In an Euclidean $\mathbb R^n$ space, is every compact set an open set? Is it possible to have sets that are both open and bounded? I know that compact sets are the ones that are both bounded and closed (Heine-Borel Theorem), but since closed and open are not opposites, I cant see if and how a compact set, or a bounded set, can be open.
 A: Note well that the only subsets of $\mathbb R^n$ ($n\in\mathbb N$) that are both closed and open are the whole set and the empty set. (This is the consequence of a fundamental property of Euclidean spaces called connectedness, which is the topological formalization of the intuitive geometric notion of contiguity.) Since $\mathbb R^n$ is not compact (it's unbounded), the only compact open subset in this space is the empty set.
Nevertheless, a bounded set may very well be open. Just consider any open interval $(a,b)$ ($a\in\mathbb R$, $b\in\mathbb R$, $a<b$) for $n=1$, any circle without its perimeter for $n=2$, or any sphere without its surface for $n=3$.

At any rate, if one digresses from the usual Euclidean topology on $\mathbb R^n$, interesting things can happen. Endow $\mathbb R^n$ with the discrete topology, in which every subset is declared to be open. But then the complement of every subset is open too, so every subset is closed as well. Consequently, every subset of $\mathbb R^n$ is both open and closed in the discrete topology! Moreover, one can show that a subset is compact with respect to this topology if and only if it contains only finitely many points, so the Heine–Borel theorem (formulated for the usual Euclidean topology) no longer applies to the discrete topology!
A: Try the empty set. Try also the interior of a compact set like the unit disc. 
A: No compact subset of $\mathbb R^n$ is open except $\varnothing$.  Since $\mathbb R^n$ is a connected space, its only subsets that are both open and closed are $\varnothing$ and $\mathbb R^n$, and the latter is not compact.
Lots of sets are both open and bounded.  Every open ball is both open and bounded.
