if Ā = ∩ F then show that A ⊆ Ā and Ā is the smallest closed set that contains A?
my question is:
The closure of A is denoted by Ā. Ā = ∩ F, where the intersection is taken
over all the closed sets F that contain A.
1- A ⊆ Ā
2- Ā is the smallest closed set that contains A
- Anonymous8 years agoFavorite Answer
The first one on the question above is correct. A ⊆ Ā if for an arbitrary element a of A, a is also in Ā. If a is in A then it is in every set in F so it is in ∩ F = Ā.
The second one is not actually a proof. You proved that there exists a set B in F containing A, not that A is the smallest closed set that contains A.
2. Every closed set that contains A is in F (by definition of F). Ā is closed because it is the intersection of closed sets. Now assume there exists a closed set B smaller than Ā (B ⊂ Ā) containing A. You need to show that, in fact, B = Ā, a contradiction, which implies a smaller set cannot exist.
To show 2 sets are equal you need B ⊂ Ā and Ā ⊂ B. The first is true by assumption. The second is true because B is in F and Ā is in every set in F (because Ā = ∩ F). Hence B = Ā and Ā is the smallest closed set containing A.
- 8 years ago
1. Assume a is in A. Then a is in each F, so it is in the intersection of all the F. So, a is in Ā and hence A ⊆ Ā .
2. Let B be any closed set containing A. Then, B must be one of the F sets, so ∩ F ⊆B. Therefore, Ā ⊆B, and Ā is contained in every closed set that contains A. Finally, Ā itself is closed because it is the intersection of closed sets.
- 5 years ago
very confusing step. search on to bing and yahoo. that could actually help!