# show that if a family of sets is closed under union, then the complements of those sets form a family that is closed under intersection

I am reading a survey on Frankl's Conjecture. It is stated without commentary that the set of complements of a union-closed family is intersection-closed. I need some clearer indication of why this is true, though I guess it is supposed to be obvious.

• Are you familiar with De Morgan's Laws? Mar 28 '19 at 4:19

This is because we can rewrite $$(A \cup B)^C = A^C \cap B^C$$ using DeMorgan's laws. Now, if $$\mathbb{S}$$ is the family of sets closed under union and $$A, B \in \mathbb{S}$$, then $$A \cup B \in \mathbb{S}$$ (since it's closed under union). Therefore, $$(A\cup B)^C \in \mathbb{S}^C$$ (by the definition of the complement family). Hence, $$A^C \cap B^C = (A \cup B) ^C \in \mathbb{S}^C$$
It is not clear if you are talking about union of two sets or arbitrary unions. In either case the result is true and it is an easy consequence of DeMorgan's Law: $$(\cup_i A_i)^{c}=\cap_i A_I^{c}$$.