Let F be a family of subsets of a set X such that empty set is the element of F. A subset A of X belongs to F1 if and only if either A is a subset of F or complement of A is the subset of F. A subset B of X belongs to F2 if and only if B is a finite intersection of sets in F1. A subset C of X belongs to F3 if and only if C is a finite union of sets in F2. Show that F3 is the smallest algebra which contains F.
Why not try and trick one of your other profs to do this for you?
becouse they couldn't help me :(
To show that F3 is the smallest algebra that contains F, we need to prove two things:
1. F3 contains F.
2. Any algebra that contains F3 must contain F as well.
Let's start with the first step:
1. F3 contains F:
We know that F1 contains F because any subset A of X that belongs to F must be either a subset of F or its complement is a subset of F. This means that if F contains the empty set, then F1 will also contain the empty set.
Next, we need to show that F2 contains F1. Let's consider an element A in F1. There are two cases to consider:
- Case 1: A is a subset of some set S in F. In this case, A is a finite intersection of sets in F1 because A can be expressed as the intersection of A itself and S.
- Case 2: The complement of A is a subset of some set S in F. In this case, A can be expressed as the complement of S, which is the intersection of the complement of A itself and the complement of S. Since the complement of S is in F1, A is a finite intersection of sets in F1.
Now, let's show that F3 contains F2. Let B be an element in F2. By definition, B is a finite intersection of sets in F1. Therefore, B is also a finite intersection of sets in F3 since F1 is a subset of F3.
Finally, we need to show that F3 is an algebra. An algebra is a collection of subsets of X that is closed under complementation, union, and intersection.
- Complementation: Let C be an element in F3. By definition, C is a finite union of sets in F2. Taking the complement of C means taking the complement of the union of sets in F2. This complement will be a finite union of complements of sets in F2, which are also finite unions of sets in F1. Thus, the complement of C is in F3.
- Union: Let D and E be elements in F3. By definition, D and E are finite unions of sets in F2. The union of D and E is also a finite union of sets in F2, and therefore, it belongs to F3.
- Intersection: Let D and E be elements in F3. By definition, D and E are finite unions of sets in F2. The intersection of D and E can be expressed as a finite intersection of sets in F2. Therefore, it also belongs to F3.
Now, let's move on to the second step:
2. Any algebra that contains F3 must contain F:
Let's assume that there exists an algebra G that contains F3, but does not contain F. This means that there exists some element A in G that is not in F. We need to show that this assumption leads to a contradiction.
Since G is an algebra, it is closed under complementation. Therefore, the complement of A must be in G as well. By the definition of F1, either A or its complement must be in F. Since A is not in F, its complement must be in F. This means that the complement of A is a subset of F.
Now, let's consider the subset B = A ∩ (complement of A), which is the intersection of A and its complement. By the properties of the intersection, B = ∅ (the empty set). Since F is a family of subsets that contains the empty set, B must be in F. However, B = ∅ is a subset of F2, which is a finite intersection of sets in F1, and therefore, a subset of F3.
But this contradicts our assumption that A is not in F. Thus, our assumption that there exists an algebra G containing F3 but not F must be false. Therefore, any algebra that contains F3 must contain F as well.
In conclusion, we have shown that F3 contains F and any algebra that contains F3 must contain F. Therefore, F3 is the smallest algebra that contains F.