Leaving aside that Cradarc still can’t spell my name even a month after I corrected her or him on

the exact same mistake, I think there’s a lot of talking past one another going on here.

Define “point x has color y” to be the ordered pair (x, y). In other words, treat color as a full-fledged dimension in its own right. Now do all the coloring in half-open intervals, eg. [0, ½). And while we’re at it, let “white” = 0 and “red = 1”.

Initially, we have the set S

_{0} consisting of the points on the line y=0 with 0≤x<½ and the points on the line y=1 with ½≤x<1. After one iteration, we have the set S

_{1} consisting of the points on the line y=0 with 0≤x<¼ ⋃ ½≤x<¾ and the points on the line y=1 with ¼≤x<½ ⋃ ¾≤x<1. Subsequent iterations yield S

_{k} as expected.

If we take the liminf of these sets, meaning the union of the intersections of the tails, we end up with the dyadic x-values at y=0 and nothing else.

If we take the limsup of these sets, meaning the intersection of the unions of the tails, we end up with the entire interval 0≤x<1 at y=0 and that same interval except for the dyadics at y=1.

However, we can do better. Instead of having a discrete sequence of sets, we can make a continuous transition from one to the next. In other words, let the subscript of the sets become a real parameter, represented by a full-fledged dimension of its own. We can call it z.

Now we don’t have a sequence at all, just a set of points in 3 dimension. At z=0 we have the points (x, y, 0) where (x, y) is in S

_{0}, and so on: for each natural number k we have the points (x, y, k) where (x, y) is in S

_{k}.

There are, of course, uncountably-many ways to connect the “slices” at each z=k to one another. Perhaps the simplest way is to hold each x constant and let y go linearly with z to make a straight line from (x, y, k) to (x, y, k+1). If we let z rise without bound then the y-values will keep oscillating between 0 and 1 everywhere except dyadic x-values, so the limit will only be points of the form (d, 0, ∞) where d is dyadic.

Another simple way to connect the “slices” is to hold each y constant and let the x’s from the nearer half of each interval go linearly with z where necessary to connect with the next slice up. If we let z rise without bound then the x-values will move less and less, so the limit will be points of the form (x, 0, ∞) for 0≤x<1 and points of the form (x, 1, ∞) for non-dyadic x in the same interval.

Even though each dyadic x-value d only has (d, 1, z) in the set when z is less than some finite bound (which depends on d), there is nonetheless for any given dyadic d some other x-value, call it v, such that the path beginning at (v, 1, ∞) and following the continuous mapping from the prior paragraph approaches (d, 1, ∞). In this “path-limit” sense, the limit consists of all points (x, y, ∞) for 0≤x<1 and y∈{0, 1}.

As it happens, the specific description of the situation given in the original post makes it abundantly clear that the pairs (x, y) can change their x-value, but not their y-value. And the mechanism by which the x-values change is consistent with the continuous piece-wise linear paths I described two paragraphs up. Moreover, the end-state is defined by the position that each point (x, y, 0) approaches along its path as z→∞.

With this understanding, the problem yields immediately to elementary analysis. Each point (x, y, 0) maintains its initial y-value, and its x-value changes by excising the first digit after the decimal point of its binary representation, as Gmalivuk described. Algebraically, when 0≤x<½ the limit is 2x, and when ½≤x<1 the limit is 2x−1. In short, we have (x, y, 0) → (2x%1, y, ∞).

Therefore the set of points that exist in the limit contains precisely (x, 0, ∞) ⋃ (x, 1, ∞), for 0≤x<1. That is two horizontal line segments. In the language of the original problem, every physical location along the edge of the square ends up with two slips of paper: one red, and one white. No point is pink, instead every point is both white and red.

This is one of the things we have to get used to when treating color as a dimension: just as a shape can contain points with multiple y-values for the same x, so too can an object have more than one color at the same location. In fact, the case at hand is quite simple, what with there being two discrete colors. But we must be prepared to deal with things which are a whole spectrum of colors all at once. In other words, objects that extend along the color dimension for some interval—or even forever!