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Statement

Monotonic $k$-term arithmetic progressions are defined on the Theorem 4.1 page.

Remark 1 (p. 4). Fix $k\ge2$. Define the ordering $C_n$ of the integer interval $[-(k-1)k^{n-1},k^{n-1}-1]$, $n\ge1$, by $C_1=\langle0,-1,-2,\dots,-(k-1)\rangle$ and $C_{n+1}=(kC_n)(kC_n+1)\cdots(kC_n+k-1)$, $n\ge1$; for $k=2$ this is Definition 2.1 (see the Theorem 2.2 page for the notation). The paper states that the arguments of Sections 2, 3 and 4 "can now be repeated with little change" (p. 4), giving a linear ordering of $\mathbb R$ with monotonic $k$-term arithmetic progressions and no monotonic $(k+1)$-term arithmetic progression. The same claim appears in the introduction (p. 1): for every $k\ge2$ there is such a linear ordering of $\mathbb R$.

Source. Hayri Ardal, Tom Brown and Veselin Jungić, Chaotic orderings of the rationals and reals, Amer. Math. Monthly 118 (2011), no. 10, 921--925, doi:10.4169/amer.math.monthly.118.10.921, read in the author copy identified on the source card, paginated 1--5.

Read depth. Claims checked for the statement only: the remark was read clause by clause on the page image. The paper gives no proof beyond the sentence quoted above, and none was written or checked here. Nothing here is independently reviewed.

Proof pointer

P. 4, by reference only: the paper points to the proofs of Lemma 2.1 and Theorems 2.2, 3.1 and 4.1 with $2$ replaced by $k$, and writes out neither the absence of monotonic $(k+1)$-term progressions nor the presence of monotonic $k$-term ones.

Bears on

• Problem 194: the problem's negative answer for every $k\ge3$ already follows from Theorem 4.1. The remark, as stated by the paper without written proof, adds that for each $k\ge2$ some ordering of $\mathbb R$ has monotonic $k$-term progressions but no monotonic $(k+1)$-term one, so the longest monotonic progression length can be any prescribed $k\ge2$.