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How to prove or disprove that the Cantor set does not include any arithmetic progression of length 5?

Markiyan Hirnyk
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2 Answers2

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Hint: Because of the linear self similarity maps $C\cap[0,1/3] \rightarrow C$ and $C\cap[2/3,1] \rightarrow C$, given any sequence entirely in either $C\cap[0,1/3]$ or $C\cap[2/3,1]$, you can transform it to a sequence with larger delta still contained in $C$.

aschepler
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  • I don't understand it at all. Could you elaborate this? – Markiyan Hirnyk Feb 22 '14 at 21:18
  • How about using the representation of numbers of the Cantor set $C$ in base $3$, to try to form the progression {$c_1,c_2,..c_5$}? The terms $c_i:=0.c_{i1}c_{i2}.....c_{in}....$ have $c_{in}$ in {$0,2$}. Now, start with a term $c_j$ in the Cantor set. What can the $r$ term ( the common difference between consecutive terms ) be ? (note we need $|r| \leq0.2$ to have five terms). If $r:=0.r_1r_2....$ has some $r_j=1 $, can $c_i+r$ be in $C$? Note then $r_j$ cannot have the $1$ in a position where $c_j=0$, or the sum will not be in $C$. – user99680 Feb 22 '14 at 22:22
  • But if $r_j=1; c_j=2$ , the sum will not be in $C$ either, since the sum will carry. – user99680 Feb 22 '14 at 22:23
  • I got that: a good idea was unclearly stated. Thank you. – Markiyan Hirnyk Feb 24 '14 at 17:45
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It is apparently known that the longest arithmetic progressions in the Cantor set are of length $4$, i.e. there are none of length 5 (or more). See here for the abstract of a talk mentioning this result.

A fill-out based on ashlepper's hint: Note that there cannot be an AP with at least one term in each of $[0,1/3]$ and $[2/3,1]$, since the removed interval $(1/3,2/3)$ would then force the common difference of the AP to be at least $1/3$ and there are four gaps in an AP of length 5, giving total length at least $4 \cdot (1/3)=4/3$ which cannot happen as the cantor set is contained in $[0,1]$.

Using the two maps $x \to 3x$ and $x \to 3(1-x)$ (which preserve the cantor set when applied to the intervals $[0,1/3]$ and $[2/3,1]$ respectively), one can successively start with any five term AP and keep mapping it until it is no longer completely contained in either $[0,1/3]$ or in $[2/3,1]$, and then arrive at the above contradiction.

coffeemath
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  • I couldn't find a link to a proof of the "at most 4 length" for AP in the cantor set, just the above which is only an abstract of a talk. If anyone can find a link it would be great! – coffeemath Feb 23 '14 at 12:33
  • Both Mike Trenfield and Dylan Airy are absent in zbMATH. – Markiyan Hirnyk Feb 23 '14 at 14:14
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    It is an easy fact (if you follow aschepler's hint). Presumably, that talk also merely mentions that it is an easy fact, then goes on to something more interesting. – GEdgar Feb 23 '14 at 14:37
  • @ GEdgar: Are you serious? Could you give the answer? – Markiyan Hirnyk Feb 23 '14 at 15:07
  • @MarkiyanHirnyk I just added an explanation based on the hint of ashlepper's answer of why no 5 term AP. [Note I didn't find the proof in a paper, but the hint mentioned was enough to give a simple proof, as noted by GEdgar in his comment. – coffeemath Feb 23 '14 at 21:28