Proof that if $np - \log n \rightarrow c$ the probability that $G(n,p)$ is connected goes to $e^{-e^{-c}}$ using Janson's inequalities I am interested in showing that in the Erdõs-Rényi random graph $G(n,p)$ for $p = p(n)$ satisfying $np - \log n \rightarrow c \in \mathbb{R}$ the probability of $G(n,p)$ being connected goes to $e^{-e^{-c}}$ as $n \rightarrow \infty$.
I found in (1) Bollobás' Random Graphs and in (2) this document by Spencer proofs using that $\mathbb{P}(G(n,p) \text{ is connected}) = \mathbb{P}(\delta(G(n,p)) \geq 1) + o(1)$, because for this $p$, $G(n, p)$ is composed by a giant component and isolated vertices, and also that the probability of $G(n,p)$ having no isolated vertices goes to $e^{-e^{-c}}$.
For the last step (2) shows that $X$ (the random variable counting the number of isolated vertices) approaches $Poisson(\mu)$ because its moments approaches those of $Poisson(\mu)$, but I don't understand the proof well.
I want to try and use Janson's inequalities in this proof, and my attempt was on the $\mathbb{P}(\delta(G(n,p) \geq 1) \rightarrow e^{-e^{-c}}$ part.
Let $X_i$ be the indicator random variable that the vertex $v_i$ is isolated and $X = \sum_{i \in [n]} X_i$. Then $\mathbb{P}(X_i = 1) = (1-p)^{n-1}$ and $$\mu = \mathbb{E}(X) = \sum_{i \in [n]} \mathbb{P}(X_i = 1) = n(1-p)^{n-1}.$$
$$\Delta = \sum_{i \sim j}\mathbb{P}(X_i \cap X_j) = n(n-1)(1-p)^{2n-3}.$$
Where $i \sim j$ if $X_i$ and $X_j$ are not mutually independent, and the sum is over ordered pairs.
Let $M = \prod_{i \in [n]} 1 - \mathbb{P}(X_i = 1)$ (the probability that $X = 0$ if all of the events $X_i$ were independent). I know that $M = \prod_{i \in [n]}1 - (1 - p)^{n-1} = (1 - (1 - p)^{n-1})^{n}$.
Janson's inequalities show that $M \leq \mathbb{P}(X = 0) \leq e^{- \mu + \Delta/2}$.
However, only using $1-x \leq e^{-x}$ I can't seem to show that $\mathbb{P}(X = 0)$ gets sandwiched to $e^{-e^{-c}}$ as $n \rightarrow \infty$ (in fact, I don't think that $\Delta = o(1)$).
I don't know if my approach can work, the problem is stated as an exercise in a chapter of a book where the Janson's inequalities are presented, and so I thought I could use them somewhere.
Any help, tips or reading suggestions would be greatly appreciated!
 A: Janson's inequalities are not strong enough to use here. As $n \to \infty$, we replace terms like $n-1, 2n-3$ with $n, 2n$, replace $1-p$ with $e^{-p}$, and the resulting $np$ with $c + \log n$, and end up getting $\mu \to e^{-c}$ and $\Delta \to e^{-2c}$. For large $c$, $e^{-\mu + \Delta/2}$ ends up being a useful estimate (and it can tell us that if $np-\log n \to \infty$, then the probability that $\delta(G(n,p)) \ge 1$ goes to $1$). But it is not going to get us the $e^{-\mu}$ we wanted.
In Alon and Spencer's Probabilistic Method, Janson's inequalities are covered at the beginning of Chapter 8. Look at section 8.3 (and Theorem 8.3.1) of the same chapter for more detail on the "asymptotically $\text{Poisson}(\mu)$" approach you saw to the problem.

Here is some intuition on what's going on and why Janson's inequality doesn't work well here. Many sources, including Alon and Spencer's book, describe it as a tool when events are "mostly independent", but there's a lot of things that could mean, and not all of them lend themselves to Janson's inequality.
In the sum $$\Delta = \sum_{i \sim j} \Pr[B_i \land B_j],$$ every $\Pr[B_i \land B_j]$ is bigger than $\Pr[B_i] \cdot \Pr[B_j]$: overlapping events are more likely to happen together, because there's some coinflips that help both events. If the sum were over every pair of events, then we'd get
$$
   \sum_{i \in I} \sum_{j \in I} \Pr[B_i \land B_j] \ge \sum_{i \in I} \sum_{j \in I} \Pr[B_i] \cdot \Pr[B_j] = \left(\sum_{i \in I}\Pr[B_i]\right)^2 = \mu^2.
$$
When $\Delta$ is close to $\mu^2$, neither form of Janson's inequality can help us. So we see that we can only win in Janson's inequality if many pairs of events are completely independent from each other.
This is not the setting we're in with isolated vertices, where in fact any two events are dependent. Here, the reason for Poisson-like behavior is that most (all, actually) small subsets of the events are only a tiny bit dependent. Although that also sounds like "mostly independent", it is "mostly independent" in a different way that Janson's inequality doesn't handle.
The "method of moments" described in section 8.3 of the Probabilistic Method is more flexible here, and can accommodate events that are "mostly independent" in either sense.
