In Euclidean space, the boundary of a subset (Closure - Interior) is always closed, but I would like to know when it's compact as well. Is it when the subset is bounded?
Community wiki answer-made-of-comments:
The boundary of a bounded set is itself bounded. By the Heine-Borel theorem, if a subset of the Euclidean space is bounded then its boundary is compact. – A. De Luca
Everything that others have said is true. I just want to add that the boundary of the complement of a bounded set is also bounded. After all, everything far out is then in the interior, and thus the boundary is still bounded + closed, and hence by Heine-Borel also compact. – Jyrki Lahtonen
In $\mathbb R^n$ with $n\ge 2$, the above two cases are exhaustive: the boundedness of $\partial A$ implies that either $A$ or $A^c$ is bounded. Not so in $n=1$. The key difference is the connectedness of $\mathbb R^n \setminus B(0,R)$. – Leonid Kovalev