A sparse graph without a large sparse pair has a large nearly covered sparse pair

Because $G$ is not $2y^4$-sparse, some vertex $v$ has degree at least $2y^4\mathrm{\mid }V(G)\mathrm{\mid }$. Let $N:=N_G(v)$. Then $\mathrm{\mid }N\mathrm{\mid }\ge 2y^4\mathrm{\mid }V(G)\mathrm{\mid }$.

Let $A^{\mathrm{\prime }}$ be the set of vertices in $V(G)\setminus (N\cup \{v\})$ with at least $\frac{1}{2}{y}^2\mathrm{\mid }N\mathrm{\mid }$ neighbours in $N$. Averaging over the edges between $A^{\mathrm{\prime }}$ and $N$, some vertex of $N$ has at least $\frac{1}{2}{y}^2\mathrm{\mid }{A}^{\mathrm{\prime }}\mathrm{\mid }$ neighbours in $A^{\mathrm{\prime }}$. Since every vertex has degree at most $y^3\mathrm{\mid }V(G)\mathrm{\mid }$ by [L1], we obtain $\frac{1}{2}{y}^2\mathrm{\mid }{A}^{\mathrm{\prime }}\mathrm{\mid }\le y^3\mathrm{\mid }V(G)\mathrm{\mid }$, so $\mathrm{\mid }{A}^{\mathrm{\prime }}\mathrm{\mid }\le 2y\mathrm{\mid }V(G)\mathrm{\mid }$.

Define $A:=V(G)\setminus (N\cup A^{\mathrm{\prime }}\cup \{v\})$. Since $G$ is $y^3$-sparse, [L1] gives $\mathrm{\mid }N\mathrm{\mid }\le y^3\mathrm{\mid }V(G)\mathrm{\mid }$, and because $\mathrm{\mid }V(G)\mathrm{\mid }\ge y^{-4}$ we have $1\le y\mathrm{\mid }V(G)\mathrm{\mid }$. Therefore $\mathrm{\mid }A\mathrm{\mid }\ge \mathrm{\mid }V(G)\mathrm{\mid }-(y^3\mathrm{\mid }V(G)\mathrm{\mid }+2y\mathrm{\mid }V(G)\mathrm{\mid }+1)\ge (1-3y)\mathrm{\mid }V(G)\mathrm{\mid }$. Also, every vertex of $A$ has fewer than $\frac{1}{2}{y}^2\mathrm{\mid }N\mathrm{\mid }$ neighbours in $N$ by definition of $A^{\mathrm{\prime }}$.

Let $N^{\mathrm{\prime }}\subseteq N$ be the set of vertices with at most $x^2\mathrm{\mid }A\mathrm{\mid }$ neighbours in $A$, and put $B:=N\setminus N^{\mathrm{\prime }}$. The number of edges between $A$ and $N^{\mathrm{\prime }}$ is at most $x^2\mathrm{\mid }A\mathrm{\mid }\mathrm{\mid }{N}^{\mathrm{\prime }}\mathrm{\mid }$, so at most $x\mathrm{\mid }A\mathrm{\mid }$ vertices of $A$ have more than $x\mathrm{\mid }{N}^{\mathrm{\prime }}\mathrm{\mid }$ neighbours in $N^{\mathrm{\prime }}$. Hence at least $\mathrm{\mid }A\mathrm{\mid }-x\mathrm{\mid }A\mathrm{\mid }\ge (1-3y-x)\mathrm{\mid }V(G)\mathrm{\mid }\ge (1-4y)\mathrm{\mid }V(G)\mathrm{\mid }$ vertices of $A$ have at most $x\mathrm{\mid }{N}^{\mathrm{\prime }}\mathrm{\mid }$ neighbours in $N^{\mathrm{\prime }}$.

If $\mathrm{\mid }{N}^{\mathrm{\prime }}\mathrm{\mid }\ge y^4\mathrm{\mid }V(G)\mathrm{\mid }$, then step 4.1 gives a set $Y\subseteq A$ with $\mathrm{\mid }Y\mathrm{\mid }\ge (1-4y)\mathrm{\mid }V(G)\mathrm{\mid }$ such that every vertex of $Y$ has at most $x\mathrm{\mid }{N}^{\mathrm{\prime }}\mathrm{\mid }$ neighbours in $N^{\mathrm{\prime }}$. By [L2], the pair $(X,Y):=(N^{\mathrm{\prime }},Y)$ is then a forbidden large sparse pair, contradicting the hypothesis. Therefore $\mathrm{\mid }{N}^{\mathrm{\prime }}\mathrm{\mid }<y^4\mathrm{\mid }V(G)\mathrm{\mid }$.

Since $\mathrm{\mid }N\mathrm{\mid }\ge 2y^4\mathrm{\mid }V(G)\mathrm{\mid }$ by step 1.1 and $\mathrm{\mid }{N}^{\mathrm{\prime }}\mathrm{\mid }<y^4\mathrm{\mid }V(G)\mathrm{\mid }$ by step 5.1, we have $\mathrm{\mid }B\mathrm{\mid }=\mathrm{\mid }N\mathrm{\mid }-\mathrm{\mid }{N}^{\mathrm{\prime }}\mathrm{\mid }\ge y^4\mathrm{\mid }V(G)\mathrm{\mid }$. By definition of $B$, every vertex of $B$ has more than $x^2\mathrm{\mid }A\mathrm{\mid }$ neighbours in $A$. Also step 3.1 gives at most $\frac{1}{2}{y}^2\mathrm{\mid }N\mathrm{\mid }$ neighbours in $N$ for each vertex of $A$, while step 5.1 implies $\mathrm{\mid }B\mathrm{\mid }>\mathrm{\mid }N\mathrm{\mid }\mathrm{/}2$; hence $\frac{1}{2}{y}^2\mathrm{\mid }N\mathrm{\mid }<y^2\mathrm{\mid }B\mathrm{\mid }$, so every vertex of $A$ has at most $y^2\mathrm{\mid }B\mathrm{\mid }$ neighbours in $B$. Therefore $A$ is $y^2$-sparse to $B$.

Step 3.1 gives $A\subseteq V(G)\setminus N_G[v]$, step 1.1 gives $B\subseteq N_G(v)$, and steps 3.1 and 6.1 give the size, sparsity, and neighbourhood clauses. These are exactly the four clauses of the statement.