For anyone interested in a more technical summary of the some of the main ideas:

Problem Statement

  The Erdos conjecture that all primitive sets S satisfy f(S) <= f(PRIMES), where f(S) = sum_{s} (1/(s log s)) and PRIMES is the set of all primes.

Definitions

  Define A_q to be a primitive set such that all elements a are divisible by prime q, but no primes less than q.
   * Example: A_3 contains multiples of 3, but no even numbers.

  Define A*_q to be the set of all integers divisible by only primes >= q.
   * Example: A_q is a subset of A*_q.
   * Example: A_q is also a subset of q \* A*_q (i.e. multiply each element by q).

  Define g(a) = 1/a * Product_{p<LargestPrimeDivisor(a)}[(1 - 1/p)]. 
   * Example: if a=7, then the Product term (excluding the 1/a) is the "density" of A*_7. This is because (1-1/2) of integers aren't divisible by 2, (1-1/3) aren't divisible by 3, and (1-1/5) aren't divisible by 5.
   * Example: if a=7, then g(7) is the density of all multiples of 7 that aren't divisible by 2,3,5.

Results

  In [1], Author proves that for all A_q, with A_q \neq {q}: f(A_q) < k*\sum_{a\in A_q} g(a) <= k*g(q), for a constant k. 
   * The first half of the inequality comes from a cited result: 1/(a log(2a)) < k*g(a).
   * The second half of the inequality comes from clever construction of the sets S_a = (a \* A*_{LargestPrimeDivisor(a)}), for all a in A_q. These sets are constructed to satisfy g(a) = |S_a|/|NaturalNumbers|. So because (1) S_a and S_b are disjoint for a\neq b (2) S_a is a subset of A*_q, this implies that sum(g(a)) = sum(|S_a|) <= sum(|A*_q|) = g(q).

  In [2], Author shows that k\*g(q) <= 1/(q log q). Note that the right-hand side of the inequality is equal to f({q}). 
   * This uses a pretty clever partitioning of the primitive set A into A_2 U A_3 U A_5 U ... U A_q U .... Combining with part 1, this implies f(A_q) < f({q}) for all primes q, so f(A) < f(PRIMES).

[1] https://arxiv.org/pdf/1806.02250.pdf [2] https://arxiv.org/pdf/2202.02384.pdf