find the least $k$ such that any subset of order $k$ contains 4 terms in arithmetic progression. Question: 

Let set $A=\{1,2,\cdots,100\}$ ,find the least $k$ such that any subset of order $k$ contains 4 terms in arithmetic progression.

It seem interesting problem.
Now I have found $k$ must $k\ge 43$,because I found following set has $42$ elements and contains no 4 terms in arithmetic progression.
take$$\{1,3,6,7,9,10,14,16,17,19,20,21,26,27,29,30,33,34,35,47,50,52,53,54,57,58,59,63,64,66,72,77,80,83,87,89,90,92,96,97,98,100\}.$$
so I think $k\ge 43$?
 A: There is some discussion of this in Guy, Unsolved Problems in Number Theory, 3rd edition, section E10 (I found the following at https://books.google.com.au/books?id=t_3lBwAAQBAJ&pg=PA113&lpg=PA113&dq=%22introduced+long+years+ago+by%22&source=bl&ots=xXUDWLeUSX&sig=Wk_o0Ux0Z_zm1Gj4JThp0BR8N7U&hl=en&sa=X&ved=0ahUKEwiR6_aRyLzTAhXDNpQKHV13DTkQ6AEIKjAD#v=onepage&q=%22introduced%20long%20years%20ago%20by%22&f=false). He refers to "$r_k(n)$, introduced long years ago by Erdos and Turan: the least $r$ such that the sequence $1\le a_1<a_2<\cdots<a_r\le n$ of $r$ numbers not exceeding $n$ must contain a $k$-term arithmetic progression.... Rankin showed that $$r_k(n)>n^{1-c_s/(\log n)^{s/(s+1)}}$$ where $s$ is defined by $2^s<k\le2^{s+1}$." We're interested in $k=4$, so $s=1$. There are some other things at E10 that will be of interest. 
Very relevant is http://oeis.org/A005048 which tabulates the "minimal span of set of $n$ elements with no 4-term arithmetic progression," where "span" is the difference between the largest and smallest elements. E.g., the entry for 34 is 69, meaning that there's a size-34 subset of $\{\,0,1,\dots,69\,\}$ with no 4-term AP (but no such subset for $\{\,0,1,\dots\,68\,\}$). 
Also of interest is http://oeis.org/A005839 which is the "greedy sequence" with no 4-term AP. 
