i.e. the complement of the intersection
of any number of sets equals the union of their complements.
i.e. the complement of the union of any
number of sets equals the intersection of their complements.
i.e. the complement of the intersection
of any number of sets equals the union of their complements.
i.e. the complement of the union of any
number of sets equals the intersection of their complements.
, i.e. x is not in the intersection of all
. Then there must be at least one set
that does not contain x (because if all
would contain x then x would be in their
intersection as well). Call this set A. Since x is not in A, x is in
the complement of A. But then x is also in the union of all complements of
, because A is one of those sets.
This proves that the right hand set is contained in the left one.
. That means there is at least one
complement that contains x, or in other words, at least one of the
does not contain x. But then x is not in
the intersection of all
, and hence it must be in the complement
of that intersection. That proves the other inequality, and hence both sets must
be equal.
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