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For bounded vertex degrees, total degree $M\to \infty$, and a forbidden graph $G_0$ of bounded maximum degree, put $\lambda =M^{-1}{\sum }_i{d}_i(d_i-1)$ and $\mu =M^{-1}{\sum }_{ij\in E(G_0)}{d}_i{d}_j$. Count prescribed loop pairs, double-edge pairs, and forbidden-edge pairs in the configuration model. Their means are respectively $\lambda /2+o(1)$, ${\lambda }^2/4+o(1)$, and $\mu +o(1)$. A fixed collection of $t$ disjoint prescribed pairs occurs with probability $[(M-1)(M-3)\cdots (M-2t+1){]}^{-1}$. Bounded degrees make intersecting bad configurations contribute $O(M^{-1})$ to each fixed factorial moment. Thus the total bad-pattern count has factorial moments $(\lambda /2+{\lambda }^2/4+\mu {)}^j+o(1)$. The Bonferroni inequalities and uniformly bounded means give the displayed zero-pattern probability, without requiring the parameters to converge.

For the number $N$ of occurring events, write $S_j=\mathbb{E}\left(\genfrac{}{}{0ex}{}{N}{j}\right)$ and $T_m={\sum }_{r=0}^m(-1{)}^r(\genfrac{}{}{0ex}{}{k+r}{k})S_{k+r}$. Then odd truncations bound $\mathrm{Pr}(N=k)$ below and even truncations above. For each integer $N>k$, the truncated binomial sum equals $(-1{)}^m(\genfrac{}{}{0ex}{}{N}{k}\left)\right(\genfrac{}{}{0ex}{}{N-k-1}{m})$ when $m<N-k$, and is zero otherwise. Its sign proves the bounds on taking expectations. This is the exact-count version of the Bonferroni inequalities.

For a finite family, the random variable $N={\sum }_i{1}_{A_i}$ counts the events that occur. Linearity of expectation gives $\mathbb{E}N={\sum }_i\mathbb{P}(A_i)$, while $\mathbb{P}(N\ge 1)$ is the exact union probability. Their difference isAn outcome in exactly $k\ge 1$ events contributes $k-1$ extra counts, weighted by its probability. This nonnegative difference vanishes exactly in equality in Boole's inequality. For two events it is $\mathbb{P}(A\cap B)$, as the inclusion-exclusion principle for probabilities shows. For more events, the Bonferroni inequalities use pairwise and higher intersection information to bound the exact union probability; blindly subtracting pairwise intersections from the sum gives a lower bound, rather than a corrected upper bound.

Severe overlap can make the union bound arbitrarily pessimistic. Let all $m$ events be the same event $A$ with probability $1/m$, where $m\ge 1$ is an integer. A concrete model is the discrete uniform distribution on $\{1,\dots ,m\}$ with $A=\{1\}$. The exact union probability is $1/m$, whereas the sum and capped bound are both one. The ratio is $m$: adding more names for the same event does not create more ways to fail. The identity above describes the gap in the uncapped sum; capping at one can reduce that gap.

The parameter relations give $\mu =\log (n/c)$ andFor each fixed integer $r\ge 1$, take an $r$-vertex set $S$ and let $X_S$ count triangles meeting $S$. The event that every vertex of $S$ is counted by $T$ is exactly $X_S=0$.

The number of such triangles isThus their total expected count is ${\mu }_S=N_S{p}^3=r\mu +O_r(n{p}^3)=r\mu +o(1)$.

Distinct triangles have dependent containment events only when they share an edge. If that common edge meets $S$, there are $O_r(n)$ choices for it and $O(n^2)$ choices for the two other vertices. If the common edge avoids $S$, both other vertices must lie in $S$, giving only $O_r(n^2)$ pairs. Each pair needs five edges. Consequently its ordered Janson dependency sum isJanson inequality gives the upper bound $\mathrm{Pr}(X_S=0)\le \exp (-{\mu }_S+{\Delta }_S/2)$. For the lower bound, the avoidance of each triangle is a decreasing event. Repeated application of Harris' inequality givesThe error $n^2{p}^6=O((\log n{)}^2/n^2)$ also tends to zero. The bounds match multiplicatively:Summing over ordered distinct vertices yields the factorial moments

Here is an explicit justification that these moments determine the limit. For every integer $t\ge 0$, the finite alternating sumsare upper bounds for $\mathrm{Pr}(T=t)$ when $L$ is even and lower bounds when $L$ is odd. This follows by applying the Bonferroni inequalities to $\left(\genfrac{}{}{0ex}{}{T}{t}\right){\sum }_j(-1{)}^j(\genfrac{}{}{0ex}{}{T-t}{j})$; the full sum is precisely the indicator of $T=t$. First let $n\to \infty$ for a fixed truncation and use the factorial-moment limits. Then let the even and odd truncations grow. Both bounds tend toHence every point probability has the Poisson distribution limit, proving the Poisson threshold for vertices lying in no triangle:This is also the factorial-moment criterion for Poisson convergence, here justified rather than left as an unproved implication.

Couple the uniform random graph process by giving every possible edge an independent uniform label in $[0,1]$ and revealing them in increasing order. The graph at label threshold $p$ is a binomial random graph. Define ${\tau }_0$ as the first step with no isolated vertices and ${\tau }_c$ as the first connected graph. Deterministically ${\tau }_c\ge {\tau }_0$. We prove the reverse inequality with high probability by tracking the same process through a short interval.

Put $w=\frac{1}{4}\log \log n$, $p_{\pm }=(\log n\pm w)/n$, and $L=n{p}_{-}$. At $p_{-}$ the isolated vertex count $Z$ hasThe variance estimate follows by comparing two isolated-vertex indicators, whose joint-to-product probability ratio is $(1-p_{-}{)}^{-1}$. Thus $Z>0$ with high probability. Also the Markov inequality gives $Z\le \log n$ with probability tending to one.

At this same threshold, exclude every graph component with order $2\le s\le n/2$. For $2\le s\le n/\log n$, any such component contains a spanning tree. The Cayley formula and a union bound therefore giveIn this range the bracket is at most $A=e^{2}L{e}^w/n=o(1)$. Summing over $s\ge 2$ gives $O(n{A}^2/L)=O(L{e}^{2w}/n)=o(1)$. For $n/\log n<s\le n/2$, even the cut-absence bound is enough:The bracket tends to zero, so this range also has total probability $o(1)$. Consequently the graph at $p_{-}$ consists of one component of order greater than $n/2$ and its isolated-vertex set $S$, where $1\le |S|\le \log n$ with high probability.

Conditional on the entire graph at $p_{-}$, every unrevealed edge label remains independently uniform on $(p_{-},1)$. In particular,There are no isolated vertices at $p_{+}$ with high probability, because their expected value is asymptotic to $e^{-w}\to 0$. On the intersection of these events, each vertex of $S$ loses isolation by joining the original large component, never by forming a separate small component with another vertex of $S$. At every step between these two thresholds there is therefore one nontrivial component and some isolated vertices. The final isolated vertex disappears precisely when the graph becomes connected. This proves the connectivity hitting time equals disappearance of isolated vertices result:

For the critical-window probability, let $p=(\log n+\alpha (n))/n$ with $|\alpha (n)|\le B$. For each fixed $j$, the falling factorial moment of the isolated-vertex count isuniformly for such shifts. Apply the Bonferroni inequalities to the zero-count event. Its odd and even truncations bracket $\mathrm{Pr}(Z=0)$ by partial sums of ${\sum }_{j\ge 0}(-1{)}^j{e}^{-j\alpha (n)}/j!$. Since $e^{-\alpha (n)}\le e^B$, their remainders tend uniformly to zero. Hence $\mathrm{Pr}(Z=0)=e^{-e^{-\alpha (n)}}+o(1)$. The process equality just proved changes this probability by at most its $o(1)$ failure probability. ThereforeThe ratio follows because the denominator is bounded below by $e^{-e^B}>0$; no convergence of $\alpha (n)$ is needed.

The printed second conclusion has the complementary denominator and is false as written. For $\alpha (n)=0$, the printed ratio tends to $e^{-1}/(1-e^{-1})=1/(e-1)\ne 1$. Its denominator $1-e^{-e^{-\alpha (n)}}$ is correct for the probability of being disconnected. Thus either remove the leading $1-$ in the denominator or replace “connected” by “disconnected”; the proof supplies both corrected versions.

Put $M=2m$ and $a_i=\left(\genfrac{}{}{0ex}{}{d_i}{2}\right)$. Use the configuration model: give vertex $i$ exactly $d_i$ labelled half-edges and choose a uniform perfect matching of all $M$ half-edges. There arepairings, with $(2m{)}_m$ denoting a falling factorial. Each allowed simple graph has exactly ${\prod }_i{d}_i!$ representations: at every vertex, assign its incident graph edges bijectively to its labelled half-edges. Therefore its enumeration reduces to the probability that a pairing is simple and avoids $G_0$.

Count three kinds of prescribed bad patterns. A loop pattern consists of a pair of half-edges at one vertex. A double-edge pattern consists of two half-edges at each of two distinct vertices and one of their two cross-pairings. A forbidden-edge pattern consists of one half-edge at each endpoint of an edge of $G_0$. Let their respective counts in the random pairing be $Z_L,Z_D,Z_F$. Simplicity and avoidance are equivalent to $Z=Z_L+Z_D+Z_F=0$, including exclusion of triple edges because those contain double-edge patterns.

A prescribed collection of $t$ disjoint half-edge pairs occurs with probabilityConsequently the bad-pattern expected values areHere $\lambda =({\sum }_i{a}_i)/m$ and $\mu =M^{-1}{\sum }_{ij\in E(G_0)}{d}_i{d}_j$, as in the PDF. Since $1\le d_i\le \Delta$ and $G_0$ has bounded maximum degree, $M=\Theta (n)$ and all three means are uniformly bounded. For the double-edge estimate, ${\sum }_i{a}_i^2=O(n)$, so its contribution after division by $M^2$ is $o(1)$.

We now give the moment-counting step rather than assume independence. For each fixed $h$, expand the factorial moment $\mathbb{E}(Z{)}_h$ as the sum over ordered distinct bad patterns. Collections on disjoint vertex sets have, by the displayed pairing probability, joint probabilities equal to the product probabilities times $1+O(1/n)$. Their total contribution isDeleting collections with a shared vertex changes the independent-product sum by $O(1/n)$, since all ordinary and forbidden vertex degrees are bounded.

For actual joint probabilities of overlapping patterns, incompatible uses of a half-edge give zero. Compatible collections with no shared paired edge lose at least one free vertex choice, without reducing the number of required pairs, so contribute $O(1/n)$. A shared paired edge can occur only between two double-edge patterns or between a double-edge pattern and a forbidden-edge pattern. In the first case their union requires at least three parallel paired edges: there are $O(n^2)$ endpoint choices and probability $O(n^{-3})$. In the second case the endpoint pair must be an edge of $G_0$: there are only $O(n)$ choices and at least two required pairs, giving $O(n^{-1})$. Adding further patterns in a fixed collection does not remove this deficit: a disjoint addition contributes bounded order, while an addition to an existing cluster has only bounded half-edge choices unless it supplies new free vertices, compensated by its new required pairs. Summing the finitely many overlap types for fixed $h$ therefore gives $o(1)$. Thus, with $\Lambda =\lambda /2+{\lambda }^2/4+\mu$,The Bonferroni inequalities bracket $\mathrm{Pr}(Z=0)$ by the odd and even partial sums of ${\sum }_h(-1{)}^h\mathbb{E}(Z{)}_h/h!$. The means $\Lambda$ are bounded, so the corresponding exponential-series remainders vanish uniformly as the truncation increases. Taking $n\to \infty$ and then the truncation to infinity proves the bounded-degree pairing avoidance estimate $\mathrm{Pr}(Z=0)=e^{-\Lambda }+o(1)$. This does not require $\lambda$ or $\mu$ to converge. The exponential is bounded away from zero, so the additive error is also a relative $o(1)$ error.

Multiplying by the pairing count and dividing by the representation factor givesThe factor $(2m{)}_m$ multiplies the exponential; it is not part of its exponent. The TeX aid misgroups this factor, whereas the original PDF is unambiguous. The bounded-degree proof also covers the stated additional assumption $2m-n\to \infty$.

For events $A_1,\dots ,A_n$, the probabilistic inclusion-exclusion principle isTo prove it, fix an outcome lying in exactly $r$ events. Its total coefficient on the right isby the binomial theorem; outcomes in no event contribute zero. Taking expectations of this pointwise indicator identity proves the formula. Truncating after the pair terms gives the Bonferroni inequalities

For the final bound, let $X={\sum }_i{1}_{A_i}$ and $S=\mathbb{E}X={\sum }_i\mathbb{P}(A_i)$. The intersection assumption givesThe Cauchy-Schwarz inequality applied to $X{1}_{X>0}$ yields $S^2\le \mathbb{E}[X^2]\mathbb{P}(X>0)\le S+n-1$. Thereforeso the required universal constant may be taken as $\boxed{c=2}$. This is a second moment method estimate.