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A giant component in a sequence of random graphs on $n$ vertices has order proportional to $n$. Its existence can change at a threshold function for a monotone graph property. A giant component need not contain every vertex, and its uniqueness requires proof for the model in question.

The largest graph component is one with the most vertices. Ties may be broken arbitrarily; $L_1$ records the order, which is unaffected by the tie. For a sequence of random graphs, the size of its giant component is a central observable.

Fix $k\ge 1$, ${\gamma }_0>1/(k+1)$, and ${\gamma }_0\le \gamma \le 1-\omega (n)/\log n$ with $\omega (n)\to \infty$. A binomial random graph has one giant component of the displayed order and all other graph components are tree components of orders at most $k$, with high probability. The expected value and variance of the isolated vertex count concentrate it around $n^{1-\gamma }\to \infty$. A Cayley formula upper bound for connected sets excludes component sizes $k+1$ through a small fixed fraction of $n$; the empty-graph cut bound excludes the remaining sizes up to $n/2$. An extra-edge count excludes cyclic small graph components. The expected value of the number of vertices in small nonisolated tree components is $O(nd{e}^{-2d})=o(n{e}^{-d})$, where $d=np$. The Markov inequality completes the count of vertices outside the unique large graph component.

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.