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.
Context
Proof:
We will prove two set-inequalities to prove equality of the left and right hand sides.
Take x contained in the complement of the intersection of all sets
Aj, i.e. x is not in the intersection of all
Aj. Then there must be at least one set
Aj that does not contain x (because if all
Aj 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
Aj, because A is one of those sets.
This proves that the right hand set is contained in the left one.
Now take x contained in the union of all complements of
Aj. That means there is at least one
complement that contains x, or in other words, at least one of the
Aj does not contain x. But then x is not in
the intersection of all
Aj, and hence it must be in the complement
of that intersection. That proves the other inequality, and hence both sets must
be equal.
The proof of the second de Morgan’s law is left as an exercise.
Interactive Real Analysis, ver. 1.9.5 (c) 1994-2007, Bert G. Wachsmuth Page last modified: Mar 28, 2007
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.
Context
