I just learned about primitive sets from the recent Numberphile video, and I'm having some difficulty with my intuition for these sets. For reference, a set $S \subseteq \mathbb{N}_{\geq 2}$ is called primitive if no element of $S$ divides any other. I'm particularly concerned with infinite primitive sets.
Any infinite subset $S$ of $\mathbb{N}$ can of course be uniquely ordered as a strictly increasing sequence $(a_n)_{n \in \mathbb{N}}$, and infinite sequences can be compared using the lexicographic order. In this way, we can define a partial order on infinite subsets of $\mathbb{N}$. In particular, we can consider this to be a partial order on the set of infinite primitive sets. My question is: is the set of primes a least element in this partial order?
It seems to me that you can argue that this is true by induction. If you choose your sequence term by term, and you always pick the smallest term possible, then you'll always be picking the primes, since the elements that come before are always made up of your previous choices.
However, if this is true, then it seems to me that you could just prove the Erdös primitive set conjecture by applying the series comparison test. This conjecture (now proven) states that if $S$ is a primitive set, then $\sum_{n \in S} 1/n\log(n) \leq \sum_{p \text{ prime}} 1/p\log(p)$. If the set of primes is minimal as I asked above, then since $1/n\log(n)$ is decreasing, the above inequality should be true. But, this is far too simple to be missed as a proof for 30 years. So, if the answer to my first question is "yes", then why doesn't this work as a proof?
You're correct, as far as I can tell, that the set of primes is the least element in this order (which is total, not just partial). But it doesn't follow that if $S \ge T$ in lex order then $\sum_{n \in S} \frac{1}{n \log n} \le \sum_{n \in T} \frac{1}{n \log n}$. You probably wanted to argue that if $S \ge T$ in lex order then, writing the elements of $S$ and $T$ as ascending sequences $s_n, t_n$, we have $s_n \ge t_n$, but that just doesn't follow, because the elements of $S$ might be "bunched up" more closely than the elements of $T$.
Dropping the requirements that $S$ or $T$ be infinite, we can get an explicit counterexample by taking $S = \{ 3, 4, 5, 7, 11, 13 \}, T = \{ 2 \}$. We can bump this up to an infinite counterexample by adding to both sets the set of primes larger than $13$.