Proof.

• Let $E_n$ denote the probability event that $G$ does not satisfy the property claimed in the first item. We need to show that $\mathbb{P}(E_n) \rightarrow 0$ as $n \rightarrow \infty$ . So consider a fixed subset $X \subseteq A$ (or symmetrically, $X \subseteq B$ ) such that $|X| \ge \varepsilon n$ , and a fixed collection $Y_1,\ldots,Y_k$ of disjoint non-empty subsets of $B$ (or symmetrically, $A$ ), where $k \ge \varepsilon n$ and $\max \{|Y_1|,\ldots,|Y_k|\} \le f$ . Let $E(X,Y_1,\ldots,Y_k)$ be the probability event ‘there exists no pair $(x,j) \in X \times [k]$ such that $x$ is fully connected to the vertices in $Y_j$ ’. Fixing a vertex $x \in X$ and an index $j \in [k]$ , clearly the probability of the event that ‘ $x$ is not fully connected to $Y_j$ ’ equals $1-p^{|Y_j|} \le 1-p^{f}$ . Since these events are independent for different choices of $(x,j)$ , we conclude that

  \begin{equation*}\mathbb {P}(E(X,Y_1,\ldots,Y_k)) \le \big(1-p^{f}\big)^{|X|\cdot k} \le \big(1-p^{f}\big)^{\varepsilon ^2n^2}\le \exp\!(\!-p^{f}\varepsilon ^2 n^2)=\exp\!\left(\!-\varepsilon ^2n^{2-f\delta }\right).\end{equation*}
  With a rough estimate, there are at most
  \begin{equation*}2 \cdot 2^n\cdot (n+1)^{n} = \exp\!(\!\ln\!(2)(n+1)+n \ln\!(n+1))\end{equation*}
  ways to select the sets $X,Y_1,\ldots,Y_k$ . Hence, applying a union bound we find that
  \begin{equation*}\mathbb {P}(E_n) \le \exp\!(\!\ln\!(2)(n+1)+n \ln\!(n+1)-\varepsilon ^2n^{2-f\delta }).\end{equation*}
  The right hand side of the above inequality tends to $0$ as $n \rightarrow \infty$ , since $f\delta \lt 1$ and hence $\varepsilon ^2n^{2-f\delta }=\Omega (n^{2-f\delta })$ grows faster than $\ln\!(2)(n+1)+n \ln\!(n+1)=O(n\ln n)$ . This proves that $G$ satisfies the properties claimed by the first item w.h.p., as required.

• To show that also the property claimed by the second item holds true w.h.p., consider the probability that a fixed vertex $x \in A \cup B$ has more than $\varepsilon n$ neighbours in $G$ . Note that the degree of $x$ in $G(n,n,p)$ is distributed like a binomial random variable $B(n,p)$ , and hence its expectancy is $np=n^{1-\delta }$ , which is smaller than $\frac{\varepsilon n}{2}$ for $n$ sufficiently large in terms of $\varepsilon$ and $\delta$ . Applying Chernoff’s bound we find for every sufficiently large $n$ :

  \begin{equation*}\mathbb {P}(d_G(x)\gt \varepsilon n )\le \mathbb {P}(B(n,p)\gt 2np) \le \exp\!\left (-\frac {1}{3}np\right )=\exp\!\left (-\frac {1}{3}n^{1-\delta }\right ).\end{equation*}
  Since this bound applies to every choice of $x \in A \cup B$ , applying a union bound we find that the probability that $G$ has maximum degree more than $\varepsilon n$ is at most
  \begin{equation*}2n\exp\!\left (-\frac {1}{3}n^{1-\delta }\right )=\exp\!\left (\ln\!(2n)-\frac {1}{3}n^{1-\delta }\right )\end{equation*}
  which tends to $0$ as $n \rightarrow \infty$ , as desired (here we used that $\delta \lt 1$ and hence $n^{1-\delta }$ grows faster than $\ln\!(2n)$ ).

Proof. Let $f\,{:\!=}\,\lceil \frac{1}{\varepsilon } \rceil \in \mathbb{N}$ and $\delta \,{:\!=}\,\frac{\varepsilon }{2}$ . Then $f\delta \lt 1$ , and hence we may apply Lemma 8. It follows directly that there exists $n_0=n_0(\varepsilon ) \in \mathbb{N}$ such that for every $n \ge n_0$ there exists a bipartite graph $G$ , whose bipartition classes $A$ and $B$ are both of size $n$ , and such that

• For every subset $X \subseteq A$ such that $|X| \ge \varepsilon n$ and every collection of pairwise disjoint non-empty subsets $Y_1,\ldots,Y_k \subseteq B$ such that $k \ge \varepsilon n$ and $\max \{|Y_1|,\ldots,|Y_k|\} \le f$ , there exists a vertex $x \in X$ and some $j \in [k]$ such that $G$ contains all the edges $xy, y \in Y_j$ . The same statement holds symmetrically for the case when $X \subseteq B$ and $Y_1,\ldots,Y_k \subseteq A$ .

• $G$ has maximum degree at most $\varepsilon n$ .

We now define $H$ as the complement of $G$ (also with vertex set $A \cup B$ ). Since $G$ is bipartite, clearly $A$ and $B$ form cliques in $H$ , verifying the first item. The second item follows directly from the fact that $\Delta(G) \le \varepsilon n$ .

It hence remains to verify the last item. Towards a contradiction, suppose that there exists a number $t \in \mathbb{N}$ , $t \ge (1+2\varepsilon )n$ , such that $H$ contains $K_t$ as a minor. This implies that there exists a collection $\mathcal{Z}$ of non-empty and pairwise disjoint subsets of $V(H)$ such that $|\mathcal{Z}|=t$ and such that for every two distinct $Z, Z' \in \mathcal{Z}$ , there exists at least one edge in $H$ connecting a vertex in $Z$ to a vertex in $Z'$ . Let us denote $\mathcal{Z}_s\,{:\!=}\,\{Z \in \mathcal{Z}||Z|=s\}$ , and $z_s\,{:\!=}\,|\mathcal{Z}_s|$ , for every $s \ge 1$ . We clearly have

\begin{equation*}2n \ge \sum _{s\ge 1}{sz_s} \ge 2(t-z_1)+z_1=2t-z_1 \ge 2n+4\varepsilon n -z_1.\end{equation*}

Rearranging yields that $z_1 \ge 4\varepsilon n$ . From this, we may conclude that either $A$ or $B$ contains at least $2 \varepsilon n$ singletons from $\mathcal{Z}$ . By symmetry (possibly by renaming $A$ and $B$ ), we may assume w.l.o.g. that $B$ contains at least $2 \varepsilon n$ singletons from $\mathcal{Z}$ , and denote the set of these singletons by $X$ . Let us now define $\mathcal{Z}_A\,{:\!=}\,\{Z \in \mathcal{Z}|Z \subseteq A\}$ and $\mathcal{Z}_B\,{:\!=}\,\{Z \in \mathcal{Z}|Z \cap B \neq \emptyset \}$ . Since the sets in $\mathcal{Z}$ are pairwise disjoint, we can see that $|\mathcal{Z}_B| \le |B|=n$ , and therefore $|\mathcal{Z}_A|=t-|\mathcal{Z}_B| \ge t-n \ge 2\varepsilon n$ . Since $|A|=n$ , the latter implies that there are at least $\varepsilon n$ distinct sets in $\mathcal{Z}_A$ which have size at most $\frac{1}{\varepsilon }$ . Let $Y_1,\ldots,Y_k$ with $k \ge \varepsilon n$ be an enumeration of the sets in $\mathcal{Z}_A$ of size at most $\frac{1}{\varepsilon } \le f$ . By the above, there exists $j \in [k]$ and a vertex $x \in X$ such that $xy \in E(G)$ for every $y \in Y_j$ . Since $H$ is the complement of $G$ , this means that $\{x\}$ and $Y_j$ are distinct sets in $\mathcal{Z}$ , which do not have any connecting edge. This is a contradiction to our initial assumptions, and hence we have shown that the third item claimed in the lemma is also satisfied. This concludes the proof.

Proof of Theorem 6. Let a fixed $\varepsilon \in (0,1)$ be given. Pick some $\varepsilon ' \in (0,1)$ such that $\frac{2-\varepsilon '}{1+2\varepsilon '} \ge 2-\frac{\varepsilon }{2}$ . Let $n_0=n_0(\varepsilon ') \in \mathbb{N}$ be as in Lemma 9, and define $t_0\,{:\!=}\,\max \{\lceil (1+2\varepsilon ')n_0\rceil,\left \lceil \frac{4}{\varepsilon }\right \rceil \}$ . Now, let $t \ge t_0$ be any given integer. Define $n\,{:\!=}\,\left \lfloor \frac{t}{1+2\varepsilon '}\right \rfloor \ge n_0$ . Applying Lemma 9, we find that there exists a graph $H$ whose vertex set is partitioned into two non-empty sets $A$ and $B$ of size $n$ , such that both $A$ and $B$ form cliques in $H$ , every vertex in $H$ has at most $\varepsilon ' n$ non-neighbours, and $H$ is $K_t$ -minor-free (since $t \ge (1+2\varepsilon ')n$ , by definition of $n$ ).

For every possible assignment $c \in [2n-1]^A$ of colours from $[2n-1]$ to vertices in $A$ , denote by $H(c)$ an isomorphic copy of $H$ , such that the vertex set of $H(c)$ decomposes into the sets $A$ and $B(c)$ of size $n$ . More precisely, the distinct copies $H(c), c \in [2n-1]^A$ of $H$ share the same set $A$ but have pairwise disjoint sets $B(c)$ . Since $A$ forms a clique of size $n$ in the $K_t$ -minor-free graph $H(c)$ for every colouring $c\,{:}\,A \rightarrow [2n-1]$ , it follows by repeated application of Lemma 10 that the graph $\mathbf{G}$ with vertex set $A \cup \bigcup _{c \in [2n-1]^A}{B(c)}$ , obtained as the union of the graphs $H(c), c \in [2n-1]^A$ , is $K_t$ -minor-free.

Now, consider an assignment $L\,{:}\,V(\mathbf{G}) \rightarrow 2^{\mathbb{N}}$ of colour lists to the vertices of $\mathbf{G}$ as follows: For every vertex $a \in A$ , we define $L(a)\,{:\!=}\,[2n-1]$ , and for every vertex $b \in B(c)$ for some colouring $c \in [2n-1]^A$ of $A$ , we define $L(b)\,{:\!=}\, [2n-1] \setminus \{c(a)|a \in A, ab \notin E(H(c))\}$ . Note that since every vertex in $B(c)$ has at most $\varepsilon ' n$ non-neighbours in $H(c)$ , we have $|L(v)| \ge 2n-1-\varepsilon ' n$ for every vertex $v \in V(\mathbf{G})$ .

We now claim that $\mathbf{G}$ does not admit a proper colouring with colours chosen from the lists $L(v), v \in V(\mathbf{G})$ , which will then prove that $\chi _\ell (\mathbf{G}) \ge 2n-\varepsilon ' n$ . Indeed, suppose towards a contradiction there exists a proper colouring $c_{\mathbf{G}}\,{:}\,V(\mathbf{G}) \rightarrow \mathbb{N}$ of $\mathbf{G}$ such that $c_{\mathbf{G}}(v) \in L(v)$ for every $v \in V(\mathbf{G})$ . Let $c$ be the restriction of $c_{\mathbf{G}}$ to $A$ , and consider the proper colouring of $H(c)$ obtained by restricting $c_{\mathbf{G}}$ . Since $H(c)$ has order $2n$ and $c_{\mathbf{G}}(v) \in [2n-1]$ for every $v \in V(H(c))$ , there must exist two (necessarily non-adjacent) vertices in $H(c)$ which have the same colour with respect to $c_{\mathbf{G}}$ . Concretely, there exist $a \in A$ , $b\in B(c)$ such that $ab \notin E(H(c))$ and $c_{\mathbf{G}}(a)=c_{\mathbf{G}}(b)$ . This however yields a contradiction, since $c_{\mathbf{G}}(b) \in L(b)$ and by definition $c(a)=c_{\mathbf{G}}(a)$ is not included in the list of $b$ .

We conclude that indeed, $\mathbf{G}$ is a $K_t$ -minor-free graph which satisfies

\begin{align*}\chi _\ell (\mathbf {G}) \ge (2-\varepsilon ')n&=(2-\varepsilon ')\left \lfloor \frac {t}{1+2\varepsilon '}\right \rfloor \gt (2-\varepsilon ')\left (\frac {t}{1+2\varepsilon '}-1\right )\\[3pt] &\ge \left (2-\frac {\varepsilon }{2}\right )t-(2-\varepsilon ') \ge (2-\varepsilon )t,\end{align*}

where for the last inequality we used $t \ge t_0 \ge \frac{4}{\varepsilon }$ .