If A⊆X, we define the boundary of A by the equation ∂A=¯A∩¯X−A. (a)Show that A∘ and ∂A are disjoint, and ¯A=A∘∪∂A. (b) Show that ∂A=ϕ⇔A is both open and closed. (c) Show that U is open ⇔ ∂U=¯U−U.
I’m assuming, we know ¯X−A=X−A∘ fact. If you want, you can prove it.
My attempt: (a) A∘∩∂A=A∘∩(¯A∩¯X−A)=(¯A∩A∘)∩(X−A∘), by associative law and above fact. Since A∘⊆¯A, we have ¯A∩A∘=A∘. Thus, A∘∩∂A=ϕ.
A∘∪∂A=A∘∪(¯A∩¯X−A)=(A∘∪¯A)∩(A∘∪¯X−A)=¯A∩X=¯A, by De Morgan's law and above fact.
(b) Suppose A is both open and closed. So ¯A=A and ¯X−A=X−A. Thus ∂A=A∩(X−A)=ϕ.
Now suppose ¯A∩¯X−A=ϕ. If A is open, then X−A is closed. So ¯X−A=X−A. Since ∂A is empty, we have ¯A⊆X−(X−A)=A. Thus ¯A=A, A is closed. Similarly result holds, if A is closed. Is this proof correct?
(c) first suppose U is open. Then ∂U=¯U∩¯X−U=¯U∩(X−U), since U is open. By Ex 2(g) of section 1, (¯U∩X)−(¯U∩U)=¯U−U. Our desired result.
Conversely ¯U−U=∂U=¯U∩¯X−U=¯U∩(X−U∘)=¯U−(¯U∩U∘). So ¯U∩U∘=U=U∘. Thus U is open. Is this proof correct?