suppose that each $S_j$ is either empty or infinite. We will show that there exists a set $J\subseteq [a]$ , of size $b$ , such that $\bar d(\cup _{j\in J}S_j)\geq b/a'\bar d(W)$ .

By definition of density, there exists a sequence $n_1\lt n_2\lt \dots$ of positive integers such that $|V(W)\cap [n_i]|/n_i\rightarrow \bar d(W)$ . For each $i$ there exists a subset $J_i\subseteq [a]$ of $b$ indices such that

\begin{equation*}\frac {|(\cup _{j\in J_i}S_j)\cap [n_i]|}{n_i}\geq \frac {b}{a'}\frac {|V(W)\cap [n_i]|}{n_i}.\end{equation*}

For infinitely many $i$ , the set $J_i$ is the same, which we denote $J$ . By taking an appropriate subsequence of $n_1, n_2, \dots$ , we can suppose without loss of generality that $J_i=J$ for all $i$ and that $n_{i+1}/n_i\rightarrow \infty$ . Let $\mathcal{F}_i$ be the union of components from $W$ contained in some $S_j$ with $j\in J$ , which contain a vertex from $[n_i]$ but no vertex from $[n_{i-1}]$ . Let $\mathcal{I}=\{I_1, I_2, \dots \}$ be the family of doubly independent sets. We can suppose that the elements in $\mathcal{I}$ are such that the sets $I_i\cup N(I_i)$ are pairwise disjoint. Indeed, because $H$ is locally finite, each $I_i\cup N(I_i)$ intersects finitely many sets $I_j\cup N(I_j)$ , so we can find an infinite subfamily $\mathcal{I}'$ by including in it only the sets $I_i$ such that $I_i\cup N(I_i)$ does not intersect a set $I_j\cup N(I_j)$ for some $j\lt i$ with $I_j\in \mathcal{I'}$ . This does not change the components in which $\mathcal{I}$ is concentrated.

Let $J'\subseteq J$ be the set of indices in $j$ for which $S_j$ is non-empty. We assign to each component $\mathcal C\subseteq H$ a number $\kappa (\mathcal{C})\in J'$ , in such a way that for every $j\in J'$ there are infinitely many sets $I_i$ in components with $\kappa (\mathcal{C})=j$ . Indeed, if finitely many components intersect $\mathcal{I}$ , there are at least $b$ components that contain infinitely many elements of $\mathcal{I}$ , so give different values of $\kappa (\mathcal{C})$ to $|J'|\leq b$ of them, whereas if there are infinitely many components that intersect $\mathcal{I}$ , we can assign each value of $J'$ to infinitely many of them. The purpose of $\kappa (\mathcal{C})$ will be to identify the shade of red to be used in the vertices while embedding $\mathcal{C}$ in the red edges of $\chi$ .

We will define an injective graph homomorphism $\Phi\;:\;H\rightarrow K_{\mathbb{N}}$ which maps edges to red edges, whose image contains $\mathcal{F}_i$ for infinitely many $i$ . This is enough to prove Theorem 3.1, because for infinitely many large enough values of $i$ we have

\begin{align*} \frac{|\Phi (V(H))\cap [n_i]|}{n_i}\geq & \frac{|V(\mathcal{F}_i)\cap [n_i]|}{n_i}\geq \frac{b}{a'}\frac{|V(W)\cap [n_i]|}{n_i}-\frac{(r+s)n_{i-1}}{n_i}\\[5pt]\geq &\frac{b}{a'}\bar d(W)-o(1). \end{align*}

We will define $\Phi$ in steps. Recall that $\Psi$ here denotes the proper vertex colouring of $H$ from Theorem 3.1. On every step, we will define the image of finitely many vertices of $H$ . After every step, the following conditions must hold. Let $u,v$ be two adjacent vertices in some component $\mathcal{C}$ of $H$ , such that $\Phi (v)$ is defined and $\Phi (u)$ is not. Then:

• If $\kappa (\mathcal{C})\neq a$ , then $\Phi (v)\in R_{\kappa (\mathcal{C})}$ .

• If $\kappa (\mathcal{C})=a$ and $\Psi (v)=a$ , then $\Phi (v)\in R_a$ .

• If $\kappa (\mathcal{C})=a$ and $\Psi (v)\neq a$ , then $\Phi (v)\in B_{\Psi (v)}$ and $\Psi (u)\lt \Psi (v)$ .

The algorithm will consist of two operations that alternate: defining the image of a vertex $v\in V(H)$ and adding some $\mathcal{F}_i$ to the image of $\Phi$ . If we identify $V(H)$ with $\mathbb{N}$ and always apply the first operation to the least vertex $v$ with undefined $\Phi (v)$ , at the end of the algorithm $\Phi (v)$ will be defined for every vertex in $V(H)$ .

Define the image of a vertex $v\in V(H)$ : Suppose first that $v\in \mathcal{C}$ and $\kappa (\mathcal{C})=k\neq a$ . Let $w_1, \dots, w_q$ be the neighbours of $v$ which have $\Phi (w_i)$ defined. By our invariant, the vertices $\Phi (w_i)$ all have shade $R_k$ , and therefore there are infinitely many vertices $x$ in shade $R_k$ which are connected to every $\Phi (w_i)$ through a red edge. Select one such $x$ which is not yet in the image of $\Phi$ and set $\Phi (v)=x$ .

Now suppose that $\kappa (\mathcal{C})=a$ . Let $\Psi (v)=k$ . For every edge $uw$ of $H$ , define an orientation $\overrightarrow{uw}$ such that $\Psi (u)\lt \Psi (w)$ . Let $T$ be the set of vertices that can be reached from $v$ in this orientation. Because $T$ is connected, does not contain a path of length greater than $a$ , and the degree of every vertex is finite, $T$ is finite. Also, $T$ does not have an oriented cycle. Observe that, by our invariant, if $\overrightarrow{uw}$ is an edge and $\Phi (u)$ is defined, then $\Phi (w)$ is defined.

Now define $\Phi (w)$ for every $w\in T$ for which the image is still undefined, in decreasing order of $\Psi (w)$ . If $\Psi (w)=a$ , choose an arbitrary vertex $x\in R_a$ which is not yet the image of any vertex and set $\Phi (w)=x$ . If $\Psi (w)=k\lt a$ , then for every $w'\in N^+(w)$ the image $\Phi (w')$ is defined and in $B_{k+1}\cup \dots \cup B_{a-1}\cup R_a$ . By the properties of $a$ -good colourings, there are infinitely many verticesFootnote ⁴ $x\in B_k$ which are connected to every $\Phi (w')$ through a red edge. Choose one which is not yet in the image of $\Phi$ and set $\Phi (w)=x$ .

Add some set $\mathcal{F}_i$ to the image: Select some $\mathcal{F}_i$ which is so far disjoint with the image of $\Phi$ . For each $K_{r,s}^R$ component $Z\subseteq \mathcal{F}_i\cap S_j$ , choose a doubly independent set $I\in \mathcal{I}$ in a component $\mathcal{C}$ with $\kappa (\mathcal{C})=j$ , such that no vertex from $I\cup N(I)\cup N(N(I))$ has a defined image. If $V(Z)=X\cup Y$ with $|X|=r$ blue and $|Y|=s$ red, then set $\Phi$ to be bijective from $I$ to $X$ , and injective from $N(I)$ to $Y$ . The vertices $v$ of $\mathcal{F}_i\cap S_j$ that remain outside of the image at this point all have shade $R_{j}$ . For each of them, choose a vertex $w$ with $\Psi (w)=a$ in a component $\mathcal{C}$ with $\kappa (\mathcal{C})=j$ (there are infinitely many of these vertices), such that no vertex in $w\cup N(w)$ has a defined image and set $\Phi (w)=v$ . After doing this for every isolated vertex in $\mathcal{F}_{i}\cap S_j$ for every $j\in J'$ , the set $\mathcal{F}_i$ is contained in the image.

After both steps are applied alternatingly infinitely many times, the image of $\Phi$ is a monochromatic red graph $H'\subseteq K_{\mathbb{N}}$ which contains infinitely many sets $\mathcal{F}_i$ , and therefore $\bar d(H')\geq b/a'\bar d(W)$ .

Proof. Let $\Psi\;:\;V(H)\rightarrow [a]$ be a proper colouring, in which in every component of $H$ the most common colour is $a$ . Without loss of generality suppose that $C$ is red. As in the proof of Lemma 3.5, there exists $J'$ with $|J'|\leq b$ , such that $\bar d(\cup _{j\in J'}R_j)\geq b/a'$ , and all $R_j$ with $j\in J'$ are infinite. Let $\mathcal{F}=\cup _{j\in J'}R_j$ . Define a function $\kappa$ from the components of $H$ to $J'$ for which the pre-image of every value contains infinitely many vertices. The algorithm now alternates between define the image of a vertex $v\in V(H)$ , as above, and add a vertex of $\mathcal{F}$ to the image. At the end of the procedure, we obtain a red $H'\subseteq K_{\mathbb{N}}$ isomorphic to $H$ which contains $\mathcal{F}$ and thus has density at least $\bar d(\mathcal{F})\geq b/a'\bar d(R)$ .

Add a vertex of $\mathcal{F}$ to the image: Let $v\in \mathcal{F}$ be a vertex in $R_j$ . Choose a vertex $w$ in a component $\mathcal{C}$ with $\kappa (\mathcal{C})=j$ , such that no vertex in $w\cup N(w)$ has a defined image and with $\Psi (w)=a$ and set $\Phi (w)=v$ .

Proof. The upper bound follows from Theorem 1.6. We will show that, for every $\epsilon \gt 0$ , we have $\rho (H)\geq f(\lambda )-\epsilon$ . Our goal is to show that $H$ satisfies the condition of Theorem 3.1 for $a=2$ , $b=1$ , a certain colouring $\Psi$ and some doubly independent sets $I'_i$ . Let $\Psi\;:\;V(H)\rightarrow \{1,2\}$ be a proper colouring. Choose $\lambda '\gt \lambda$ such that $f(\lambda ')\gt f(\lambda )-\epsilon$ (it exists by continuity of $f$ ). There exist infinitely many pairwise disjoint independent sets $I_i$ , all with the same size, such that $\frac{|N(I_i)|}{|I_i|}\leq \lambda '$ (by the condition from the statement). Partition each set $I_i$ into non-empty sets $I_{i,1}, \dots, I_{i,k_i}$ , where each vertex $v$ is classified according to its colour by $\Psi$ and the component it belongs to. If two vertices $v,w$ have a common neighbour, then they are in the same component and $\Psi (u)=\Psi (v)$ . For this reason, $|N(I_i)|=\sum _{j=1}^{k_i}|N(I_{i,j})|$ . There exists some $\tau _i$ such that

\begin{equation*}\frac {|N(I_{i,\tau _i})|}{|I_{i,\tau _i}|}\leq \frac {\sum _{j=1}^{k_i}|N(I_{i,j})|}{\sum _{j=1}^{k_i}|I_{i,j}|}=\frac {|N(I_i)|}{|I_i|}\leq \lambda '.\end{equation*}

Set $I'_i=I_{i,\tau _i}$ . Set $r_i=|I'_i|$ and $s_i=|N(I'_i)|$ . There is a pair $(r,s)$ satisfying $(r,s)=(r_i,s_i)$ for infinitely many values of $i$ . Considering only the values of $i$ for which this equality holds, we have our set of independent sets. Note that, because $H$ is bipartite, $N(I'_i)$ is monochromatic and thus independent, meaning that $I'_i$ is doubly independent. If $\Psi (N(I'_i))=2$ does not hold for infinitely many $i$ , replace $\Psi$ with $\bar \Psi =3-\Psi$ . We can now apply Theorem 3.1 to obtain $\rho (H)\geq f(s/r)\geq f(\lambda ')$ .

Proof of Theorem 1.7. Let $\lambda =\liminf \limits _{n\rightarrow \infty }\frac{\mu (H,n)}{n}$ . Fix $\lambda '\gt \lambda$ . We will show that, in both cases, there exist infinitely many pairwise disjoint independent sets $I_1, I_2, \dots$ , all with the same size, with $\frac{|N(I_i)|}{|I_i|}\leq \lambda '$ .

For graphs with infinite orbits: Choose $n$ such that $\frac{\mu (H,n)}{n}\lt \lambda '$ . Let $I$ be an independent set of size $n$ with $|N(I)|=\mu (H,n)$ . We will show that there are infinitely many automorphisms $\sigma _i\in \textrm{Aut}(H)$ such that the sets $\sigma _i(I)$ are pairwise disjoint. Then we can take $I_i=\sigma _i(I)$ to conlude the proof. We proceed by induction on $n$ . For $n=1$ , if $I=\{v\}$ , this is equivalent to the orbit of $v$ being infinite.

Suppose that the result is true for $n-1$ . Suppose that we have already found $\sigma _1, \sigma _2, \dots, \sigma _k$ such that the sets $\sigma _i(I)$ are pairwise disjoint. Let $X=\cup _{i=1}^k\sigma _i(I)$ . We will construct $\sigma _{k+1}\in \textrm{Aut}(H)$ such that $\sigma _{k+1}(I)$ is disjoint from $X$ . Choose $v\in I$ . By the induction hypothesis, there is an infinite family $\{\tau _i\}_{i=1}^\infty \subseteq \textrm{Aut}(H)$ such that the sets $\tau _i(I-v)$ are pairwise disjoint. If $\tau _i(v)\not \in X$ for some $i$ , then we can take $\sigma _{k+1}=\tau _i$ , and we are done. Therefore, assume that $\tau _i(v)\in X$ for every $i$ . By pigeonhole principle, there exists $w$ such that $\tau _i(v)=w$ for infinitely many $i$ . Choose $\phi \in \textrm{Aut(H)}$ such that $\phi (w)\not \in X$ (it exists because the orbit of $w$ is infinite). The set $\phi ^{-1}(X)$ intersects finitely many sets $\tau _i(I-v)$ , therefore there exists some $i$ with $\tau _i(I-v)$ disjoint from $\phi ^{-1}(X)$ and $\tau _i(v)=w$ . Putting this together, $\phi (\tau _i(I))$ is disjoint from $X$ , as we wanted.

For forests: The following lemma will be used to find independent sets of bounded size with bounded expansion within larger independent sets:

Knowing this lemma, choose $\lambda ''\lt \lambda '''\in (\lambda,\lambda ')$ and set $M=M(\lambda ''',\lambda ')$ . Suppose that we have already constructed pairwise disjoint independent sets $I_1, I_2, \dots, I_k$ with $|I_i|\leq M$ and $|N(I_i)|\leq \lambda '|I_i|$ . We will find a new set $I_{k+1}$ , disjoint from the others. Let $S=\cup _{i=1}^kI_i$ . There exists $n$ large enough such that $\frac{n}{n-|S|}\leq \frac{\lambda '''}{\lambda ''}$ . By definition of $\liminf$ and $\mu (H,n)$ , there exists an independent set $I$ with $|I|\geq n$ and $|N(I)|\leq \lambda ''|I|$ . Then

\begin{equation*}|N(I\setminus S)|\leq |N(I)|\leq \lambda ''|I|\leq \lambda ''(|I\setminus S|+|S|)\leq \lambda '''|I\setminus S|.\end{equation*}

By our claim, there exists $I_{k+1}\subseteq I\setminus S$ such that $|I_{k+1}|\leq M$ and $|N(I_{k+1})|\leq \lambda ' |I_{k+1}|$ . Once we have constructed an infinite family of independent sets $I_1, I_2, \dots$ , simply take a pair $(r,s)$ which is equal to $(|I_i|, |N(I_i)|)$ for infinitely many $i$ (which is possible because this pair can only take finitely many values), and we are done.

Proof of Lemma 4.2. Let $\delta =\delta (\lambda,\lambda ')\gt 0$ be small enough, which we will fix later. Let $F$ be the forest with vertex set $I\cup N(I)$ and containing only the edges between $I$ and $N(I)$ in our original graph. It is enough to prove our result in $F$ . Denote $J=N(I)$ . For every component of $F$ take a vertex of $I$ as the root.

There exists a set $S\subseteq V(F)$ with $|S|\leq \delta |V(F)|$ , satisfying that every component of $F\setminus S$ has size at most $\delta ^{-1}$ . Indeed, start with $S=\emptyset$ and consider the set $U$ of vertices whose component in $F\setminus S$ contains at least $\delta ^{-1}$ vertices. The rooted forest structure in $F$ induces a rooted forest structure in $F\setminus S$ . Let $U'$ be the set of vertices in $F\setminus S$ which have at least $\delta ^{-1}-1$ descendants. If $U\neq \emptyset$ then $U'\neq \emptyset$ , because the root of the largest component will be in $U'$ . Select a minimal vertex $v$ in $U'$ and add it to $S$ . This removes $v$ and all its descendants from $U$ and thus reduces the size of $U$ by at least $\delta ^{-1}$ . After at most $\delta |V(F)|$ steps, $U$ will be empty.

Let $X$ be the union of $S$ and the parents of the vertices of $S\cap J$ . This set has $|X|\leq 2|S|\leq 2\delta |V(F)|$ , and every component $T$ of $F\setminus X$ (which has the structure of a rooted tree) is adjacent to at most one vertex in $X\cap J$ , in which case that vertex is the parent (in $F$ ) of the root of $T$ . This is because a vertex $v$ in $T$ cannot have a child in $X\cap J$ , as that child would be in $S\cap J$ and $v$ would be its parent, and hence in $X$ . As a consequence, every component of $F\setminus (X\cap I)$ contains at most one vertex from $X\cap J$ .

Let $\mathcal{C}=\{C_1, \dots, C_k\}$ be the components of $F\setminus (X\cap I)$ . Then

\begin{equation*}\frac {\sum _{j=1}^k|C_i\cap J|}{\sum _{j=1}^k|C_i\cap I|}= \frac {|J|}{|I|-|X\cap I|}\leq \frac {|N(I)|}{|I|-2\delta (|I|+|N(I)|)}\leq \frac {\lambda }{1-2\delta (1+\lambda )}\;=\!:\;\lambda ''.\end{equation*}

There exists some component $C_i$ such that $|C_i\cap J|\leq \lambda ''|C_i\cap I|$ . If $C_i\cap I$ has size not greater than $M\;:\!=\;2\delta ^{-1}$ , then set $I'=C_i\cap I$ and we are done, because $N(I')\subseteq C_i\cap J$ . Otherwise $C_i$ has size greater than $2\delta ^{-1}$ , hence it must contain a (unique) vertex $v\in X\cap J$ . Let $C'_{\!\!1}, C'_{\!\!2}, \dots, C'_{\!\!r}$ be the components obtained from $C_i$ by removing $v$ , labelled in decreasing order of $|C'_{\!\!j}\cap J|/|C'_{\!\!j}\cap I|$ . Consider the minimum integer $t$ such that $\sum _{j=1}^t|C'_{\!\!j}\cap I|\geq \delta ^{-1}$ . Because every component in $F\setminus X$ has size at most $\delta ^{-1}$ , we have $\sum _{j=1}^t|C'_{\!\!j}\cap I|\leq \sum _{j=1}^{t-1}|C'_{\!\!j}\cap I|+\delta ^{-1}\leq 2\delta ^{-1}=M$ . Set $I'=\cup _{j=1}^t(C'_i\cap I)$ . Then

\begin{equation*}\frac {|N(I')|}{|I'|}=\frac {1+\sum _{j=1}^t|C'_{\!\!j}\cap J|}{\sum _{j=1}^t|C'_{\!\!j}\cap I|}\leq \delta +\frac {\sum _{j=1}^r|C'_{\!\!j}\cap J|}{\sum _{j=1}^r|C'_{\!\!j}\cap I|}\leq \delta +\lambda ''.\end{equation*}

This proves Lemma 4.2, for $\delta \gt 0$ small enough such that $\delta +\lambda ''\lt \lambda '$ .

Proof of Corollary 1.8. We will show that $\mu (T_k,n)=kn$ . For every independent set $I$ of size $n$ , the set of children of the vertices of $I$ has size $kn$ and is contained in $N(I)$ , thus $|N(I)|\geq kn$ . Equality can hold, for example for $I=\{v_1, \dots, v_n\}$ where $v_1$ is the root of $T_k$ and $v_{i+1}$ is a grandchild of $v_i$ . We therefore have $\mu (T_k,n)=kn$ . Since $T_k$ is a forest, Theorem 1.7 applies and $\rho (T_k)=f(k)$ .

Proof of Corollary 1.9. Let $I$ be an independent set. The set $I+(1,0,\dots, 0)$ is contained in $N(I)$ , so $|N(I)|\geq |I|$ and $\mu (\mathrm{Grid}_d,n)\geq n$ for all $n$ . On the other hand, let $I_k$ be the set of vertices in $[2k]^d$ with odd sum of coordinates. $I_k$ is an independent set of size $(2k)^d/2$ , and $I_k\cup N(I_k)$ is contained in $[2k+2]^d-(1,1,\dots, 1)$ . Since $I_k$ and $N(I_k)$ are disjoint,

\begin{equation*}\frac {|N(I)|}{|I|}=\frac {|I\cup N(I)|}{|I|}-1\leq \frac {(2k+2)^d}{(2k)^d/2}-1,\end{equation*}

which tends to 1 as $k\rightarrow \infty$ . We have $\liminf \limits _{n\rightarrow \infty }\frac{\mu (\mathrm{Grid_d},n)}{n}=1$ . The graph $\mathrm{Grid}_d$ is vertex-transitive, so by Theorem 1.7 we have $\rho (\mathrm{Grid}_d)=f(1)$ .

Proof of Corollary 1.10. The graph $\omega \cdot F$ satisfies that every orbit of the automorphism group on $V(\omega \cdot F)$ is infinite (because it intersects every copy of $F$ ), so we are in the setting of Theorem 1.7. We need to show that $\liminf \limits _{n\rightarrow \infty }\frac{\mu (\omega \cdot F,n)}{n}=\min \limits _{I\subseteq V(F)\text{ indep.}}\frac{|N(I)|}{|I|}$ .

Let $I$ be an independent set in $F$ that minimises $\frac{|N(I)|}{|I|}$ , and let $J\subseteq V(\omega \cdot F)$ be an independent set of size $n$ . Partition $J$ into independent sets $J_1, J_2,\dots, J_m$ , according to the component in which the vertices are contained. Then

\begin{equation*}\frac {|N(J)|}{|J|}=\frac {\sum _{i=1}^m|N(J_i)|}{\sum _{i=1}^m|J_i|}\geq \min \frac {|N(J_i)|}{|J_i|}\geq \frac {|N(I)|}{|I|}.\end{equation*}

This implies that $\frac{\mu (\omega \cdot F,n)}{n}\geq \frac{|N(I)|}{|I|}$ . Equality holds infinitely many times, since for all $n$ divisible by $I$ we can take the union of the sets $I$ in $n/|I|$ different copies of $F$ . Therefore $\rho (\omega \cdot F)=f\left (\liminf \limits _{n\rightarrow \infty }\frac{\mu (\omega \cdot F,n)}{n}\right )=f\left (\frac{|N(I)|}{|I|}\right )$ .

In an even cycle $C_{2k}$ , each independent set satisfies $|N(I)|\geq |I|$ , because $C_{2k}$ contains a perfect matching. Since each chromatic class in the bipartition satisfies $|N(I)|=|I|$ , we have $\rho (\omega \cdot C_{2k})=f(1)$ . For $1\leq a\leq b$ , in $K_{a,b}$ , every independent set has size at most $b$ , and its neighbourhood has size at least $a$ . Both inequalities are sharp if $I$ is the side of the bipartition with size $b$ . Thus $\frac{|N(I)|}{|I|}\geq \frac ab$ , and $\rho (\omega \cdot K_{a,b})=f(a/b)$ .

Proof of Theorem 1.11. Let $a=|V(F)|$ , and let $b=a-1$ . Let $\Psi\;:\;V(F)\rightarrow [a]$ be a colouring that assigns the value $a$ to every vertex in $N(I)$ , and where the remaining vertices in $F$ all get different values in $[a-1]$ . Because $I$ is doubly independent, this is a proper colouring. $\Psi$ extends to a colouring of $\omega \cdot F$ , by colouring all copies of $F$ equally.