Proof. To see this let $\varepsilon =(16C)^{-1}, t_1=C\log |V_1|$ , $t_2=C\log |V_2|$ and assume that $t_1\leq t_2$ . It is enough to show that our graph cannot have density larger than $1-\varepsilon$ , as the other statement follows by looking at the (bipartite) complement $\overline{G}$ .

For the sake of contradiction, suppose the density $d(G) \gt 1-\varepsilon$ , and let us count in two ways the number $M$ of stars $K_{t_1,1}$ which are formed by taking a single vertex in $V_2$ and $t_1$ of its neighbours. On one hand, each vertex $v\in V_2$ contributes $\binom{d(v)}{t_1}$ stars, giving us:

(1) \begin{align} M\geq \sum _{v\in V_2}\binom{d(v)}{t_1}\geq |V_2|\cdot \binom{e(G)/|V_2|}{t_1}\geq |V_2|\cdot \binom{(1-\varepsilon )|V_1|}{t_1} \geq |V_2| \cdot (1-2\varepsilon )^{t_1} \binom{|V_1|}{t_1}. \end{align}

Here we have used Jensen’s inequality for the map $x\to \binom{x}{t_1}$ in the second inequality, and that $t_1 \leq \varepsilon |V_1|$ for the final step, as $|V_1| \geq n_C$ .

On the other hand, if $G$ is $K_{t_1,t_2}$ -free then for each subset $S\subset V_1$ of $t_1$ vertices there are at most $t_2-1$ vertices in $V_2$ that can form a star with $S$ , and so $M\leq (t_2-1)\binom{|V_1|}{t_1}$ . Combined with (1) this gives $|V_2| \leq (t_2-1) (1-2\varepsilon )^{-t_1} \leq |V_2|^{1/2} e^{4\varepsilon t_1}$ , using that $|V_2| \geq n_C$ and that $1 - x \geq e^{-2x}$ for $x \in [0, 1/2]$ . It follows that $|V_2| \leq e^{8\varepsilon t_2} = |V_1|^{8C\varepsilon } = |V_2|^{1/2}$ , a contradiction.

Proof. Suppose for the sake of contradiction that the conclusion is not true. Let $\varepsilon \,:\!=\,1/32C$ and suppose at most $2|V_1|/3$ vertices $v\in V_1$ have degrees between $\varepsilon |V_2|$ and $(1-\varepsilon )|V_2|$ . Then, by the pigeonhole principle, without loss of generality we can assume that there is a set $U_1 \subset V_1$ with $|U_1| \geq |V_1|/6$ such that all vertices in $U_1$ have degree less than $\varepsilon |V_2|$ . But then the edge density of the induced bipartite graph $G[U_1,V_2]$ is less than $\varepsilon$ . On the other hand, it is easy to see that if $G$ is $C$ -bipartite-Ramsey, since $6|U_1| \geq |V_1| \geq n_C$ , the graph $G[U_1,V_2]$ is $2C$ -bipartite-Ramsey. This is a contradiction by Lemma 2.1.

Proof. By Corollary 2.2 we can pick $u_1\in V_1$ such that $d(u_1)\in [\varepsilon _1|V_2|,(1-\varepsilon _1)|V_2|]$ . Our requirement is clearly satisfied for $i=1$ . Now let us assume that we have found $u_1,u_2,\ldots,u_i$ with $i \lt L$ that satisfy our requirements and let us look for a vertex $u_{i+1}$ .

Let $S_i\,:\!=\,V_1\setminus \{u_1,u_2,\ldots,u_i\}$ and $T_i\,:\!=\,V_2\setminus \underset{j\leq i}{\cup } N(u_j)$ . Note that $|S_i| \geq |V_1| - L \geq |V_1|^{1/2}$ and that $|T_i| \geq (64C)^{-L}|V_2| \geq |V_2|^{1/2}$ since $|V_1|, |V_2| \geq n_0$ . Therefore, the subgraph $G[S_i,T_i]$ is $(2C)$ -bipartite-Ramsey. Thus, by Corollary 2.2, it follows that we can find a vertex $u_{i+1}\in S_1$ such that $d^{T_i}_G(u_{i+1})\in [\varepsilon _1|T_i|,(1-\varepsilon _1)|T_i|]$ . But this is precisely the vertex we were looking for, as then we have $\Big|N(u_{i+1})\setminus \underset{j\leq i}{\cup }N(u_j)\Big|\geq d^{T_i}_G(u_{i+1})\geq \varepsilon _1|T_i|\geq \varepsilon _{i+1}|V_2|$ and we also get that $\Big|V_2\setminus \underset{j\leq i+1}{\cup } N(u_j)\Big|\geq \Big|T_i\setminus N^{T_i}_{G}(u_{i+1})\Big|\geq \varepsilon _1|T_i| \geq \varepsilon _{i+1}|V_2|$ .

By repeating this step $L$ times we reach our conclusion.

Proof. It is enough to prove that the bipartite-richness condition holds when $i = 1$ . So set $\varepsilon \,:\!=\,(200C)^{-1}$ and suppose there is a set $U_1\subset V_1$ with $|U_1|\geq \delta |V_1|$ which contradicts the bipartite-richness condition – more precisely that there is a set $W_2\subset V_2$ with $|W_2|\geq |V_2|^{1/5}$ such that $|N(v)\cap U_1|\lt \varepsilon |U_1|$ or $|\overline{N(v)}\cap U_1|\lt \varepsilon |U_1|$ for all $v\in W_2$ . Without loss of generality, we can assume that there is a subset $U_2\subset W_2$ of size $|V_2|^{1/5}/2$ such that $|N(v)\cap U_1|\lt \varepsilon |U_1|$ for all $v\in W_2$ . But this means that the edge density of $G[U_1,U_2]$ is less than $\varepsilon$ .

On the other hand, as $|U_i| \geq |V_i|^{1/5}/2 \geq |V_i|^{1/6}$ as $|V_i| \geq n_C$ , and since $G = G[V_1,V_2]$ is $C$ -Ramsey, it follows $G[U_1, U_2]$ is $6C$ -Ramsey. But by Lemma 2.1 such a graph must have edge density at least $(196C)^{-1}\gt \varepsilon,$ which is a contradiction, thus proving our result.

Proof. It is enough to show that each property holds when taking $i = 1$ .

(i) For each $v\in V_1$ either $|N(v)| \geq |V_2|/2$ or $|\overline{N(v)}|\geq |V_2|/2$ . In the former case, all but at most $|V_1|^{1/5}$ vertices $w\in V_1$ we have $|N(v)\cap \overline{N(w)}|\geq \varepsilon |N(v)|\geq \varepsilon/2\cdot |V_2|$ , and in the latter, for all but at most $|V_1|^{1/5}$ vertices $w\in V_1$ we have $|\overline{N(v)}\cap{N(w)}|\geq \varepsilon |\overline{N(v)}|\geq \varepsilon/2\cdot |V_2|$ . In either case, there are at most $|V_1|^{1/5}$ vertices $w\in V_1$ with $|\text{div}_G(v,w)| \lt \varepsilon/2\cdot |V_2|$ , as desired.

(ii) Suppose now there is an ordered pair $\boldsymbol{{p}}\,:\!=\,\{x,y\}\in \binom{V_1}{2}$ and a collection $Y$ of $|V_1|^{1/5}$ vertex disjoint pairs $\boldsymbol{{p}}_i\,:\!=\,(x_i,y_i) \in \binom{V_1}{2}$ such that $|\text{divb}(\boldsymbol{{p}})\setminus N(x_i)|\leq \alpha \varepsilon/2\cdot |V_2|$ for each $i$ . Note that if $|\text{div}(\boldsymbol{{p}})|\geq \alpha |V_2|$ then $|\text{divb}(\boldsymbol{{p}})|\geq \alpha/2\cdot |V_2|$ , as $\text{divb}(\boldsymbol{{p}})$ is simply the largest of the two sets $N(x)\setminus N(y)$ and $N(y)\setminus N(x)$ , whose union is $\text{div}(\boldsymbol{{p}})$ . But in this case there are $|V_1|^{1/5}$ distinct vertices $z_i \in \{x_i, y_i\}$ such that $|\text{divb}(\boldsymbol{{p}})\cap \overline{N(z_i)}|\leq \alpha \varepsilon/2\cdot |V_2|\leq \varepsilon |\text{divb}(\boldsymbol{{p}})|$ , which contradicts the richness of the set $\text{divb}(\boldsymbol{{p}})$ .

Proof of Proposition 2.10. We can view choosing $X$ as going through each number from $1$ to $n$ and independently tossing a fair coin for each to decide whether it is an element of $X$ or not. Thus $|X|\ (\text{mod }d)$ can be viewed as a Markov chain on $\{0,1,2,\ldots,d-1\}$ starting at $0$ and with transition probabilities $p_{i,i}=p_{i,i-1}=0.5$ for each $0\leq i\lt d$ , where indices ar taken modulo $d$ . This chain is aperiodic as $p_{i,i}^{(1)}=0.5$ , and it is also irreducible as we can reach state $i$ from state $j$ in $i-j(\text{mod }d)$ steps with positive probability. It is easy to see that the stationary distribution $\pi$ is given by $\pi _i=1/d$ . The conclusion now follows from Theorem 2.11.

Proof. We will prove the statement under the assumption that $A_i \subset [0,M]$ for all $i\in [K]$ . Note that the general case follows immediately by taking translations of the sets, i.e. replacing $A_i$ with $A^{\prime}_i = A_i-c_i$ for $c_i = \min \{a\,:\, a\in A_i\}$ . Indeed if (2) holds for $A^{\prime}_1,\ldots, A^{\prime}_K$ (which may now lie in $[0,2M]$ instead of $[{-}M,M]$ ) then (2) also holds for $A_1,\ldots, A_K$ (possibly with a different value of $a$ ). The same argument also allows us to assume that $0 \in A_i$ for all $i \in [K]$ .

Next, we double count pairs $(m,j)\in [M]\times [K]$ for which $m$ is among the largest $\delta M/2$ elements in $A_j$ . Counting by each set $A_j$ , we deduce that there are $\delta KM/2$ such pairs. Therefore, there is $M^\prime \in [M]$ which appears in at least $\delta K/2$ of these pairs. Let $J\,:\!=\,\{j\in [K]\,:\,M^\prime \in A_j\}$ , so that $|J|\geq \delta K/2$ , and note that $|A_j\cap [0,M^\prime ]|\geq \delta M/2$ as $M^\prime$ is one of the largest $\delta M/2$ elements in $A_j$ . Recalling that also $0\in A_j$ , by restricting all the sets to $[0,M^{\prime}]$ , then eventually reordering them and slightly adjusting the parameter $C$ , we can now assume that $A_i\subset [0,M]$ for all $i\in [K]$ , but with $\{0,M\} \subset A_i$ as well.

Given a set $S \subset{\mathbb Z}$ , we will let $\overline{S} \,:\!=\, \{a \in [0,M-1]\,:\, a \equiv s \mod M \mbox{ for some } s \in S\}$ . For each $i\in [K]$ let $S_i \,:\!=\, \overline{A_1 + \cdots + A_i}$ . By reordering the sets $A_1,\ldots, A_K$ , we may assume that $|S_{i+1}| \gt |S_{i}|$ for $i \leq K^{\prime}$ and that $S_i = S_{K^{\prime}}$ for all $i \geq K^{\prime}$ . Observe that $S_{K^{\prime}}\subset [0,M-1]$ and that $|S_{K^{\prime}}| \geq |S_1| + (K^{\prime}-1) \geq K^{\prime}+ 1$ , therefore $K^{\prime} \lt M$ .

Now, as $0 \in A_i$ for all $i\in [K]$ , we have $0 \in S_{K^{\prime}} \subset \overline{S_{K^{\prime}} + A_{K^{\prime}+1}} = S_{K^{\prime}+1}$ . As $|S_{K^{\prime}+1}| = |S_{K^{\prime}}|$ , it follows that the set $S_{K^{\prime}}$ contains the subgroup of ${\mathbb Z}_M$ generated by $A_{K^{\prime}+1}$ . However, recalling that $|S_{K^{\prime}}| = |S_{K^{\prime}+1}| \geq |A_{K^{\prime}+1}| \geq \delta M$ , we obtain that there is some $d \leq d_0(\delta )$ with $d|M$ such that $\{id \,:\, 0\leq i \leq M/d\} \subset S_{K^{\prime}} \subset S_M$ .

To proceed with the last step, recall that $\{0,M\} \in A_i$ for all $i\in [K]$ . Using this, it is easy to see that as $\{id \,:\, 0\leq i \leq M/d\} \subset S_{K^{\prime}} \subset S_M$ , we have $\{M^2 + id\,:\, 0 \leq i \leq M/d\} \subset A_1 + \cdots + A_{2M}$ . More generally, as $d | M$ , we have:

\begin{equation*} \big \{M^2 + id + jM\,:\, 0 \leq i \leq M/d, \mbox { } 0 \leq j \leq K-2M \big \} \subset A_1 + \cdots + A_{K}, \end{equation*}

and so (2) holds by taking $a = M^2$ and $K \geq CM \geq (B + 2)M$ . This completes the proof.

Proof. Let us start by dividing the interval $[0,|V_2|]$ into $\ell \,:\!=\,{|V_2|}^{1/2}$ disjoint intervals $I_1,I_2,\ldots I_l$ of length ${|V_2|}^{1/2}$ and let, for each interval $I_j$ , $n_j$ denote the number of vertices in $V_1$ whose degree lie in $I_j$ . Now any vertices $x, y \in I_j$ with $d(x) \geq d(y)$ give an ordered pair $\boldsymbol{{p}} = (x,y)$ with $\operatorname{deg-diff}^{}({{\bf p}}) \leq{|V_2|}^{1/2}$ . Obviously $n_1+n_2+\ldots + n_{\ell }=|V_1|$ , so by Jensen’s inequality the collection $\mathcal{S}$ of such ordered pairs $\boldsymbol{{p}} = (x,y) \in \binom{V_1}{2}$ has size:

\begin{align*} |{\mathcal{S}}| \geq \sum _{j=1}^{\ell }\binom{n_j}{2}\geq \ell \cdot \binom{|V_1|/\ell }{2} \geq \frac{|V_1||\big (|V_1|/\ell -1\big )}{2} \geq \frac{|V_1|^{1.5}}{4}, \end{align*}

using $\ell = |V_2|^{0.5} \leq (2|V_1| )^{0.5}$ and $|V_1| \geq n_C$ . By Lemmas 2.8 and 2.5, we deduce that our graph $G$ is $\varepsilon _0$ -diverse, where $\varepsilon _0\,:\!=\,(400C)^{-1}$ . Therefore, at most $|V_1|^{1.2}=o(|\mathcal{S}|)$ pairs from $\mathcal{S}$ fail to be ${\varepsilon }_0$ -diverse. By removing such pairs, we obtain a collection $\mathcal{S}_0$ of size at least $|V_1|^{1.5}/8$ such that $\operatorname{deg-diff}\!({\bf p}) \leq |V_2|^{1/2}$ and $|\text{div}(\boldsymbol{{p}})|\geq \varepsilon _0|V_2|$ for all $\boldsymbol{{p}}\in \mathcal{S}_0$ .

Next, we view the elements of $\mathcal{S}_0$ as the (unordered) edges of a graph $H$ on $V(G)$ . We claim that either $H$ has a matching of size $m\,:\!=\, |V_1|^{0.75}/4$ or a set of $m+1$ edges that have a common vertex. To see why this is true, suppose that neither of these two events hold and let $M$ be a largest matching in $H$ . Then $|M|\lt m$ and so all the edges in $e(H)\setminus M$ must be adjacent to an edge of $M$ . But because of the other condition, each edge of $M$ has at most $2m$ other edges adjacent to it. Thus, $e(H)\lt 2m^2 \leq |\mathcal{S}_0|$ , which is a contradiction.

Let us now observe that this set of edges will create our pair-star or pair-matching. Indeed, assume first that we have obtained $m+1$ edges $x_0x_1,x_0x_2,\ldots,x_0x_{m+1}$ in $\mathcal{S}_0$ . Again, by diversity, each vertex $x_j$ has at most $|V_1|^{0.2}$ other vertices $x_i$ such that $|\text{div}(x_j,x_i)|\lt \varepsilon _0|V_2|$ . Thus, by Turán theorem there is a set $I\subset [0, m+1]$ , with $0 \in I$ , of size at least $(m+1)/(1+|V_1|^{0.2}) \geq |V_1|^{0.5}$ such that $|\text{div}(x_j,x_i)|\geq \varepsilon _0|V_2|$ for all $i\neq j$ in $I$ . Taking $x_0$ as the root gives our pair-star.

In the other case, assume that we have obtained $m$ disjoint ordered pairs ${\bf p}_1,\ldots,{\bf p}_m$ in ${\mathcal{S}}_0$ , with ${\bf p}_i = (x_i, y_i)$ . Then, by Lemmas 2.5 and 2.8 our graph $G$ is $((\varepsilon _0/2)^2, \varepsilon _0/2)$ -pair-diverse, so for each pair $(x_j,y_j)$ there are at most $|V_1|^{0.2}$ other pairwise disjoint edges $x_iy_i$ such that $|\text{divb}(x_j,y_j)\setminus N(x_i)|\lt (\varepsilon _0/2)^2 |V_2|$ or $|\text{divb}(x_j,y_j)\setminus N(y_i)|\lt (\varepsilon _0/2)^2 |V_2|$ . Setting $\varepsilon = (\varepsilon _0/2)^2$ , by Turán, there is a set $I\subset [m]$ of size at least $m (1+|V_1|^{0.2} )^{-1} \geq |V_1|^{0.5}$ such that $|\text{divb}({\bf p}_j)\setminus N(x_i)|\geq \varepsilon |V_2|$ and $|\text{divb}({\bf p}_j)\setminus N(y_i)|\geq \varepsilon |V_2|$ for all $i\neq j$ in $I$ . This represents our pair-matching.

Proof. (i) We will use a classical randomness exposure argument. Suppose we reveal the random set $W$ on $V_2\setminus \text{div}(x,y)$ . Then, given such a choice, the difference $d^{W}(x)-d^{W}(y)$ becomes $d^{U}(x)-d^{U}(y)+\text{constant}$ , where $U \,:\!=\,W \cap \text{div}(x,y)$ . Now, $d^{U}(x)-d^{U}(y)\,:\!=\,\sum _{v\in \text{div}(x,y)}\theta _vX_v$ , where for each $v\in \text{div}(x,y)$ we have $\theta _v\in \{-1,1\}$ and $X_v\sim \text{Bern}(0.5)$ . Since $|\text{div}(x,y)|\geq \delta |V_2|$ , by Theorem 2.9, the random variable $d^{W^{\prime}}(x)-d^{W^{\prime}}(y)$ hits any particular value with probability $O_\delta (|V_2|^{-1/2} )$ . The conclusion follows from the law of total probability.

(ii) The argument here is similar, but a little more involved. By the choice of the ordering we have $\mbox{divb}({\bf p}_1) = N(x_1) \setminus N(y_1)$ . Note that the condition in the lemma now gives that the set $T = \mbox{divb}({\bf p}_1) \setminus N(x_2) = N(x_1) \setminus ( N(x_2) \cup N(y_1) )$ satisfies $|T| \geq \delta |V_2|$ . Letting $X$ denote the random variable $X = \operatorname{deg-diff}^{W}\!({{\bf p}_1}) - \operatorname{deg-diff}^{W}\!({{\bf p}_2})$ we have:

\begin{align*} X = d^{W}(x_1)-d^{W}(y_1) -d^{W}(x_2) + d^{W}(y_2). \end{align*}

Now note that if we expose the random set $W \cap (V_2 \setminus T)$ , then the random variable $X$ reduces to $d^{W \cap T}(x_1)+d^{W \cap T}(y_2)-C$ , where $C$ is some constant depending only on $W\cap (V_2 \cap T)$ – crucially, here we use that $y_1$ and $x_2$ have no neighbour in $T$ . But $d^{W \cap T}(x_1)+d^{W \cap T}(y_2)=\sum _{v\in T}\theta _vX_v$ where $\theta _v\in \{1,2\}$ is a constant for each $v\in T$ and $X_v\sim \text{Bern}(0.5)$ so that $\{X_v\}_{v\in T}$ are independent. By Theorem 2.9, since $|T|\geq \delta |V_2|$ , the random variable $d^{W \cap T}(x_1)+d^{W\cap T}(y_2)$ takes any particular value with probability at most $O_\delta (|V_2|^{-0.5} )$ . The conclusion again follows from the law of total probability.

Proof. To begin, pick $j\in [|V_2|^{0.5}]$ and set $D_j \,:\!=\, d^W(x_j) - d^W(x_0)$ . By Chebyshev’s inequality, we have $ |D_j-\mathbb E[D_j] |\leq 2{|V_2|}^{0.5}$ with probability at least $15/16$ . As $ |\mathbb E[D_j] |\leq |V_2|^{1/2}/2$ , it follows by triangle inequality that $\mathbb P (|d^{W}(x_0)-d^{W}(x_j)|\leq 3|V_2|^{0.5} )\geq 15/16$ .

Next, call a vertex $x_j$ with $j\in [|V_2|^{0.5}]$ bad if $|d^{W}(x_0)-d^{W}(x_j)|\leq 3|V_2|^{0.5}$ and good otherwise. Note $\mathbb P(x_j\text{ is bad})\leq 1/16$ and so $\mathbb E [ |\{j\in [|V_1|^{0.5}]\,:\,x_j\text{ is bad}\} | ]\leq |V_2|^{0.5}/16$ . If $U$ denotes the set of good vertices, then it follows by Markov that $\mathbb P (|U|\geq |V_2|^{0.5}/2 )\geq 7/8$ .

Now we build a graph $H$ on $\{x_1,\ldots, x_{|V_2|^{1/2}}\}$ where we join two vertices by an edge in $H$ if their degrees in $W$ are equal. By Lemma 3.2 (i) there is an edge in $H$ between two vertices $x_i$ and $x_j$ with probability $O_{\varepsilon } (|V_2|^{-1/2} )$ . Therefore, $\mathbb E[e(H)]=O_{\varepsilon } ((|V_2|^{0.5})^2\cdot |V_2|^{-0.5} )$ and by Markov we easily deduce that, with probability at least $7/8$ , one has $e(H)=O_{\varepsilon } (|V_2|^{0.5} )$ .

Combining our estimates, by applying the union bound we find that $|U| \geq |V_2|^{0.5}/2$ and that $e(H[U]) \leq e(H)=O_{\varepsilon } (|V_2|^{0.5} )$ with probability at least $3/4$ . But then the average degree of $H$ is of order $O_{\varepsilon } (1 )$ , hence by Turán $H$ contains an independent set of size $\Omega _{\varepsilon } ({|V_2|^{0.5}} )$ . This set though gives us precisely the vertices with pairwise distinct degrees that we sought.

Proof. Suppose that $W \subset V_2$ is selected as in the lemma. To begin, we fix a pair $(k,m)$ so that $0 \leq k \leq m$ and $2 \leq m\leq d$ . For $i \in [L]$ let $E_i$ denote the event $E_i \,:\!=\, \left\{d^W_G(u_i) \not \equiv k \mod m\right\}$ . Our first aim is to upper bound ${\mathbb{P}} \left( \cap _{i\in [L]} E_i \right)$ . To do this, we can observe that:

\begin{align*} \mathbb P\big (\cap _{i=1}^L E_i\big )=\prod _{i=1}^L \mathbb P\!\left (E_i|\cap _{j\lt i}E_j\right ). \end{align*}

Now the event $\cap _{j\lt i}E_i$ is entirely determined by the choice of $W^{\prime} = W \cap (V_2 \setminus S_i)$ . Thus, to upper bound $\mathbb P \!\left(E_i|\cap _{j\lt i}E_j \right)$ , it suffices to upper bound $\mathbb P\! \left(E_i|W^{\prime} = W_0 \right)$ for each choice of $W_0$ . However, given such a choice of $W_0$ , the conditional probability $\mathbb P\! \left(E_i|W^{\prime}=W_0 \right)$ becomes $\mathbb P\! \left(d^{W\cap S_i}_G(u_i)\not \equiv k_i\ (\text{mod }m) \right)$ for some $0\leq k_i\lt m$ . Besides this, from our hypothesis we know that $W \cap S_i \sim \mbox{Bin}(|S_i|,0.5)$ and $|S_i| \geq D$ , so by Proposition 2.10 this gives $\mathbb P\! \left(E_i|\cap _{j\lt i}E_j \right) \leq \max _{W_0}{\mathbb{P}}\left(E_i| W^{\prime} = W_0\right) \leq d(d+1)^{-1}$ . Combining all these, we obtain:

\begin{align*} \mathbb P\big (\cap _{i=1}^L E_i\big ) \leq \prod _{i=1}^L \mathbb P\!\left (E_i|\cap _{j\lt i}E_j\right ) \leq \Big ( 1 - \frac{1}{d+1} \Big ) ^{L}\leq \exp \!\left ({-}L(d+1)^{-1}\right ). \end{align*}

To finish the proof, let $F_{(k,m)}$ denote the event that $\{d_G^W(u_i) \not \equiv k \mod m \mbox{ for all } i\in [L]\}$ . We have shown that ${\mathbb{P}}(F_{(k,m)}) \leq \exp\!\left( -L(d+1)^{-1} \right)$ . It follows that the probability that some congruence is not obtained is ${\mathbb{P}}\! \left( \cup _{(k,m)} F_{k,m} \right) \leq \sum _{(k,m)}{\mathbb{P}}\!\left(F_{k,m}\right) \leq d^2 \exp\!\left({-}L(d+1)^{-1} \right) \leq \varepsilon$ , provided that $L$ (and $D$ from above) are sufficiently large.

Proof. We start by fixing parameters $1\geq C^{-1} \gg \varepsilon \gg \delta \gg d_0^{-1} \gg L^{-1} \gg C_0^{-1} \gg c \gg n_C^{-1} \gt 0$ . Without loss of generality, we may assume that $|V_1| \geq |V_2|$ . We also fix an arbitrary set $V^{\prime}_2 \subset V_2$ with $|V^{\prime}_2| = c|V_2|$ and a partition $V_1 = U_1 \cup U_2 \cup U_3$ , where $|U_i| \geq |V_1|/4$ for $i\in [3]$ .

To begin, we claim that there are ${\mathcal{P}}_1,\ldots,{\mathcal{P}}_K$ with $K \,:\!=\, C_0|V_2|^{0.5}$ such that each ${\mathcal{P}}_i$ is either an $\varepsilon$ -pair-star or a $\varepsilon$ -pair-matching in $G\left[U_1, V^{\prime}_2\right]$ of size $|V^{\prime}_2|^{0.5}$ and so that the ${\mathcal{P}}_i$ ’s are all vertex disjoint. To see this, suppose that we have found ${\mathcal{P}}_1,\ldots,{\mathcal{P}}_i$ and now seek ${\mathcal{P}}_{i+1}$ . Let $U^{\prime}_1$ denote the subset of $U_1$ with the vertices of $\cup _{j\leq i}{\mathcal{P}}_j$ removed and apply Lemma 3.1 to $G\!\left[U^{\prime}_1, V^{\prime}_2\right]$ , noting that as $|U^{\prime}_1| \geq |U_1| - 2 K |V^{\prime}_2|^{0.5} \geq |U_1| - 2(C_0|V_2|^{0.5}) (c|V_1|)^{0.5} \geq |U_1|/2 \geq |V_1|/8$ , making $G\!\left[U^{\prime}_1, V^{\prime}_2\right]$ a $2C$ -Ramsey graph. By Lemma 3.1, we can find a $\varepsilon$ -star-pair or a $\varepsilon$ -pair-matching ${\mathcal{P}}_{i+1}$ of size $|U^{\prime}_1|^{0.5} \geq |V^{\prime}_2|^{0.5}$ , which gives the desired set (perhaps after removing some of its elements). Thus such a collection ${\mathcal{P}}_1,\ldots,{\mathcal{P}}_K$ exists.

Our next step is to apply Lemma 2.3 to $G\!\left[U_2, V^{\prime}_2\right]$ , which is again $2C$ -Ramsey, to find a set $S_2 = \{s_1,\ldots, s_L\} \subset U_2$ such that $\big|N(s_i) \setminus ( \cup _{j\lt i}N(s_j) )\big| \geq |V_2|^{0.5}$ for all $i\in [L]$ . It is possible to do this since $C^{-1} \gg L^{-1} \gg n_C^{-1}$ .

The final step of preparation is to select a set $W \subset V^{\prime}_2$ uniformly at random. For each $i\in [K]$ let $A_i \,:\!=\, A_{{\mathcal{P}}_i}^W$ . Now consider the following events:

• Let ${\mathcal{E}}_1$ denote the event that, for at least $K/2$ values $i\in [K]$ , we have $|A_i| \geq \delta |V^{\prime}_2|^{1/2}$ . By Lemmas 3.3 and 3.4, and using that $\varepsilon \gg \delta$ , we see that for any $A_i$ we have $|A_i| \geq \delta |W|^{1/2}$ with probability at least $3/4$ , so a simple application of Markov gives ${\mathbb{P}}({\mathcal{E}}_1) \geq 1/2$ ;

• Secondly, let ${\mathcal{E}}_2$ denote the event that, for every $0 \leq k \leq m$ and $2 \leq m \leq d_0$ , there exists $s \in S_2$ with $d^W_G(s) \equiv k \mod m$ . By the choice of $S_2$ , Lemma 3.5 gives ${\mathbb{P}}({\mathcal{E}}_2) \geq 7/8$ ;

• Lastly, let ${\mathcal{E}}_3$ denote the event that $|W| \geq |V^{\prime}_2|/4$ . By Chebyshev’s inequality ${\mathbb{P}}({\mathcal{E}}_3) \geq 7/8$ .

It follows from the union bound that ${\mathbb{P}}({\mathcal{E}}_1 \cap{\mathcal{E}}_2 \cap{\mathcal{E}}_3) \geq 1/2$ . Fix a choice of $W \subset V^{\prime}_2 \subset V_2$ for which all three events occur. By reordering, we get that $|A_i| \geq \delta |V^{\prime}_2|^{1/2} \geq \delta |W|^{1/2}$ for $i\in [K/2]$ .

We are now ready to complete the proof. Take $V_1^{\mbox{init}}$ to be the union of the heads of ${\mathcal{P}}_i$ , i.e. $V_1^{\mbox{init}} \,:\!=\, \cup _{i \leq L/2} H({\mathcal{P}}_i)$ , and $e_0 \,:\!=\, e(G[V_1^{\mbox{init}}, W])$ . Take $a \in A_i$ and note the following:

1. (i) If ${\mathcal{P}}_i = \{x_0,\ldots, x_{|V^{\prime}_2|^{0.5}}\}$ is a pair-star rooted at $x_0$ , then by definition $d^W(x_j) - d^W(x_0) = a$ for some $j\in \left[\big|V^{\prime}_2\big|^{1/2}\right]$ . Removing $x_0$ from $V_1^{\mbox{init}}$ and adding $x_j$ changes the number of edges in the resulting graph by exactly $a$ .

2. (ii) Similarly, if ${\mathcal{P}}_i$ is a pair-matching, then there is ${\bf p} = (x,y) \in{\mathcal{P}}_i$ such that $\operatorname{deg-diff}^{W}\!({{\bf p}}) = a$ . Removing $x$ from $V_1^{\mbox{init}}$ and adding $y$ to it changes the number of edges by exactly $a$ .

3. (iii) The edits from different ${\mathcal{P}}_i$ ’s can be performed independently of one another, resulting in the same changes to the edge size given in (i) and (ii).

From observations (i) $-$ (iii) we can immediately deduce that:

\begin{equation*}\{e_0\} + A_1 + A_2 + \cdots + A_{K/2} \subset \big \{ e(G[U,W])\,:\, U \subset U_1 \big \}.\end{equation*}

By definition of $A_i\,:\!=\,A_{{\mathcal{P}}_i}^W$ , we have $A_i \subset \left[{-}3|V^{\prime}_2|^{1/2}, 3|V^{\prime}_2|^{1/2}\right] \subset \left[{-}6|W|^{1/2}, 6|W|^{1/2}\right]$ and $|A_i| \geq \delta |W|^{1/2}$ . By taking $M = 6|W|^{1/2}$ and $B = 1$ in Lemma 2.12, as $\delta,1 \gg C_0^{-1}, d_0^{-1}$ , it follows that there is $a \in{\mathbb Z}$ and $1 \leq d \leq d_0$ such that (2) holds, which gives:

(3) \begin{align} \left\{e_0 + a + id\,:\, 0 \leq i \leq 3|W|\right\} \subset \{e_0\} + \left\{a + id\,:\, 0 \leq i \leq M^2\right\} & \subset \{e_0\} + A_1 + A_2 + \cdots + A_{L/2} \nonumber \\ & \subset \left\{e(G[U, W])\,:\, U \subset U_1 \right\}. \end{align}

Furthermore, as $d \leq d_0$ , by the choice of $S_2$ there are $s_{i_0}, s_{i_1},\ldots, s_{i_{d-1}} \in S_2$ with $d^W(s_j) \equiv j \mod d$ for each $j\in [0,d-1]$ . Combined with (3), setting $e_1 = e_0 + a + |W|$ , this gives:

(4) \begin{align} \big [e_1, e_1 + |W| \big ] \subset \big \{e(G[U \cup \{s\}, W])\,:\, U \subset U_1, s \in S_2 \big \}. \end{align}

To finish the proof, note that $|U_3| \geq |V_1|/4 \geq |V_1|^{1/2}$ and $|W| \geq |V^{\prime}_2|/4 \geq |V_2|^{1/2}$ . As $G$ is $C$ -Ramsey, we get that $G[U_3, W]$ is $(2C)$ -Ramsey. It follows from Corollary 2.2 that at least $2|U_3|/3$ vertices in $U_3$ have degrees between $(64C)^{-1}|W|$ and $(1-(64C)^{-1})|W|$ in $G[U_3, W]$ . In particular, $\{e(G[U^{\prime},W]) \,:\, U^{\prime} \subset U_3\}$ contains an element from each interval $[|W|i, |W|(i+1)]$ with $i\in [0,(96C)^{-1}|U_3|]$ . However, $e(G[U \cup \{s\} \cup U^{\prime}, W]) = e(G[U \cup \{s\}, W]) + e(G[ U^{\prime}, W])$ for any $U \subset U_1, s \in S_2$ and $U^{\prime} \subset U_3$ . Therefore, we deduce from (4) that:

\begin{align*} \big [ e_1 + |W|, e_1 + |W| + (96C)^{-1}|U_3||W| \big ] \subset \big \{ e(G[U \cup \{s\} \cup U^{\prime}, W])\,:\, U \subset U_1, s \in S_2, U^{\prime} \subset U_3 \big \}. \end{align*}

We can finally claim that the lemma holds with $a_C \,:\!=\, c/4000C$ . To see this, note that:

\begin{align*} e_1 + |W| = e_0 + a + |W| & \leq (K/2)|V^{\prime}_2|^{0.5}|W| + (K/2)(6|W|)^{0.5}|W| + |W|\\ &\leq 4K|W|^{1.5} \leq 4(C_0|V_2|)^{0.5}(c|V_2|)^{1.5} = 4c^{1.5}(C_0)^{0.5}|V_2|^2 \leq a_C|V_1||V_2|, \end{align*}

where we have used that $|W| \leq c|V_2|$ , that $c \ll C_0^{-1}, C^{-1}$ , and that $|V_2| \leq |V_1|$ .

Since $a_C|V_1||V_2| = (c/4000C)|V_1||V_2| \leq (200C)^{-1}|U_3||W|$ , it is quite easy to observe that $[a_C|V_1||V_2|, 2a_C|V_1||V_2|] \subset \{e(G[U])\,:\, U \subset V(G)\}$ , as required. This completes the proof.

Proof of Theorem 1.2. By making $\alpha$ small enough we may assume, without losing generality, that $|V_1|\geq |V_2|$ and that $|V_2|$ is large enough so that our estimates below hold. We can also assume that $C\gt 1$ since the Ramsey condition still holds when we increase $C$ .

First note that as $G = (V_1, V_2, E)$ is $C$ -Ramsey, there is a vertex $v \in V_2$ with degree at least $(32C)^{-1}|V_1|$ in $V_1$ , by Corollary 2.2. But then the induced subgraphs of the form $G[W, \{v\}]$ , where $W \subset V_1$ , give all edge sizes in $[0,(32C)^{-1}|V_1|]$ (and so in particular the analogue of the Alon $-$ Krivelevich $-$ Sudakov theorem from [Reference Alon, Krivelevich and Sudakov1] is easier in the bipartite Ramsey context).

On the other hand, observe that by fixing any $W_1 \subset V_1$ and $W_2 \subset V_2$ with $|W_i| \geq |V_i|^{1/3}$ for $i \in \{1,2\}$ , the graph $H = G[W_1, W_2]$ is $(3C)$ -Ramsey, as $G$ is $C$ -Ramsey. It follows by applying Lemma 3.6 to $H$ that there are induced subgraphs of each size in $[a_{3C}|W_1||W_2|, 2a_{3C}|W_1||W_2|]$ . It follows, taking $M = \{(m_1, m_2)\,:\, |V_i|^{1/3} \leq m_i \leq |V_i| \mbox{ for } i \in \{1,2\}\}$ , that $G$ contains a subgraph of each size in $\cup _{(m_1,m_2) \in M} [a_{3C}m_1m_2, 2a_{3C}m_1m_2]$ . It can be easily seen that these sets together cover the interval $[a_C(|V_1||V_2|)^{1/3}, 2a_C |V_1||V_2|]$ .

Finally, note that the intervals $[0,(32C)^{-1}|V_1|]$ and $[a_{3C}(|V_1||V_2|)^{1/3}, 2a_{3C} |V_1||V_2|]$ together cover $[0, 2a_{3C} |V_1||V_2|]$ , since $|V_1| \geq |V_2|$ and $|V_2|$ is large enough, thus completing the proof of the theorem.