Proof. Choose any constant $D\gt \max \{1,3p^{-(1/\delta ^2)}\}$ . Let $A, B$ be the vertex bipartition of $G$ with $|A| = |B| = n$ , let $H$ be an $n$ -vertex graph and let $k, l \ge \delta n$ . There are at most $n^n$ choices of distinct vertices $x_1, \ldots, x_k, y_1, \ldots, y_l \in V(H)$ and at most $n^{2n}$ choices of disjoint vertex sets $X_1, \ldots, X_k \subseteq A, \ Y_1, \ldots, Y_l \subseteq B$ . Consider a fixed such choice satisfying the premises in Definition 10 and the random event that for every pair $(i,j) \in [k]\times [l]$ such that $x_iy_j \in E(H)$ , not all of the potential edges between $X_i$ and $Y_j$ are included in $G$ . The probability that this holds is $\prod _{x_iy_j \in E(H)}(1-p^{|X_i||Y_j|}) \leq (1-p^{(1/\delta ^2)})^{Dn\log n}$ , where we used the premises that the sets $X_i, Y_j$ are of size at most $\frac{1}{\delta }$ and that there are at least $s\ge Dn\log n$ edges of the form $x_iy_j\in E(H)$ . Using a union bound over the choices described above, we have

\begin{align*}\mathbb {P}(G\text { does not satisfy property }\textsf {P}(H,\delta,s)) &\le n^{3n} (1-p^{(1/\delta ^2)})^{Dn\log n} \\[5pt] &\le \exp\!(3n\log n - p^{(1/\delta ^2)} Dn\log n).\end{align*}

We have $3-p^{(1/\delta ^2)}D\lt 0$ and thus the above expression tends to $0$ as $n\rightarrow \infty$ . Setting $q_n\;:\!=\;1-\exp\!((3-p^{(1/\delta ^2)}D)n\log n)$ then concludes the proof of the lemma.

Proof. Define $p\;:\!=\;\frac{\varepsilon }{2}$ and $\delta \;:\!=\;\varepsilon ^2$ . By Lemma 9 there is a sequence $p_n=1-o(1)$ such that $G(n,n;p)$ has maximum degree at most $2pn=\varepsilon n$ with probability at least $p_n$ , and by Lemma 11 there exists an absolute constant $D\gt 0$ and a sequence $q_n=1-o(1)$ such that for every $n$ -vertex graph $H$ the probability that $G(n,n;p)$ satisfies property $\textsf{P}(H, \delta, \lceil Dn\log n\rceil )$ is at least $q_n$ . Let $n_1$ be such that $p_n, q_n\gt \frac{1}{2}$ for every $n \ge n_1$ . Moreover, let $n_2 \in \mathbb{N}$ be chosen large enough such that the inequality $\delta ^2 n^2\ge D n \log n$ holds for every $n \ge n_2$ . Finally, we put $N\;:\!=\;\max \{n_1,n_2\}$ and let $n \geq N$ be arbitrary. By our choice of $N$ , there then exists at least one bipartite graph $G$ with bipartition $\{A',B'\}$ such that $|A'|=|B'|=n$ , $G$ has maximum degree at most $\varepsilon n$ , and $G$ satisfies property $\textsf{P}(H,\delta, \lceil Dn\log n\rceil )$ . Let $A \subseteq A', B \subseteq B'$ be chosen (arbitrarily) such that $|A|=\lfloor (1-2\varepsilon )\kappa (H)\rfloor$ , $|B|=\lfloor (1-2\varepsilon )n\rfloor$ . Note that this is possible as $\kappa (H)\lt{\textrm{v}}(H)=n$ . We now define $F$ as the graph complement of the induced subgraph $G[A \cup B]$ of $G$ . Since $A$ and $B$ are independent sets in $G$ , they form cliques in $F$ . Thus the first item of the lemma is satisfied. To verify the second item, it suffices to note that since $G$ has maximum degree at most $\varepsilon n$ , the same is true for $G[A \cup B]$ , and thus every vertex in $F$ can have at most $\varepsilon n$ non-neighbors in $F$ .

It thus remains to prove that $F$ is indeed $H$ -minor-free. Towards a contradiction, suppose that there exists an $H$ -minor model $(Z_h)_{h \in V(H)}$ in $F$ . Let $X_A, X_B, X_{AB}$ be the partition of $V(H)$ defined as follows: $X_A\;:\!=\;\{h \in V(H)|Z_h \subseteq A\}$ and $X_B\;:\!=\;\{h \in V(H)|Z_h \subseteq B\}$ contain those branch sets which are subsets of $A$ or of $B$ , respectively, and $X_{AB}\;:\!=\;\{h \in V(H)|Z_h \cap A \neq \emptyset \neq Z_h \cap B\}$ contains the branch sets which overlap with both $A$ and $B$ .

Our goal now is to find at least $\delta n$ vertices in $X_A$ and in $X_B$ whose corresponding vertex subsets of $A$ or of $B$ have size at most $\frac{1}{\delta }$ and which have at least $D n \log n$ edges between them. We will then be able to use property $\textsf{P}(H,\delta,\lceil Dn\log n\rceil )$ to complete the proof.

Note that we have $|X_B|+|X_{AB}|\le |B|\le (1-2\varepsilon )n$ as the sets in $(Z_h)_{h \in V(H)}$ are pairwise disjoint. Given that $|X_A|+|X_B|+|X_{AB}|={\textrm{v}}(H)=n$ , this implies that $|X_A| \ge 2\varepsilon n$ . Since the sets $(Z_h)_{h \in X_A}$ are disjoint and since $|A|\le (1-2\varepsilon )\kappa (H)\lt (1-2\varepsilon )n\lt n$ , there cannot be more than $\delta n$ sets of size greater than $\frac{1}{\delta }$ in the collection $(Z_h)_{h \in X_A}$ . Hence, there exists $k \ge 2\varepsilon n-\delta n\ge \delta n$ and $k$ distinct vertices $x_1, \ldots,x_k \in X_A$ such that $|Z_{x_i}| \le \frac{1}{\delta }$ for $i=1,\ldots,k$ . Note that $H$ has minimum degree at least $\kappa (H)$ , for otherwise one could separate a vertex in $H$ from the rest of the graph by deleting fewer than $\kappa (H)$ vertices. Using this, we have

\begin{equation*}|N_H(x_i) \cap X_B|\ge \deg _H(x_i)-|X_A \cup X_{AB}| \ge \delta (H)-|A|\end{equation*}

\begin{equation*}\ge \kappa (H)-(1-2\varepsilon )\kappa (H) =2\varepsilon \kappa (H) \ge 2\varepsilon ^2n=2\delta n\end{equation*}

for every $i=1,\ldots,k$ , where in the last step we used that $\kappa (H)\ge \varepsilon n$ by assumption. Consider for any fixed index $i \in [k]$ the set collection $(Z_h)_{h \in N_H(x_i) \cap X_B}$ . Since the sets are pairwise disjoint and contained in the set $B$ of size at most $n$ , as above it follows that at most $\delta n$ sets in this collection can be of size greater than $\frac{1}{\delta }$ . Consequently, for each $i \in [k]$ there exists a subset $N_i \subseteq N_H(x_i) \cap X_B$ of size at least $2\delta n-\delta n=\delta n$ such that $|Z_h| \le \frac{1}{\delta }$ for every $h \in N_i$ and $i \in [k]$ . Let $y_1,\ldots,y_l \in X_B$ be distinct vertices such that $\{y_1,\ldots,y_l\}=\bigcup _{i=1}^{k}{N_i}$ . Then clearly, $l \ge |N_1|\ge \delta n$ . Furthermore, we have

\begin{equation*}{\textrm {e}}_H(\{x_1,\ldots,x_k\},\{y_1,\ldots,y_l\})\ge \sum _{i=1}^{k}{|N_i|} \ge k\cdot \delta n\ge \delta ^2 n^2\ge D n\log n,\end{equation*}

where in the last step we used our assumption that $n \ge N \ge n_2$ .

We can now use that $G$ satisfies property $\textsf{P}(H,\delta,\lceil Dn\log n\rceil )$ , which directly implies that there exists a pair $(i,j) \in [k]^2$ such that $x_iy_j \in E(H)$ and $G$ contains all edges of the form $xy$ where $(x,y) \in Z_{x_i}\times Z_{y_j}$ . However, by definition of $F$ this means that there exists no edge in $F$ which has endpoints in both $Z_{x_i}$ and $Z_{y_j}$ . This is a contradiction to our initial assumption that $(Z_h)_{h \in V(H)}$ form an $H$ -minor model in $F$ . Thus, $F$ does not contain $H$ as a minor, which establishes the third item of the lemma and concludes the proof.

Proof. Let $F_1,\ldots,F_K$ be an ordering of the copies of $F$ in the pasting graph $F^{(K)}$ , and let $B_1,\ldots,B_K$ be the corresponding copies of $B$ . Let $f : [|A|+|B|-1]^A \rightarrow [K]$ be an arbitrary bijection and let $c_1,\ldots,c_K : A \rightarrow [|A|+|B|-1]$ be the ordering of colour assignments to $A$ that satisfies $f(c_i) = i$ for all $i \in [K]$ . Consider the list assignment $L : V(F^{(K)}) \rightarrow 2^{[|A|+|B|-1]}$ defined as follows:

• $L(a) \;:\!=\; [|A|+|B|-1]$ for all $a \in A$

• $L(b) \;:\!=\; [|A|+|B|-1] \setminus \{c_i(a) \mid a \in A \setminus N_{F_i}(b) \}$ for all $b \in B_i$ for all $i \in [K]$

Given that every vertex in $B$ by assumption has at most $d$ non-neighbors in $F$ , we have $|L(v)| \geq |A|+|B|-1-d$ for all $v \in V(F^{(K)})$ . Now assume towards a contradiction that $F^{(K)}$ admits a proper $L$ -colouring $c$ and let $i \in [K]$ be the unique index satisfying $c_i(a) = c(a)$ for all $a \in A$ . Then let $c_{F_i} : A \cup B_i \rightarrow [|A|+|B|-1]$ be the colouring $c$ restricted to the graph $F_i$ . Since ${\textrm{v}}(F_i) = |A|+|B|$ , there exist by the pigeonhole principle vertices $u,v \in V(F_i)$ with $c(u) = c(v)$ . Since $c$ is proper and $A$ , $B$ are cliques, we have $uv \notin E(F_i)$ and $u \in A$ , $v \in B$ without loss of generality. However, $c(u) \notin L(v)$ by the construction of $L$ , a contradiction.

Proof of Theorem 5. Let a constant $\varepsilon \gt 0$ be given choose $\tilde{\varepsilon }\in (0, \frac{\varepsilon }{4})$ . Let $N=N(\tilde{\varepsilon })$ be as in Lemma 12. We now set $n_0\;:\!=\;\max \{N,\lceil \frac{4}{\varepsilon ^2}\rceil \rceil \}$ and claim that Theorem 5 holds for this choice of $n_0$ .

Let $H$ be a graph on $n \ge n_0$ vertices. We have to prove that $f_\ell (H)\ge (1-\varepsilon )(n+\kappa (H))$ . If $\kappa (H)\lt \varepsilon n$ , then this follows directly from the trivial lower bound via

\begin{equation*}f_\ell (H) \ge {\textrm {v}}(H)-1=n-1\ge (1-\varepsilon ^2)n=(1-\varepsilon )(n+\varepsilon n)\gt (1-\varepsilon )(n+\kappa (H)).\end{equation*}

Thus, we may now assume $\kappa (H)\ge \varepsilon n$ , in particular, $\kappa (H) \ge \tilde{\varepsilon } n$ . Using $n \ge N$ and Lemma 12 we now find that there exists an $H$ -minor-free graph $F$ whose vertex set is partitioned into two cliques $A, B$ such that $|A|=\lfloor (1-2\tilde{\varepsilon })\kappa (H)\rfloor \lt \kappa (H)$ and $|B|=\lfloor (1-2\tilde{\varepsilon })n\rfloor$ , and such that every vertex in $B$ has at most $\tilde{\varepsilon }n$ non-neighbors in $F$ . Let $d\;:\!=\;\lfloor \tilde{\varepsilon }n \rfloor$ and $K\;:\!=\;(|A|+|B|-1)^{|A|}$ . Let $F^{(K)}$ denote a $K$ -fold pasting of $F$ at the clique $A$ . Since every vertex in $B$ has at least $|A|-d$ neighbours in $A$ , we can apply Lemma 15 to find that

\begin{equation*}\chi _\ell (F^{(K)}) \ge |A|+|B|-d \ge (1-2\tilde {\varepsilon })(\kappa (H)+n)-2-\tilde {\varepsilon }n\end{equation*}

\begin{equation*}\ge (1-3\tilde {\varepsilon })(n+\kappa (H))-2\ge (1-\varepsilon )(n+\kappa (H)),\end{equation*}

using $n \ge n_0 \ge \frac{4}{\varepsilon ^2}$ in the last step. In addition, since $|A|\lt \kappa (H)$ , the graph $F^{(K)}$ is $H$ -minor-free by repeated application of Lemma 13. We conclude that $f_\ell (H)\ge (1-\varepsilon )({\textrm{v}}(H)+\kappa (H))$ , as desired.

Proof. Assume $G^\complement$ contains $H[U]$ as a minor for some $U \subseteq V(H)$ with $|U| \geq (1-\delta )n$ . Let $(Z_h)_{h \in U}$ be an $H[U]$ -minor model in $G^\complement$ and define $X_A \;:\!=\; \{h \in U \mid Z_h \subseteq A\}$ , $X_B \;:\!=\; \{h \in U \mid Z_h \subseteq B\}$ , and $X_{AB} \;:\!=\; \{h \in U \mid Z_h \cap A \neq \emptyset \neq \ Z_h \cap B\}$ . We have $|X_A| + |X_{AB}| \leq |A|$ , $|X_B| + |X_{AB}| \leq |B|$ , and $|X_A| + |X_B| + |X_{AB}|=|U| \geq (1-\delta )n$ , which implies $|X_A|, |X_B| \ge (1-\delta )n-(1-3\delta )n=2\delta n$ .

Since the branch sets $(Z_h)_{h \in X_A}$ in $A$ and the branch-sets $(Z_h)_{h \in X_B}$ in $B$ are pairwise disjoint, at most $\delta (1-3\delta ) n\lt \delta n$ branch sets in each of $(Z_h)_{h \in X_A}$ and $(Z_h)_{h \in X_B}$ can be larger than $\frac{1}{\delta }$ . Thus, there are at least $2\delta n-\delta n=\delta n$ branch sets of size at most $\frac{1}{\delta }$ in $(Z_h)_{h \in X_A}$ as well as in $(Z_h)_{h \in X_B}$ . Thus for $k\;:\!=\;l\;:\!=\; \lceil \delta n \rceil$ , there exist distinct vertices $x_1, \ldots, x_{k} \in X_A$ , $y_1,\ldots,y_l \in X_B$ such that $|Z_{x_i}|, |Z_{y_j}| \le \frac{1}{\delta }$ for all $1 \le i, j \le k=l$ . Since $H\in \mathcal{Q}_n(\delta,D)$ , we have ${\textrm{e}}_H(\{x_1,\ldots,x_k\},\{y_1,\ldots,y_l\})\ge \lceil D n \log n\rceil =s$ . Next we use our assumption that $G$ satisfies property $\textsf{P}(H, \delta,s)$ . It implies that there exists an edge $x_i y_j \in E(H)$ with $(i,j) \in [k]\times [l]$ such that $G$ contains all the edges $xy$ with $(x,y) \in Z_{x_i} \times Z_{y_j}$ . Then, however, there is an edge between vertices $x_i$ and $y_j$ in $H$ , but no edge between the corresponding branch sets $Z_{x_i}$ and $Z_{y_j}$ in $G^\complement$ , a contradiction.

Proof. In the following, let $F_1, \ldots, F_K$ denote the copies of $F$ such that $F^{(K)}=\bigcup _{i=1}^{K}{F_i}$ .

Suppose $F^{(K)}$ has an $H$ -minor and fix an $H$ -minor model $(Z_h)_{h\in V(H)}$ in $H$ . Let us denote $X_W \;:\!=\; \{h \in V(H) \mid Z_h \cap W \neq \emptyset \}$ and $\xi _W\;:\!=\;|X_W|$ , and $X_i \;:\!=\; \{h \in V(H) \mid Z_h \subseteq V(F_i) \setminus W\}$ and $\xi _i\;:\!=\;|X_i|$ for every $i \in [K]$ . Note that since every branch-set $Z_h$ induces a connected subgraph of $F$ , every vertex $h \in V(H)$ appears in exactly one of the sets $X_W, X_1,\ldots,X_K$ , i.e., they form a partition of $V(H)$ . In particular, we have $\xi _W+\sum _{i=1}^{K}{\xi _i}={\textrm{v}}(H)=n$ .

We have $\xi _W \leq |W|\le n-3\delta n$ and thus $\sum _{i=1}^{K} \xi _i=n-\xi _W \geq 3 \delta n$ . In the following, let us w.l.o.g. assume $[K]$ is ordered such that $\xi _1 \geq \xi _2 \geq \cdots \geq \xi _K$ . We claim that $\xi _1\ge (1-\delta )n-\xi _W$ . Towards a contradiction, suppose in the following that $\xi _1 \lt (1-\delta )n - \xi _W$ . We first note that using this assumption, we have that $\sum _{i=2}^K \xi _i=n-(\xi _W+\xi _1) \gt n-(1-\delta )n=\delta n$ .

Now suppose for a first case that $\xi _1 \ge \delta n$ . Then the two disjoint sets of vertices $X_1$ and $\bigcup _{i=2}^{K}{X_i}$ in $H$ are both of size at least $\delta n$ . By property $\textsf{Q}(\delta,D)$ this implies that ${\textrm{e}}_H(X_1,\bigcup _{i=2}^{K}{X_i})\ge Dn\log n\gt 0$ . In particular there exists $2 \le i \le K$ and an edge $uv\in E(H)$ for some $u \in X_1$ and $v \in X_i$ . This implies that there must exist an edge in $F^{(K)}$ connecting a vertex in $Z_u\subseteq V(F_1)\setminus W$ to a vertex in $Z_v\subseteq V(F_i)\setminus W$ . However, by construction of $F^{(K)}$ no such edges exist, and so we arrive at the desired contradiction in this first case.

For the second case, suppose that $\xi _1\lt \delta n$ (and thus in particular $\xi _i\lt \delta n$ for all $i \in [K]$ ). Let $j \in [K]$ be the smallest index such that $\sum _{i=1}^{j}{\xi _i}\gt \delta n$ (this is well-defined, since $\sum _{i=1}^K{\xi _i}\ge 3\delta n$ , see above). By the minimality of $j$ , we have $\sum _{i=1}^{j}{\xi _i}=\xi _j+\sum _{i=1}^{j-1}{\xi _j}\le \delta n+\delta n=2\delta n$ . This implies that $\sum _{i=j+1}^K{\xi _i}=\sum _{i=1}^K{\xi _i}-\sum _{i=1}^j{\xi _i}\ge 3\delta n - 2\delta n =\delta n$ . In consequence, we find that the two disjoint vertex sets $\bigcup _{i=1}^{j}{X_i}, \bigcup _{i=j+1}^{K}{X_i}$ in $H$ are both of size at least $\delta n$ . Hence, using property $\textsf{Q}(\delta,D)$ we have ${\textrm{e}}_H(\bigcup _{i=1}^{j}{X_i},\bigcup _{i=j+1}^{K}{X_i})\ge Dn\log n\gt 0$ . Similar as above, this implies the existence of two indices $i,i'$ with $1 \le i\le j\lt i'\le K$ such that there exists an edge between $V(F_i)\setminus W$ and $V(F_{i'})\setminus W$ in $F^{(K)}$ . As this is impossible by construction of $F^{(K)}$ , a contradiction follows also in the second case. Thus our initial assumption $\xi _1\lt (1-\delta )n-\xi _W$ was false.

We therefore have $|X_1 \cup X_W|=\xi _1+\xi _W \geq (1-\delta )n$ . Let $U\;:\!=\;X_1 \cup X_W$ . For every $h \in U$ , let $Z_h'\;:\!=\;Z_h$ if $h \in X_1$ and $Z_h'\;:\!=\;Z_h \cap V(F_1)$ if $h \in X_W$ . We now show that $(Z_h')_{h \in U}$ is an $H[U]$ -minor model in $F_1$ , which will then conclude the proof of the lemma.

First of all, note that $F_1[Z_h']$ is a connected graph for every $h \in U$ . If $h \in X_1$ , then $F_1[Z_h']=F^{(K)}[Z_h]$ is connected since $(Z_h)_{h \in V(H)}$ is an $H$ -minor model. And if $h \in X_W$ , then the connectivity of $F_1[Z_h']=F^{(k)}[Z_h \cap V(F_1)]$ follows since (1) $F^{(K)}[Z_h]$ is connected and (2) every path connecting two vertices in $Z_h'$ that is contained in $F^{(K)}[Z_h]$ can be shortened to a path whose vertex set is completely contained in $V(F_1)$ by short-cutting every segment of the path that starts and ends in the clique $W$ by the direct connection between its endpoints.

Let us now consider any edge $uv \in E(H[U])$ . Then there must exist an edge $xy \in E(F^{(K)})$ with $x \in Z_u, y \in Z_v$ . If we have $x, y\in V(F_1)$ , then this witnesses the existence of an edge between $Z_u'$ and $Z_v'$ in $F_1$ , as desired. If on the other hand at least one of $x,y$ lies outside of $V(F_1)$ , then we necessarily must have $Z_u \cap W \neq \emptyset \neq Z_v \cap W$ , and thus there exists an edge in the clique induced by $W$ (and thus also in $F_1$ ) that connects a vertex in $Z_u'$ to a vertex in $Z_v'$ . All in all, this shows that $F_1$ contains $H[U]$ as a minor. Since $|U| \ge (1-\delta )n$ , this concludes the proof.

Proof of Theorem 7. Let a constant $\varepsilon \in (0,1)$ be given. Let $\delta \gt 0$ be chosen small enough such that $7\delta \lt \varepsilon$ , set $p\;:\!=\;\frac{\delta }{2}$ , let $D=D(\delta,p)\gt 1$ be the constant given by Lemma 11, and let $C \;:\!=\; \frac{D^2}{\delta ^2}$ .

For every $n \in \mathbb{N}$ , put $s=s(n)=\lceil Dn\log n\rceil$ . By Lemma 17, a random graph $H = G(n; \lceil Cn \log n\rceil )$ chosen uniformly from all $n$ -vertex graphs with $\lceil Cn\log n\rceil$ edges satisfies property $\textsf{Q}(\delta,D)$ w.h.p. as $n \rightarrow \infty$ . Now assume the graph $H$ satisfies property $\textsf{Q}(\delta,D)$ . By Lemmas 9 and 11, w.h.p. as $n \rightarrow \infty$ , the random bipartite graph $G = G(\lfloor (1-3\delta )n\rfloor, \lfloor (1-3\delta )n\rfloor ; p )$ has maximum degree at most $2p\lfloor (1-3\delta )n\rfloor \le \delta n$ and satisfies property $\textsf{P}(H, \delta,s)$ . Now fix $n$ large enough and consider a graph $G$ with bipartition $\{A,B\}$ , $|A|=|B|=\lfloor (1-3\delta n)\rfloor$ satisfying these two properties. By Lemma 18, $G^\complement$ does not contain any induced subgraph $H[U]$ as a minor for any $U \subseteq V(H)$ with $|U| \geq (1-\delta ) n$ . Let $K \;:\!=\; (|A|+|B|-1)^{|A|}$ and let $(G^\complement )^{(K)}$ be a $K$ -fold pasting of $G^\complement$ at $A$ . Then by Lemma 19, $(G^\complement )^{(K)}$ does not contain $H$ as a minor. Moreover, by Lemma 15, applied with $d=\lfloor \delta n\rfloor$ , we find that $(G^\complement )^{(K)}$ has list chromatic number at least $|A|+|B|-d\gt 2(1-3\delta )n - \delta n -2 \gt (2-\varepsilon )n$ for $n$ large enough. This shows that w.h.p. the random graph $H=G(n;\lceil Cn \log n\rceil )$ satisfies $f_\ell (H)\ge (2-\varepsilon )n$ , which concludes the proof.

Proof of Theorem 3. We start by fixing an integer $d \in \mathbb{N}$ as guaranteed by Theorem 21, i.e. such that every graph of minimum degree at least $d$ contains $F$ as a minor. We now define $k_0(F)\;:\!=\;\min \{d+1,9\cdot{\textrm{v}}(F)^3\}$ . Let $k \ge k_0(F)$ be any given integer. Let $H$ denote the graph obtained from $F$ by adding $k$ isolated vertices. We will now show that every $H$ -minor-free graph is $({\textrm{v}}(H)-2)$ -degenerate, which then easily implies $f_\ell (H)={\textrm{v}}(H)-1$ .

Towards a contradiction, suppose that there exists an $H$ -minor-free graph $G$ which is not $({\textrm{v}}(H)-2)$ -degenerate, and let $G$ be chosen such that ${\textrm{v}}(G)$ is minimised. Note that the minimality assumption on $G$ immediately implies that $\delta (G)\ge{\textrm{v}}(H)-1={\textrm{v}}(F)+k-1$ . Observe that since $\delta (G)\ge k\gt d$ , the graph $G-x$ for some $x \in V(G)$ has minimum degree at least $d$ and thus must contain $F$ as a minor. Let $X \subseteq V(G)$ be chosen of minimum size subject to $G[X]$ containing $F$ as a minor. Note that from the above it follows that $|X|\le{\textrm{v}}(G)-1$ and hence that $V(G)\setminus X \neq \emptyset$ . Let $(Z_f)_{f \in V(F)}$ be an $F$ -minor model in $G[X]$ . By minimality of $X$ , we have that $(Z_f)_{f \in V(F)}$ forms a partition of $X$ . With the goal of bounding the number of edges in $G-X$ , we present our next argument as a separate claim. We will later use this bound and Turán’s theorem to show the existence of an $F$ -subgraph in $G-X$ .

It follows immediately from Claim 22 that $|N(v) \cap X| \le \sum _{f \in V(F)}{|N(v) \cap Z_f|} \lt 9{\textrm{v}}(F)^2$ for every $v \in V(G)\setminus X$ . Additionally recalling that $\delta (G) \ge{\textrm{v}}(F)+k-1\ge k$ , we find that for every $v \in V(G)\setminus X$ , we have $\text{deg}_{G-X}(v)=|N(v) \setminus X|=\text{deg}(v)-|N(v) \cap X|\gt k-9{\textrm{v}}(F)^2$ . Having established $V(G)\setminus X \neq \emptyset$ at the beginning of the proof, it now follows that $G-X$ is a graph of minimum degree greater than $k-9{\textrm{v}}(F)^2$ . Also note that since $G[X]$ contains $F$ as a minor, we are not able to find $k$ distinct vertices in $V(G)\setminus X$ as these could be used to augment the $F$ -minor in $G[X]$ to an $H$ -minor in $G$ , contradicting our assumptions. We thus have ${\textrm{v}}(G-X)\lt k$ . Using our choice of $k_0$ and $k \ge k_0$ , it now follows that

\begin{equation*}\delta (G-X) \gt k-9{\textrm {v}}(F)^2\gt \left (1-\frac {1}{{\textrm {v}}(F)-1}\right )k \gt \left (1-\frac {1}{{\textrm {v}}(F)-1}\right ){\textrm {v}}(G-X).\end{equation*}

Therefore, $G-X$ has more than $\bigl (1-\frac{1}{{\textrm{v}}(F)-1}\bigr )\frac{{\textrm{v}}(G-X)^2}{2}$ edges and thus Theorem 20 implies the existence of a clique on ${\textrm{v}}(F)$ vertices in $G-X$ . In particular, $G-X$ and thus $G$ contain a subgraph isomorphic to $F$ . Let $K \subseteq V(G)$ be the vertex set of such a copy of $F$ . Then, since ${\textrm{v}}(G)\ge \delta (G)+1\ge{\textrm{v}}(F)+k$ , there are at least $k$ vertices outside of $K$ in $G$ , which can be added to the copy of $F$ on vertex set $K$ to create a subgraph of $G$ that is isomorphic to $H$ . In particular, this means that $G$ contains $H$ as a minor, a contradiction. All in all, we find that our initial assumption, namely regarding the existence of a smallest counterexample $G$ to our claim, was wrong. This concludes the proof that all $H$ -minor-free graphs are $({\textrm{v}}(H)-2)$ -degenerate.

It is a well-known fact and easy to prove by induction that for every $a \in \mathbb{N}$ all $a$ -degenerate graphs are $(a+1)$ -choosable. Thus what we have proved also implies that every $H$ -minor-free graph is $({\textrm{v}}(H)-1)$ -choosable, as desired. All in all, it follows that $f_\ell (H)={\textrm{v}}(H)-1$ , concluding the proof of the theorem.