Constructing Intersection of Sets for De Morgans: Get Union of Sets

[b0it0i]

Homework Statement

Given A1 superset of A2 superset of A3 superset of A4 ... and so on
how can i construct sets B1, B2, ...
so that each Bi's are disjoint.

The goal is to get

the infinite intersection of Ai = the infinite union of Bi

Homework Equations

De morgans law:
(AUB)^c = (A^c N B^c)
(ANB)^c = (A^c U B^c)

which can be applied to infinite unions and intersections

The Attempt at a Solution

somewhere along the way, i understand that i need to involve de morgans law to turn the intersection into the union, but each Bi that I try gives me something strange and i can't come up with the result i want

The book suggests to use
Let
B1 = A1
B2 = A2\B1
.
.
.
Bk = Ak\Bk-1

All the Bk's are disjoint, but...

what exactly is B2=A2\A1... i mean everything in A2, is technically in A1,
since A2 subset of A1. So an element in B2 is an element in A2, yet is not an element in A1, which then implies it is not an element of A2...

so I'm totally lost

 

[Dick]

No, B1 and B2 don't have any common elements. B1=A1, B2=A2\B1. If B2 is the difference between something and B1, it's hardly possible for B1 and B2 to have common elements, isn't it?

 

[b0it0i]

No, B1 and B2 don't have any common elements. B1=A1, B2=A2\B1. If B2 is the difference between something and B1, it's hardly possible for B1 and B2 to have common elements, isn't it?

i see. however, i should have worded that part better, noting that A1 super set of A2 superset of A3...

what exactly is B2=A2\A1... i mean everything in A2, is technically in A1,
since A2 subset of A1. So an element in B2 is an element in A2, yet is not an element in A1, which then implies it is not an element of A2...

so something is in A2, yet not in A2... isn't that a contradiction

 

[Dick]

Ok, so it's saying A1 contains A2 contains A3 etc, right? In that case saying B2=A2\A1 may simply be a typo. That would be empty. I see your point. I guess you should just assume they muffed the notation on the hint and fix it for them.

 

[b0it0i]

Ok, so it's saying A1 contains A2 contains A3 etc, right? In that case saying B2=A2\A1 may simply be a typo. That would be empty. I see your point. I guess you should just assume they muffed the notation on the hint and fix it for them.

this problem actually hard two parts

one dealt with when A1 C A2 C A3 C ...

the other was when A1 contains A2 contains A3 contains

but the book only gave that single hint. It didn't say it strictly applied to one or the other, and I'm assuming it meant that it should work for both cases. And the hint works when A1 C A2 C A3 C ...
but, the problem is in the second case, the one i stated in my problem.

are there any other possible Bi's that would work for the original problem. That's the main problem I'm having, coming up with such Bi's that would work