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Source: published version, p. 238, Theorem 1.4 and Remark 1.5; proof in Section 2, pp. 238--242.

Statement with rounding made explicit

There are universal constants A,C,c>0A,C,c>0 with the following property. Let ℓ≥2\ell\ge2, let H\mathcal H be a nonempty ℓ\ell-bounded hypergraph on an nn-element set XX, and let p∈[0,1]p\in[0,1]. If H\mathcal H is not pp-small, then a uniformly random mm-subset XmX_m, where

m=min⁡{n,⌈Apnlog⁡ℓ⌉},m=\min\{n,\lceil Apn\log\ell\rceil\},

satisfies

P(Xm∈⟨H⟩)≥1−C(log⁡ℓ)−c.(1)\mathbb P(X_m\in\langle\mathcal H\rangle) \ge1-C(\log\ell)^{-c}. \tag{1}

In particular the right side is 1−oℓ→∞(1)1-o_{\ell\to\infty}(1), which is the form of Theorem 1.4. The inverse-polylogarithmic rate is Remark 1.5. The paper suppresses integer rounding and absorbs all universal factors into its constant LL.

Full proof from the fragment chain

If ∅∈H\varnothing\in\mathcal H, then ⟨H⟩=2X\langle\mathcal H\rangle=2^X and (1) is immediate. If p=0p=0 and no edge is empty, H\mathcal H covers itself with zero cost, contrary to the hypothesis. We may therefore assume p>0p>0 and that all edges are nonempty.

The sample-size calculation on the shrinking-fragment iteration page gives a prospective batch total M0≤A0pnlog⁡ℓM_0\le A_0pn\log\ell. Choose A≥A0A\ge A_0. If ⌈Apnlog⁡ℓ⌉≥n\lceil Apn\log\ell\rceil\ge n, the target set is XX and belongs to ⟨H⟩\langle\mathcal H\rangle because H\mathcal H is nonempty. Otherwise M0<nM_0<n, so every successive batch fits in the remaining ground set and the iteration is defined.

Run the shrinking-fragment iteration. It produces a uniformly random M0M_0-subset W=⋃iWiW=\bigcup_iW_i, with

M0≤A0pnlog⁡ℓ,(2)M_0\le A_0pn\log\ell, \tag{2}

and a family U=⋃iUi\mathcal U=\bigcup_i\mathcal U_i satisfying

E[∑U∈Up∣U∣]≤C0(log⁡ℓ)−c0.(3)\mathbb E\left[\sum_{U\in\mathcal U}p^{|U|}\right] \le C_0(\log\ell)^{-c_0}. \tag{3}

Let EE be the event that U\mathcal U covers H\mathcal H. Because H\mathcal H is not pp-small, on EE one necessarily has

∑U∈Up∣U∣>12.\sum_{U\in\mathcal U}p^{|U|}>\frac12.

Markov's inequality and (3) give

P(E)≤2E[∑U∈Up∣U∣]≤2C0(log⁡ℓ)−c0.(4)\mathbb P(E) \le2\mathbb E\left[\sum_{U\in\mathcal U}p^{|U|}\right] \le2C_0(\log\ell)^{-c_0}. \tag{4}

By Proposition 2.3, every outcome for which EE fails has W∈⟨H⟩W\in\langle\mathcal H\rangle. Hence

P(W∈⟨H⟩)≥1−P(E),\mathbb P(W\in\langle\mathcal H\rangle) \ge1-\mathbb P(E),

and (4) proves the desired estimate at level M0M_0. Couple this uniform M0M_0-subset with a uniform mm-subset by taking the first M0M_0 and first mm points of a random permutation. The event S∈⟨H⟩S\in\langle\mathcal H\rangle is increasing, so its probability cannot decrease under this enlargement. Adjusting the absolute constants in (4) proves (1).

Bears on

  • Problem 202, through later uses of the Kahn--Kalai theorem derived from this result.