Let us say f({1,2,3,4,5}) = 5, for instance. Then I claim that for every B ⊆ A such that B contains 5, we must have f(B) = 5. The reason is that B has a complement B’, and f({1,2,3,4,5}) must equal either f(B) or f(B’). But f(B’) cannot be 5, so f(B) must be 5.

Now let us say f({1,2,3,4}) = 4, for instance. Then similar reasoning shows that f(B) = 4 for every B containing 4 but not 5. We may continue in this manner. Of course the selection of 5 and 4 above was arbitrary; what it comes to is that there must be some ranking of the numbers 1, 2, 3, 4, 5, such that f(B) is the highest-ranked element of B. The number of such rankings is 5! and each gives rise to a unique selector function. Thus there are 5! selector functions.

]]>That would be a heck of a superhero name!

]]>While playing with neodymium magnet balls, I did all sorts of spheres by removing some shapes that paves the plane, and then instead of removing I started adding more “fabric” of space and got hyperbolic surfaces.

Like for crochet, the easiest way is to make concentric circles of length greater than 6n, for example 7n or 8n… this is the equivalent to adding stiches every rank. ]]>