Thread: conclude that the closure of a union is the union of the closures.

For this question, we are allowed to assume the following is true and has been proven:

Let A,B R and p R. Prove that p is an accumulation point of A B if and only if p is an accumulation point of A or p is an accumulation point of B: (A B)'=A' B'.


Now since we assumed the above is true,

Conclude that the closure of a union is the union of the closures:
(A B)^cl=A^cl B^cl. How would I go about proving this?

The closure of a set is the ‘smallest’ closed that contains the set.
We know that .
The union of two closed sets is a closed set.
That forces . Why and how?

Can you do the converse?

After working on this for a little, is this proof correct?

Assume p A^cl and that p B^cl. That is, there exists >0 such that A has no point in (p- ,p+ _1) and there exists >0 such that B has no point in (p- ,p+ _2). Let =min{ , }, and clearly >0 and satisfies the requirement.

Then Assume p A^cl B^cl.
case 1: p A^cl which means that every neighborhood of p contains a point of A. Hence every neighborhood contains a point of A B .

case 2: p B^cl which means that every neighborhood of p contains a point of B. Hence every neighborhood contains a point of A B .
Thus, (A B)^cl=A^cl B^cl