another topology question?

Question

Let A be a fixed subset of the set X
Let T={U, a member of power set of X, (U is a subset of X): U is
an empty set or A is a subset of U}

I proved that T is topology on X
and for A and B (A is a proper subset of B, B is a proper subset
of X), I guess intB = B, clB = empty set, FrB = empty set.
(int = interior, cl = closure, Fr = boundary) Is this a right guess?

Answer 1

I'm assuming A is not empty. Then, by the definition of the topology T, it follows A is in T, so that A is open and intB = B. Right.

clB = B? No! B is a subset of clB and since B is not empty, for it contains non-empty A, it follows clB is not empty. Wrong.

Now, see. In the topology T, a set S is closed if it's complement is open. Therefore, S is closed if S = X or if the complement of S contains A, which, in this case, implies S doesn't intersect A,. clB is the intersection f all closed sets containing B. So, the only closed set that contains B is X, so that clB = X.

We have that Fr(B) = clB - intB = X - B, so that Fr(B) is the complement of B. therefore, unless B = X, your guess is wrong.

Just check, I may have made a mistake.