Each number in a subset $S\subseteq \{1,\ldots,2n\}$ does not divide another one. Then $\max |S|$?

Question

This problem comes from a seemingly innocuous question from a professor during a lesson for a Math Olympiad course. [A part of this question is really a classic of number theory/combinatorics]

Problem: Let $S$ be a a non-empty subset of $\{1,2,\ldots,2n\}$. It is clear that if $|S|$ is "enough large" then, independently from the choice of $S$, there are for sure some distinct $a,b \in S$ such that $a$ divides $b$.

Then, what is the maximal $k$ such that, for some $S$ with cardinality $k$, if $a,b$ are distinct members of $S$ then $a$ does not divide $b$?

[Ps. I know the solution]

If you know the solution, then what are you looking for here? — Ben Grossmann, Aug 25 '15 at 15:51
With $100$ instead of $1000$, a golden oldie that has appeared more than once on MSE. — André Nicolas, Aug 25 '15 at 15:54
@Omnomnomnom One should not post question necessarily when they do not know the answer, am I wrong? — Paolo Leonetti, Aug 25 '15 at 15:55
@AndréNicolas 100 or 1000 does not make so much difference; do you have the link of thread you are talking? — Paolo Leonetti, Aug 25 '15 at 15:57
@PaoloLeonetti "quizzing" or "puzzling" the other users with questions "for fun" is frowned upon on this site. — Ben Grossmann, Aug 25 '15 at 15:58
@Omnomnomnom: I've noticed that, and so I have refrained from asking questions that I know the answers to. But I find that to be a puzzling (no pun intended) restriction. What is the harm to the site and its users of these questions that is commensurate with (say) asking questions that are not mathematical, or are avenues to cheating? (I guess I should be asking this on meta.) — Brian Tung, Aug 25 '15 at 16:10
Not quite. though a small twist on that solution will do it. — André Nicolas, Aug 25 '15 at 16:11
@BrianTung I've found a few threads on the meta site already (here, here, and here). The answers are a little inconsistent, but it seems that it's okay to ask such a question as long as the asker is either actually interested in the answers or answers the question himself. — Ben Grossmann, Aug 25 '15 at 16:15

score 3 · Accepted Answer · answered Aug 25 '15 at 17:27

Let $S=\{n+1,n+2,\ldots,2n\}$. The claim is that $|S|$ is maximal.

To prove this, form descending sequences starting with each term in $S$, where each sequence is finite and ends on an odd term and the term in any sequence is the previous term divided by 2. Denote each sequence by its starting term, e.g. $$\begin{align} S^{(2n)}&=\{2n,n,\ldots\} \\ S^{(2n-1)}&=\{2n-1\} \\ &\quad\vdots \\ S^{(n+1)}&=\{n+1,\ldots\} \end{align}$$

Every number in $\{1,2,\ldots,2n\}$ appears in exactly one of the $S^{(k)}$ because:

Surjective. Every number inclusively between $1$ and $n$ has a multiple of a power of two inclusively between $n+1$ and $2n$. If not, $2 \ge \frac{2n+1}{n}$, which is absurd.
Injective. No number can appear in more than one sequence. If this was untrue we could find two distinct terms in $\{n+1,n+2,\ldots,2n\}$, one of which is a multiple (of a power of two) of the other.

It is clear that every pair of terms in a given $S^{(k)}$ has the smaller term dividing the larger one, so we are free to choose at most one term from each $S^{(k)}$, i.e. $|S|\le n$ for any prospective set $S$. It is also clear that choosing $S$ as originally proposed gives $|S|=n$, so this is maximal.

Each number in a subset $S\subseteq \{1,\ldots,2n\}$ does not divide another one. Then $\max |S|$?

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