Given $n$ and ${\rm v}>0$, assume no such $\epsilon$ exists. Then there exists a sequence of spaces $(M^{n}_{j},g_{j},p_{j})$ such that $|{\rm Ric}_{M^{n}_{i}}|\leq\epsilon_{j}\to 0$, ${\rm Vol}(B_{1}(p_{j}))>{\rm v}>0$ and

$\displaystyle d_{GH}\big(B_{2}(p_{j}),B_{2}(0_{j})\big)<\epsilon_{j}\to 0\,,$

(6.4)

where $0_{j}\in\mathds{R}^{n-3}\times C(Y_{j})$ is a vertex but $r_{h}(p)<1$. After possibly passing to a subsequence,we have

$\displaystyle B_{2}(p_{j})\to B_{2}(0)\,,$

(6.5)

where $0\in\mathds{R}^{n-3}\times C(Y)\equiv X$ is a vertex. But if $C(Y)$ has any point with $r_{h}(x)=0$, then there is a set of Hausdorff codimension $3$ in $X$ which is not smooth. By the Hausdorff estimate of Theorem 1.1 this is not possible, so we must have that $C(Y)$ is smooth. Thus, $Y$ is a smooth manifold, and in fact, $C(Y)$ is itself be smooth if and only if $Y$ is the unit $2$-sphere. Thus,

$\displaystyle B_{2}(p_{j})\to B_{2}(0^{n})\subseteq\mathds{R}^{n}\,.$

(6.6)

But now, we can apply the standard $\epsilon$-regularity theorem, to conclude $r_{h}(p)\leq 1$, which is a contradiction. ∎