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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

Notation (printed p. 246): Kn∗K_n^* is the complete symmetric loopless digraph of order nn, LmL_m the transitive tournament of order mm, and r(Kn∗,Lm)r(K_n^*,L_m) the least order pp such that every digraph on pp vertices has an independent set of nn vertices (no arcs in either direction between them) or includes an LmL_m.

Lemma 4.4 (printed p. 249, quoted). "For all n≥2n\ge2, r(Kn∗,L4)≤2n3/3+n2+4n/3−4r(K_n^*,L_4)\le2n^3/3+n^2+4n/3-4."

In the problem's notation. r(Kn∗,Lm)=k(n,m)r(K_n^*,L_m)=k(n,m) in the letters of Problem 112, so the lemma is k(n,4)≤2n3/3+n2+4n/3−4k(n,4)\le2n^3/3+n^2+4n/3-4 for n≥2n\ge2. At n=2n=2 the right side is 8=k(2,4)8=k(2,4); at n=3n=3 and n=4n=4 it is 2727 and 6060.

Source. J. A. Larson and W. J. Mitchell, On a Problem of Erdős and Rado, Ann. Comb. 1 (1997), 245--252; Lemma 4.4 with its proof on printed p. 249, Lemma 4.3 on the same page, Lemma 4.2 on p. 248, and the value v(4)=8v(4)=8 with Lemma 2.1 on p. 247, read on the page images. The artifact is identified in the source digest.

Read depth. Claims checked: the statement and the proof (an induction of four lines and a three-line display) were read clause by clause on the page image and the display's algebra followed. Lemma 4.3, on which the induction step rests, is printed without proof. Nothing here is independently reviewed.

Proof pointer

Page 249. Induction on nn. The base case is the known value r(K2∗,L4)=v(4)=8r(K_2^*,L_4)=v(4)=8 (Lemma 2.1 and the values listed on p. 247). For the step, Lemma 4.3 at m=3m=3 gives r(Kn+1∗,L4)≤r(Kn∗,L4)+2r(Kn+1∗,L3)+1r(K_{n+1}^*,L_4)\le r(K_n^*,L_4)+2r(K_{n+1}^*,L_3)+1, and Lemma 4.2 bounds r(Kn+1∗,L3)r(K_{n+1}^*,L_3) by (n+1)2(n+1)^2; the identity

2n33+n2+4n3−4+2(n+1)2+1=2(n+1)33+(n+1)2+4(n+1)3−4\frac{2n^3}3+n^2+\frac{4n}3-4+2(n+1)^2+1 =\frac{2(n+1)^3}3+(n+1)^2+\frac{4(n+1)}3-4

closes the induction. Filing observations, not review verdicts: the printed basis check reads "8=2⋅23/3+22+2/3−48=2\cdot2^3/3+2^2+2/3-4" [sic], whose right side is 66; with 8/38/3 in place of 2/32/3 it is 88, the value of the statement's polynomial at n=2n=2. The printed induction hypothesis reads "2(n)3/3−(n)2+(n)/3−42(n)^3/3-(n)^2+(n)/3-4" [sic], which differs from the statement in the coefficients of n2n^2 and nn; the displayed computation that follows uses the statement's polynomial and is correct. The paper adds that the argument generalizes to a polynomial bound in nn of degree m−1m-1, which is Lemma 4.13.

Dependencies

Within the paper: Lemma 4.3 (p. 249), the recurrence r(Kn+1∗,Lm+1)≤2r(Kn+1∗,Lm)+r(Kn∗,Lm+1)+1r(K_{n+1}^*,L_{m+1})\le2r(K_{n+1}^*,L_m)+r(K_n^*,L_{m+1})+1 for n>1n>1 and m≥2m\ge2, printed without proof and sketched on Lemma 4.13; Lemma 4.2; and Lemma 2.1, r(K2∗,Lm)=v(m)r(K_2^*,L_m)=v(m), quoted from Bermond's Proposition 2.4, with v(4)=8v(4)=8 as listed on p. 247.

Bears on

  • Problem 112: an upper bound k(n,4)≤2n3/3+n2+4n/3−4k(n,4)\le2n^3/3+n^2+4n/3-4 for n≥2n\ge2, of order n3n^3 in the column m=4m=4, where the Erdős--Rado bound quoted as the paper's Theorem 2.7 is (8(n−1)4+n−2)/(2n−3)(8(n-1)^4+n-2)/(2n-3), of order 4n34n^3. It determines no value of k(n,4)k(n,4) beyond the known k(2,4)=8k(2,4)=8 at which it is tight.