Solving Set Theory Question Homework: Assume A Closed Algebra

[sammycaps]

Homework Statement

Assume I have the property that for any {Ei} (i=1 to infinity) contained in some algebra A, if E1 contained in E2 contained in E3... (infinite nesting), then Union Ei (i=1 to infinity) is also contained in A.

I simply want to show that for any {Ei} (i=1 to infinity) in A, I can construct an infinite nesting such that its Union will be in A, and this union will be exactly the same as the union of the original Ei.

The Attempt at a Solution

So basically, I choose infinite arbitrary sets from A. If each is contained in another then that's easy, and we can construct a nesting simply by rearranging. However, assume that some sets overlap, are not the same, and one is not contained in the other. If there are finite sets that all overlap with each other, are not the same, and none are contained completely in any others, then their union is contained in A because an algebra is closed under finite union. But if there are an infinite amount of these, then we have to construct an infinite nested set. Here's where I'm not sure if what I did makes sense. What I did was take a finite number of sets, take their union, and set that equal to E1. Then we choose another set, take its union with E1, and set that to E2. Then we have that E1 is contained in E2. Then I say that we can do this infinitely, and then say that U Ei (i=1 to infinity) is contained in A by assumption. I'm confused though, because I feel like I could do this last part even without the fact that for nested Ei, U Ei is contained in A, and it would still be valid (which is obviously not true because an algebra is not closed under infinite union). If someone can explain why what I did is right or wrong, that would help.

 

[mighty2000]

Your approach seems to be correct. By taking an infinite number of finite unions, you are essentially constructing an infinite nesting. And since each finite union is contained in A, the infinite union must also be contained in A.
As for your confusion about the last part, it is true that an algebra is not closed under infinite union, but in this case, we are not taking the infinite union of all the Ei's. We are taking the infinite union of the nested sets that we have constructed, and by assumption, this nested union is contained in A.
Overall, your solution is sound and shows that for any {Ei} in A, we can construct an infinite nesting whose union is also in A.