The interior of a set is open and the boundary and closure of a set is closed
So I am pretty sure my proof is correct, just want to verify for correctness and rigor.
Let . Then there exists such that . Now suppose . Since is open, there exists such that ; in other words, for any there is a ball centered at contained solely in , which means . So given there exists , namely , such that , so is open.
So now consider . Then since if is not a point such that every ball centered at intersects and , then there exists some ball centered at that is either contained solely in or . So we have that is the union of two open sets, which is open, so is closed. Similarly, if we consider , then , which is open, so is closed.
So I think this is a perfectly valid and rigorous proof, but any feedback would be appreciated (I am somewhat unsure on the closed portions of the proof, especially the claims about what the complements of the boundary of S and the closure of S actually equal). Thanks.