n$ for the final inequality.

\[rem1\] A similar calculation shows that if $U$ is $\sigma$-close to bipartite with bipartition $A,B$, we have $|A|,|B|\geq D - 2\sqrt{\sigma}n \geq (\alpha -2\sqrt{\sigma})n$.

We will use the algorithm in Theorem \[th:cheeger\] to iteratively find subgraphs of $G[U]$ that are not well connected to the rest of $U$ and remove them until this is no longer possible. If this process continues to a point where the removed parts are large enough then we can show both the removed part and the remaining part each form a $\rho'$-component. If the process stops before the removed part becomes large then we can show $G[U]$ is a robust expander.

Let $G=(V,E)$ and, in this proof, for any subset $S \subseteq U$ we will use $\overline{S}$ to denote $U \setminus S$ rather than our usual convention where it denotes $V \setminus S$.

Let $n'=|U|$ so that $n' \geq (\alpha - \sqrt{\rho})n \geq \frac{1}{2} \alpha n$ (by the previous proposition). Let $U_0$ be the vertices of degree at most $\frac{1}{2}\alpha n$ in $G[U]$ so that $|U_0| \leq  \frac{2\rho}{\alpha(\alpha-\sqrt{\rho})} n' \leq \alpha \nu n'/2$ also by the previous proposition. Note for later that $$\label{eq:volU0}
{\text{vol}}_G(U_0) \leq  n |U_0| \leq (2n' / 