Constant-density limiting center loses regularity

Fix ${\rho }_0={ϵ}_0\mathrm{/}{c}^2>0$. For integers $n\ge 2$, set $u_n=8\mathrm{/}9-1\mathrm{/}n$ and $R_n=c^2\sqrt{3u_n\mathrm{/}(8\pi G{ϵ}_0)}>0$. With $M_n=4\pi {\rho }_0{R}_n^3\mathrm{/}3$, direct substitution gives $2G{M}_n\mathrm{/}(c^2{R}_n)=8\pi G{ϵ}_0{R}_n^2\mathrm{/}(3c^4)=u_n\in (0,8\mathrm{/}9)$. Thus each $({\rho }_0,R_n)$ supplies an admissible regular star of [F1], rather than varying $u$ at fixed radius. By [F2], after reindexing $k=n-1$, $1\mathrm{/}n=1\mathrm{/}(k+1)\to 0$. Set $a_n=\sqrt{1-u_n}=\sqrt{1\mathrm{/}9+1\mathrm{/}n}>1\mathrm{/}3$; then $0<a_n-\frac{1}{3}=\frac{1\mathrm{/}n}{a_n+1\mathrm{/}3}\le \frac{3}{2n},$ For any real $\eta >0$, [F3] gives rational $q$ with $0<q<2\eta \mathrm{/}3$. By [F2], eventually $1\mathrm{/}n<q$, so $\mathrm{\mid }{a}_n-1\mathrm{/}3\mathrm{\mid }\le 3\mathrm{/}(2n)<3q\mathrm{/}2<\eta$. This proves $a_n\to 1\mathrm{/}3$ directly from the convergence definition, including all rational positive error thresholds.

For $n\ge 9$, step 1.1 gives $a_n\le 1\mathrm{/}2$, so ${ϵ}_0(1-a_n)\ge {ϵ}_0\mathrm{/}2$ and $0<3a_n-1\le 9\mathrm{/}(2n)$. Hence $p_c(u_n)\ge {ϵ}_{0}n\mathrm{/}9$. Given $M>0$, apply [F3] to $(0,{ϵ}_0\mathrm{/}(9M))$ to choose a rational $q>0$ with $q<{ϵ}_0\mathrm{/}(9M)$; by [F2], for all sufficiently large $n$, $1\mathrm{/}n<q$, so $p_c(u_n)>M$. Also $0<e^{\varphi (0)}(u_n)=(3a_n-1)\mathrm{/}2\le 9\mathrm{/}(4n)$; for any $\eta >0$, apply [F3] to $(0,4\eta \mathrm{/}9)$ to choose rational $q>0$ with $q<4\eta \mathrm{/}9$ and apply [F2] to get $e^{\varphi (0)}(u_n)<\eta$ eventually. Thus the central pressure diverges to $+\mathrm{\infty }$ and the central lapse tends to zero. At $u=8\mathrm{/}9$ the denominator is exactly zero with positive numerator, so no finite central pressure or positive central lapse is obtained by extending these formulas.