Proofs regarding the boundary of a set

Let b(A) represent the boundary of the set A.

Prove:

(i) b(A) ⊂ A iff A is closed
(ii) b(A) ∩ A = ∅ iff A is open
(iii) b(A) = ∅ iff A is both open and closed

I'll use b(A) = set of all points p such that any open set containing p always contains points in A and ~A (the complement of A) [Edit. This is equivalent to the definition you added, "points not in the interior of A or ~A," since if p is in b(A), any open set containing p is not entirely in A or ~A, and therefore must contain points in A and ~A.]

These all follow from a straightforward application of the definition of open, closed and boundary.

i. Assume b(A) ⊂ A, and that p is in ~A. Since p in not in A, neither is it in b(A) which is a subset of A. So there is an open set O containing p such that O ⊂ ~A. Therefore ~A is open, so A is closed.

Assume A is closed, and p is not in A. Then there is an open set O containing p which does not intersect A, so p is not in b(A), and b(A) must be a subset of A.

ii. Assume b(A) ∩ A = ∅, and p is in A. Point p cannot be in b(A), so there is an open set O containing p which lies entirely within A. Therefore A is open.

Assume A is open and p is in A. There is on open set O containing p which lies entirely in A. So p is not in b(A). Therefore b(A) ∩ A = ∅.

iii. Let p be in A and q be in ~A. Since b(A) is empty, there is an open set O containing p but not intersecting ~A, and an open set R containing q but not intersecting A. Therefore A and ~A are both open, and both are also closed.

Assume A is both open and closed, p is in A, and Q is in ~A. There are open sets O and R containing p and q and entirely withing A and ~A, respectively. So neither p nor q can be in b(A), which must therefore be empty.

It is hard to answer this question without knowing where you are in your topology course or is this a course in real analysis? Some context would be helpful. In particular what definition (there are several) are you using for the boundary of a set.

More info please.