#Let S = {1, 2, ..., 6}.

9 messages · Page 1 of 1 (latest)

versed shoal
#

Let S = {1, 2, ..., 6}. How many rules F, which associate a subset F(X) of S to each subset X of S, satisfy the condition F(F(A) ∪ B) = A ∩ F(B) for any subsets A and B of S?

languid chasmBOT
versed shoal
#

<@&286206848099549185>

#

Please

#

It’s either 64 or 1

honest sparrow
#

I think I'm late to this but I think I will still propose you a solution.
As, F(B) ⊆ F(F(A)∪B) and, A ∩ F(B) ⊆ F(B), if F(F(A) ∪ B) = A ∩ F(B), then:
F(F(A) ∪ B) = A ∩ F(B)=F(B) =>F(A)⊆F(B)=A. (1)
I think only the rule F(X) for any subset X of S so that F(X)={1,2,3,4,5,6}=S sastify this, or else we can always choose one or more element for set B so that F(B) ⊂ S, and then choose set F(B)' ⊂ S as set A to contradict the (1) statement

#

@brain4brain

#

@Brain4Brain

honest sparrow