# Simple analysis question

## Main Question or Discussion Point

Hello,
I know how to start but I don't know how to end that proof. It's supposed to be easy:
Let S a subset of Rn.
PROVE THAT the boundary of S is a closed set.
(I'll use d for delta, so dS is my convention for "the boundary of S").

So here I go:
dS is closed iff it contains all of its boundary points,
so dS is closed iff d(dS) is included in dS.
Let x be any point such that x belongs to d(dS).
So for any Ball B(r, x), r>0, (ie centered at x),
| B intersection dS is not empty
| and B interesection (dS)complement is not empty.
(the second line is equivalent to) B interesection (interiorOfS union exteriorOfS) is not empty

Now what's next??
Thanks for your suggestions. If you do have a suggestion, please don't skip a step or don't bypass a detail because it seems obvious (trust me, nothing is obvious to the one who doesn't know yet!)

morphism
Homework Helper
Have you tried showing that the complement of dS is open?

mathwonk
Homework Helper
the boundary is the complement of the interior which is easily shown to be open. qed. oops. my bad. that was just said.

HallsofIvy