$n$ vertices whose checkerboard graphons (\[checkerboard\]) lie in $U_\delta$ or $F_\delta$. The large deviation principle, Theorem 2.3 of [@CV], implies that: $$\limsup_{n \to \infty} \frac{\ln|F_\delta^n|}{n^2} \le -\inf_{\tilde g\in \tilde F_\delta} I(\tilde g),$$ which also equals $-\inf_{g \in F_\delta} I(g)$, and that $$\liminf_{n \to \infty} \frac{\ln|U_\delta^n|}{n^2} \ge -\inf_{\tilde g \in \tilde U_\delta} I(\tilde g),$$ which also equals $-\inf_{g \in U_\delta} I(g)$. This yields a chain of inequalities $$-\inf_{U_\delta} I(g) \le \liminf \frac{\ln|U^n_\delta|}{n^2} \le \limsup \frac{\ln|U^n_\delta|}{n^2} \le 
\limsup \frac{\ln|F_\delta^n|}{n^2} \le - \inf_{F_\delta} I(g) \le -
\inf_{U_{\delta+\delta^2}} I(g)$$ As $\delta \to 0^+$, the limits of $-\inf_{U_\delta} I(g)$ and $-\inf_{U_{\delta+\delta^2}} I(g)$ are the same, and everything in between is trapped.

So far we have proven that $$s_{e,t} = -\lim_{\delta \to 0^+}
\inf_{U_\delta} I(g).$$ Next we must show that the right hand side is equal to $- \min_{F_0} I(g)$. By definition, we can find a sequence of reduced graphons $\tilde g_\delta \in \tilde U_\delta$ such that $\lim_{\delta \to 0} I(\tilde g_\delta) = \lim \inf_{U_\delta}
I(g)$. Since $\tilde W$ is compact, these reduced graphons converge to a reduced graphon $\tilde g_0$, represented by a graphon $g_0 \in
F_0$. Since $I$ is lower-semicontinuous [@CV], $I(g_0) \le \lim
I(g_\delta)$, so $\min_{F_0} I(g) \le \lim \inf_{U_\delta} I(g)$. (We write 