Connectivity in the logarithmic random-graph window

Suppose $c_n\to c\in \mathbb{R}$ and let $X$ count isolated vertices. Applying [F4] and then [F5] with $\lambda =e^{-c}>0$ gives $\mathbb{P}(X=0)\to e^{-e^{-c}}$. For $n\ge 2$ a connected graph has no isolated vertex. If a graph is disconnected but has no isolated vertex, its smallest component has order between 2 and $n\mathrm{/}2$: there are at least two nonempty components, whose sizes sum to $n$. Let $A$ be the connectivity event, $B=\{X=0\}$, and $D$ the event that a component has order between 2 and $n\mathrm{/}2$. Then $A\subseteq B$ and $B\setminus A\subseteq D$. For the graph's outcome weights $w$, [F8] gives $0\le \mathbb{P}(B)-\mathbb{P}(A)={\sum }_{\omega \in B\setminus A}w(\omega )\le \mathbb{P}(D)=o(1)$ by [F6], since convergence of $c_n$ implies eventual boundedness. Thus $\mathbb{P}(B)-\mathbb{P}(A)\to 0$ by the convergence definition, and subtracting this null sequence from $\mathbb{P}(B)$ proves the finite-limit case by [F13], without converting from a fixed-edge model.

Put $L(C)=e^{-e^{-C}}$. The real-variable limits in this step have the meaning in [F16], with domain $\mathbb{R}$. As $C\to +\mathrm{\infty }$, [F11] gives $-e^{-C}\to 0$; continuity at zero from [F12] gives $L(C)\to e^0=1$, where $\exp (0)=\exp (\log 1)=1$ by [F9] and [F14]. As $C\to -\mathrm{\infty }$, [F11] gives $e^{-C}\to +\mathrm{\infty }$, so $-e^{-C}\to -\mathrm{\infty }$ and a second application of [F11] gives $L(C)\to 0$. To see the final implication directly, for any $\eta >0$ choose $T$ such that $t<T$ implies $e^t<\eta$; then choose $C$ sufficiently negative that $-e^{-C}<T$. These uses of composition are therefore justified at both ends.

Write $f_n(t)=\mathbb{P}(G(n,t)\text{ is connected})$. For fixed $C\in \mathbb{R}$, set $p_C=(\log n+C)\mathrm{/}n$. By [F9], for any real $M$ there is an $x_M>0$ with $\log x_M=M$; for every integer $n>x_M$, strict increase gives $\log n>M$. Hence $\log n\to +\mathrm{\infty }$ in the sense of [F15], and $p_C>0$ eventually. Also [F10] with $a=1$ and [F13] give $p_C\to 0$, since $C\mathrm{/}n\to 0$: given $\eta >0$, $n>\mathrm{\mid }C\mathrm{\mid }\mathrm{/}\eta$ implies $\mathrm{\mid }C\mathrm{\mid }\mathrm{/}n<\eta$. Consequently $p_C\in [0,1]$ eventually and step 1.1 applies to this tail, giving $f_n(p_C)\to e^{-e^{-C}}$.

By [F8], $0\le f_n(t)\le 1$ for $t\in [0,1]$, since an event sums a subset of the nonnegative weights whose total is one. Fix $\epsilon >0$. If $c_n\to +\mathrm{\infty }$, choose a fixed $C$ with $L(C)>1-\epsilon \mathrm{/}2$ by step 1.2. By [F15], eventually $c_n>C$ and hence $p\ge p_C$, so [F7] and step 2.1 give $1\ge f_n(p)\ge f_n(p_C)>L(C)-\epsilon \mathrm{/}2>1-\epsilon$. If $c_n\to -\mathrm{\infty }$, choose a fixed $C$ with $L(C)<\epsilon \mathrm{/}2$. By [F15], eventually $c_n<C$ and hence $p\le p_C$, so $0\le f_n(p)\le f_n(p_C)<L(C)+\epsilon \mathrm{/}2<\epsilon$. These are precisely convergence to one and zero. The given assumption $p\in [0,1]$ ensures that the graph exists in both regimes, and step 2.1 makes each comparison probability admissible eventually.