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Erdős–Szekeres upper bound for Ramsey numbers

Statement

For all integers s,t≥2s, t \ge 2, the Ramsey number R(s,t)R(s, t) is finite and satisfies R(s,t)≤R(s−1,t)+R(s,t−1)R(s, t) \le R(s - 1, t) + R(s, t - 1). Consequently, R(s,t)≤(s+t−2s−1)R(s, t) \le \binom{s + t - 2}{s - 1}.

Why is it true?

Fix a vertex vv: either its red neighborhood is large enough to force a red Ks−1K_{s-1} or a blue KtK_t, or else its blue neighborhood is large enough to force a red KsK_s or a blue Kt−1K_{t-1}.

Proof sketch

**Step 1 (pigeonhole split of the neighborhood of vv).** Let N=R(s−1,t)+R(s,t−1)N = R(s - 1, t) + R(s, t - 1) and fix a vertex v∈V(KN)v \in V(K_N). Partition the remaining N−1N - 1 vertices into VRV_R (red edge to vv) and VBV_B (blue edge to vv). Since ∣VR∣+∣VB∣=R(s−1,t)+R(s,t−1)−1|V_R| + |V_B| = R(s - 1, t) + R(s, t - 1) - 1, either ∣VR∣≥R(s−1,t)|V_R| \ge R(s - 1, t) or ∣VB∣≥R(s,t−1)|V_B| \ge R(s, t - 1).

Step 2 (inductive conclusion). If ∣VR∣≥R(s−1,t)|V_R| \ge R(s - 1, t), the subgraph on VRV_R contains a blue KtK_t or a red Ks−1K_{s-1} that joins vv to form a red KsK_s. Symmetrically for VBV_B.

Step 3 (binomial bound by induction). Base cases R(2,t)=tR(2, t) = t and R(s,2)=sR(s, 2) = s match (t1)\binom{t}{1} and (ss−1)\binom{s}{s-1}. For s,t≥3s, t \ge 3, Pascal's identity gives R(s,t)≤(s+t−3s−2)+(s+t−3s−1)=(s+t−2s−1)R(s, t) \le \binom{s + t - 3}{s - 2} + \binom{s + t - 3}{s - 1} = \binom{s + t - 2}{s - 1}.

Topics that use this theorem

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

  1. Marcelo Campos, Simon Griffiths, Robert Morris, Julian Sahasrabudhe (2023). An exponential improvement for diagonal Ramsey · arXiv:2303.09521
  2. Sam Mattheus, Jacques Verstraëte (2024). The asymptotics of r(4,t) · DOI:10.4007/annals.2024.199.2.8
  3. Ronald L. Graham, Bruce L. Rothschild, Joel H. Spencer (1990). Ramsey Theory (2nd ed.)