Is there a way to construct an open set whose boundary is A?

[Newtime]

Homework Statement

Prove: a set in a topological space is closed and nowhere dense if and only if it is the boundary of an open set.

Homework Equations

Basic definitions of closed, nowhere dense, open and boundary.

The Attempt at a Solution

One direction is easy. Let $$A \subset X$$ be a subset in a topological space. Since $$\partial A = \bar{A} \cap \overline{X-A}$$ and the (finite) interesection of closed sets is closed, the boundary is closed. We also have another interpretation of the boundary. Namely, $$\partial A = \bar{A} - \text{int}(A).$$ The boundary is nowhere dense since the interior is the largest open set so deleting it means the boundary contains no nonempty open set.

I'm having trouble with the other direction. It seems necessary to do it by construction since there is no general criteria to conclude some random set is a boundary of an open set other than using the definitions, which is how most proofs go in point set topology. However, we know nothing about the set a priori or even about the space, so I'm pretty confused. I think all I need is a (small) nudge in the right direction and any help is appreciated. Thanks in advance.

 

[Newtime]

Anyone?

 

[Newtime]

I figured it out. For anyone else stuck on this question: if A is a closed nowhere dense set, consider the set X-A.