Using your Head is Permitted

Suppose you want to find the factors of a number, n. An easy way to do this is to go over all x on the interval [2,floor(sqrt(n))] and to calculate for each x the value of the function y(x)=n mod x. Specifically, we are interested in finding the x values in which y(x) equals zero.

Because y(x) receives only non-negative values, every zero should be a local minimum. For simplicity, let us consider only those points that are strict local minima, meaning those points satisfying that y(x) is smaller than both y(x-1) and y(x+1). We do this by taking all (x,y) pairs and removing from them all pairs that do not represent local minima of the function.

We are left with a set of (x,y) tuples where x and y are both bounded on the interval [0,sqrt(n)]. Let us scale this down to [0,1] by dividing both axes by sqrt(n). This month's question is about the set of tuples that results from this operation.

The question is: what does it converge to, when n tends to infinity?

To be mathematically precise, we are asking here about convergence in the upper Vietoris topology. The meaning of this is as follows: The limit, S, of an infinite sequence of sets, {S₀, S₁,...}, under the upper Vietoris topology, is the set of all a for which there exists an infinite sequence of integers, {i₀, i₁,...}, diverging to infinity, and an infinite sequence {a₀, a₁,...} converging to a, with a_n in S_in for every natural n.

For brevity, we refer to "convergence" and to "limit" without specifying the topological space each time.

NB - We will only accept this month solutions in which the description of the limit set is exact.

As a bonus question, you are welcomed to consider also what is the limit, if we repeat the filter operation for a second time: instead of looking at local minima, we look in the bonus question only at local minima of local minima.

List of solvers: