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.

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