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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Statement

For finite bb and r≥2r\ge2, g(b,r)=f(b,b,r)g(b,r)=f(b,b,r) is the least aa with a→(b,b)ra\to(b,b)^r (printed p. 139). Put a∗b=aba*b=a^b and, generally, a0∗a1∗∗am=a0∗(a1∗∗am)a_0*a_1**a_m=a_0*(a_1**a_m) for 2≤m<ω2\le m<\omega, so the tower is evaluated from the right. 16.3. If 2≤r≤b<ω2\le r\le b<\omega, then

g(b,r)≤2∗(2r−1)∗(2r−2)∗∗(22)∗(2b−2r+1),g(b,r)\le2*(2^{r-1})*(2^{r-2})**(2^2)*(2b-2r+1),

"and hence g(b,r)≤2∗2∗∗2∗(krb)g(b,r)\le2*2**2*(k_rb) (rr 'factors' in all), where the positive real number krk_r depends on rr only" (p. 140). The paper attributes the estimate to Erdős and Rado [3], Proc. London Math. Soc. (3) 2 (1952), 417--439.

For r=3r=3 the factors are 22, 222^2 and 2b−52b-5, so g(b,3)≤242b−5=224b−10g(b,3)\le2^{4^{2b-5}}=2^{2^{4b-10}}: the Ramsey number R3(b)R_3(b) of Problem 564 is at most double exponential in bb.

Source. P. Erdős, A. Hajnal and R. Rado, Partition relations for cardinal numbers, Acta Math. Acad. Sci. Hungar. 16 (1965), 93--196; statement 16.3 at the foot of printed p. 139 (PDF p. 47 of the scan), the "hence" line at the head of p. 140 (PDF p. 48). The scan's text layer is garbled; both were read on the page images, the inequality glyphs on a 300 dpi crop.

Read depth. Claims checked: the definition of gg, the ∗* convention and the statement were read clause by clause on the page images. The paper gives no proof ("we omit the proofs", p. 140); the proof is in the cited 1952 paper, which is not held.

Proof pointer

None in this paper. The cited source is Erdős and Rado 1952 (not held).

Dependencies

External: Erdős and Rado 1952.

Bears on

  • Problem 564: the upper bound R3(n)≤224n−10R_3(n)\le2^{2^{4n-10}}, of the same double-exponential shape as the lower bound the problem asks for; the site writes it as 22n2^{2^n}.
  • Problem 562: the upper side of the problem for every r≥3r\ge3: the "hence" form is a tower of height rr with top krbk_rb, so log⁡r−1Rr(n)≤krn\log_{r-1}R_r(n)\le k_rn with log⁡r−1\log_{r-1} the (r−1)(r-1)-fold iterated logarithm; the problem asks for the matching lower bound.