Certain arithmetic progessions in the Cantor set How to prove or disprove  that the Cantor set does not include any arithmetic progression of  length 5?
 A: 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 $C$.
A: 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.
