Understanding σ-Fields: Exploring F and G Intersection & Union in Set Theory

[Alexsandro]

Can someone help me with this question ?

Let F and G be σ-fields os subsets of S.

(a) Let H = F intersection G be the collection of subsets of S lying in both F and G. Show that H is a σ-field.

(b) Show that F union G, the collection of subsets of S lying in either F or G, is not necessarilt a σ-field.

 

[EnumaElish]

(a) show that the intersection of two closed sets is a closed set (e.g. "closed under complements").

(b) counterexample: Let f be an element of F and g be an element of G. Suppose f U g is neither in F nor in G, therefore not in H. This means H is not closed under union.

P.S. Let S = {1,2,3}, F = {S, ø, {1}, {2,3}}, G = {S, ø, {2}, {1,3}}. Then F U G = {S, ø, {1}, {2,3}, {2}, {1,3}}. If H = F U G was σ, then it would be closed under union, which implies that {1} U {2} = {1,2} would be a distinct element of H. Since it is not, H is not σ.

 

[tacman]

(a) To show that H is a σ-field, we must prove three properties: closure under complement, closure under countable union, and non-emptiness.

Closure under complement: Let A be a subset of S lying in both F and G. Then, A is also a subset of F and G individually. Since F and G are σ-fields, they are closed under complement. Therefore, A' (complement of A) must also be a subset of both F and G, and hence, A' is in H. This proves closure under complement for H.

Closure under countable union: Let {A_n} be a countable collection of subsets of S lying in both F and G. This means that each A_n is also a subset of F and G individually. Since F and G are σ-fields, they are closed under countable union. Therefore, the countable union of A_n, denoted by ∪A_n, must also be a subset of both F and G, and hence, ∪A_n is in H. This proves closure under countable union for H.

Non-emptiness: Since F and G are σ-fields, they both contain the empty set ∅. Therefore, ∅ is also a subset of both F and G, and hence, ∅ is in H. This proves non-emptiness for H.

Hence, we have shown that H is a σ-field, as it satisfies all three properties.

(b) To show that F union G is not necessarily a σ-field, we must find a counterexample where at least one of the three properties is not satisfied.

Let F = {∅, {1}, {2,3}, {1,2,3}}, G = {∅, {1}, {2,3}, {2,4}, {1,2,3,4}}.

F union G = {∅, {1}, {2,3}, {2,4}, {1,2,3}, {1,2,4}, {1,2,3,4}}.

This collection is not closed under complement, as {2,4} is in F union G, but its complement {1,2,3} is not. Therefore, F union G is not a σ-field.