We begin with the following Claim:
Claim: If x,x+1∈A, then x≤31. Furthermore, if x,x+2∈A, then x≤43.
Proof: Easy to see from the given hypothesis ◼
Therefore, by partitioning the elements of the set X={1,2,3,…,50} in 10 groups as follows,
Group 1: X1={1,2,…,31},
Group 2: X2={32,33},
Group 3: X3={34,35},
…
Group7: X7={42,43},
Group 8: X8={44,45,46},
Group 9: X9={47,48,49} και
Group 10: X10={50},
we note that A can contain at most one element from each Xi, with 2≤i≤7, and at most one element from each Xi, with 8≤i≤9. In total, we have
|A|≤31+(1+1+…+1)+(1+1)+1=31+6+2+1=40.
An example showing that |A|=40 is feasible, is the set {1,2,…,31,32,34,36,38,40,42,44,47,50}. To check that this set works, note that for all a,b,c∈A with a<b<c, we have
1a−1c=(1a−1b)+(1b−1c)>1a−1b,
so it suffices to check the hypothesis for each two consecutive (when written in increasing order) elements. This is immediate from the Claim.