A set A is compact, is its boundary compact?

Question

I am trying to understand the concept of a boundary, and I have seen it defined $Bd(A) = \overline{A} \cap \overline{A^{\complement}}$. I was wondering three things, First how can I show that the boundary of a compact set is compact. I want to try proving with the definition above. I feel as thought it is one of those things that is inherently true but hard to prove, I know that if A is compact, it must be bounded, thus anything it intersects with is bounded, but what about the situation where A closure is not bounded.

There other two things I was wondering about was can you have a boundary that is the empty set, and a boundary that is the real numbers. For the empty set I was thinking 1/n plus 0, would produce a set that has a boundary of the empty set, not sure about the reals tho.

Answer 1

postscript: I was a bit hasty. In a Hausdorff space, compact sets are closed. end of postscript

Being closed [but see above], compact sets contain all of their boundary points. The boundary of a compact set is therefore a subset of that compact set and is closed. Closed subsets of compact sets are compact.

Answer 2

About the rest of the question, which has been skipped by Michael, a set with empty boundary is necessarily open and closed (because its closure is itself, and the closure of its complelent is the complement itself). So in R the only sets with empty boundary are the empty set and R itself.

On the other side, a set with the whole R as booundary is easy, just take any dense and co-dense set, like the rationals. Being dense, its closure is all R, and as the complement is dense too, then the boundary is actually R.