A is a nonempty set. Show boundary of A is closed. Obviously dealing in the real number space.

I've seen a couple of proofs for this, however they involve 'neighborhoods' and/or metric spaces and we haven't covered those. Here are some following terms we've define: segment, limit point, boundary point, interior point, boundary, derived set( the set of all limit points), isolated point, density, open, closed, interior, exterior, connected, disconnected.

I've started with this.. not much of a proof, just a thought process.

The boundary of A is the set of points that are both limit points of A and A^{C }. Where A^{c}is A complement.

(i.e. boundary of A is the derived set of A intersect the derived set of A^{c })

Note: boundary of A is closed if and only if every limit point of boundary of A is in boundary of A. (?or in boundary of the derived set of A is open?)

- If you could lead me in the right direction or show me how to go out the proof, that would be rad to say the least. Thanks.

Another one if you care to:

A is a subset of the interior of A. Show A is open.

All I know is a set is open if and only if for all x in A there exists a segment S containing x such that S is a subset of A.

Thank you