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For a fixed set $L$ of $\lfloor \sqrt{n}\rfloor$ vertices, let ${\tau }_L$ be the first step at which they all belong to one graph component. In the uniform random graph process, ${\tau }_L$ lies between $n(\log n-\omega )/4$ and $n(\log n+\omega )/4$ with high probability for every $\omega \to \infty$. Below that scale the marked isolated vertex count has divergent expected value and small relative variance. Above it the logarithmic-regime giant component misses $o(\sqrt{n})$ vertices; exchangeability and a union bound show that none are marked. The label coupling then transfers the two assertions to the process.

It suffices to prove the assertion after replacing $\omega$ by $\min \{\omega ,\sqrt{\log n}\}$: that replacement gives a tighter interval. Thus assume $\omega \to \infty$ and $\omega =o(\log n)$, and putAt $p_{-}$, let $J$ count isolated vertices in $L$. The same indicator random variable calculation as for all isolated vertices givesBy the Chebyshev inequality, $J>0$ with high probability. Since $\ell \ge 2$ eventually, the marked vertices are then not all connected.

At $p_{+}$, the logarithmic-regime giant component result applies with $k=2$ and any fixed ${\gamma }_0\in (1/3,1/2)$. Its unique giant component misses $(1+o(1))n^{1/2}{e}^{-\omega /4}$ vertices. On the event that it misses at most twice that number, exchangeability makes the missed vertex set uniform conditional on its size. A union bound therefore givesThus all of $L$ is connected at $p_{+}$ with high probability.

For the uniform random graph process, assign uniform labels to the edges of the complete graph as independent random variables and reveal them in label order. The binomial random graph $G(n,p)$ is the prefix containing the $M(p)$ labels at most $p$, where $M(p)$ has a binomial distribution with parameters $N=\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ and $p$. Its variance at $p_{\pm }$ is $O(n\log n)$, whereas the gap between $\mathbb{E}M(p_{\pm })$ and the relevant endpoint $m_{\pm }$ is $(1+o(1))\omega n/8$. The Chebyshev inequality gives $M(p_{-})\ge m_{-}$ and $M(p_{+})\le m_{+}$ with high probability, with harmless integer rounding. The monotone graph property that all marked vertices are connected now impliesThe marked-set connectivity threshold is lower than the threshold for connecting every vertex, because only about $\sqrt{n}$ specified vertices need to avoid the small graph components.

Put $d=np=\gamma \log n$ and $\mu =n^{1-\gamma }$. The hypothesis implies $\mu \ge e^{\omega (n)}\to \infty$. For the number $I$ of isolated vertices, the isolated vertices in the Erdős-Rényi model formulas giveThus the Chebyshev inequality gives $I=(1+o(1))\mu$ with high probability, uniformly over the specified $\gamma$ range.

The Cayley formula supplies a spanning tree on any connected vertex set. Consequently the expected value $C_j$ of the number of graph components of order $j\ge 2$ is bounded byChoose a fixed $\eta \in (0,1/2)$ with ${\gamma }_0(1-\eta )(k+1)>1$. For $k+1\le j\le \eta n$, with $j\ge 2$, the binomial coefficient bound $\left(\genfrac{}{}{0ex}{}{n}{j}\right)\le (en/j{)}^j$ givesHere $ed{e}^{-d(1-\eta )}\le e\log n{n}^{-{\gamma }_0(1-\eta )}\to 0$. The geometric series and a union bound therefore exclude this entire size range, since $n[e\log n{n}^{-{\gamma }_0(1-\eta )}{]}^{k+1}=o(1)$. For $\eta n\le j\le n/2$, an empty graph cut would be necessary. Its total probability is at mostThere are consequently no graph components of orders between $k+1$ and $n/2$, with high probability.

For each fixed $2\le j\le k$, a graph component containing a graph cycle has a spanning tree and at least one extra edge. Counting that extra edge gives an expected value $O_k(d^j{e}^{-dj})=o(1)$; hence every small graph component is a tree component. The total number $R_2$ of vertices in tree components of orders $2,\dots ,k$ satisfiesThe Markov inequality gives $R_2=o(\mu )$ with high probability. When $k=1$, this sum is empty and $R_2=0$. Since $\mu =o(n)$, the remaining vertices must lie in one graph component larger than $n/2$. It is unique, and its order is $n-(1+o(1))n^{1-\gamma }$. Every other graph component is a tree with at most $k$ vertices. This is the logarithmic-regime giant component mechanism: almost all missing vertices are isolated.