Unions and Intersections

Proposition: Unions of Open Sets, Intersections of Closed Sets

Every union of open sets is again open.
Every intersection of closed sets is again closed.
Every finite intersection of open sets is again open
Every finite union of closed sets is again closed.

Proof:

Let { U _n } be a collection of open sets, and let U = U _n. Take any x in U. Being in the union of all U's, it must be contained in one specific U _n. Since that set is open, there exists a neighborhood of x contained in that specific U _n. But then that neighborhood must also be contained in the union U. Hence, any x in U has a neighborhood that is also in U, which means by definition that U is open.

To prove the second statement, simply use the definition of closed sets and de Morgan's laws.

Now let U _n, n=1, 2, 3, ..., N be finitely many open sets. Take x in the intersection of all of them. Then:

x is in the first set: there exists an with ( x - , x + ) contained in the first set
x is in the second set: there is with ( x - , x + ) contained in the second set.
....
x is in the N-th set: there is with ( x - , x + ) contained in the last set.

But then

let = min{ , , ..., }. Then ( x - , x + ) is contained in each set U _n

Hence it is contained in their finite intersection, which is therefore open, since x was arbitrary.

The last statement follows again from de Morgan's laws.

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