Prop}\,\ref{Claim1}}{\geq} \alpha n - \frac{1}{4}\alpha n - (\rho + \nu )\frac{n'}{\tau (\alpha -\sqrt{\rho})^2}  \geq \frac{1}{2}\alpha n' \geq |S|+\nu n,
            \end{aligned}$$ contradicting our choice of $S$.

Similarly if $|S| > |U| - \frac{1}{4}\alpha n$ recall that by Proposition \[Claim1\] that all but the $\gamma n'$ vertices in $U_0$ have degree at least $\frac{1}{2}\alpha n$ in $U$ and so for all $x \in  U \setminus U_0$, we have $$d_S(x) \geq \frac{1}{2}\alpha n - |U\setminus S| \geq \frac{1}{4}\alpha n \geq \nu n.$$ Hence $N\supseteq U\setminus U_0$ and so $|N|\geq |U| - |U_0| \geq (1- \nu)n' \geq |S|+\nu n'$, a contradiction. This proves the claim.

![Overview of subsets mentioned in the coming section.[]{data-label="fig"}](pic1.pdf){width="30.00000%"}

We continue with the proof of Claim \[Claim4\]. We define $Y = S\setminus N$, $X = S \cap N$, $Z = N\setminus S$, $W = U\setminus (S\cup N)$; see Figure \[fig\]. Since each vertex in $Y$ has at most $ \nu n$ neighbours in $S$ and since $G$ is $D$-regular and $U$ is a $\rho$-component we have $e_G(Y,\overline{S}) \geq D |Y| - \rho n^2 - \nu n^2$. Using this, we obtain $$\begin{aligned}
        \label{(eYZ)}
 e_G(Y,Z) = e_G(Y,\overline{S})-e_G(Y,W)            
            &\geq D |Y|- \rho n^2 - \nu n^2 - |W|\nu n \notag \\ &\geq D 