Is the boundary of a manifold always compact? Coming from a background of undergrad physics training, I know that "the boundary of a boundary is zero".
Does this mean that the boundary of a manifold is always compact(or maybe just closed)?  
Keep in mind that some of the terminology that mathematicians use might not be familiar to me due to my aforementioned background.
Thank you.  
Note: This question arises from my reading of the anti De-Sitter spacetime which has a compactified Minkowski spacetime as its boundary.  
EDIT: How about the boundary of a compact manifold?
 A: A manifold with boundary is locally like a half space (e.g. upper half plane if in dimension 2). As such, any topological properties will be inherited. Indeed, you can see that the boundary of a half space is a plane, which has empty boundary, and also is not compact.
A: No, consider $\mathbb R^n $ itself embedded in the " Standard Way" in $\mathbb R^{n+1}$ as $(x_1,x_2,..,x_n,0): x_i \in \mathbb R$. It is itself a manifold with boundary -- and every point is a boundary point -- but it is not compact, since it is not bounded -- you can even let a single coordinate grow as much as you want it to , to argue for unboundedness. 
A: In general no, but if you have the added assertion that you're manifold $M$ is compact, then it's boundary $\partial M$ is also compact.
Proof: If $M$ is a compact $n$-manifold with boundary $\partial M$. It is a well-known theorem that $\partial M$ is closed in $M$, therefore since any closed subset of a compact space is compact, it follows that $\partial M$ is compact. $\square$
As you can see, it is a well known theorem that for any manifold $M$, $\partial M$ is closed in $M$ equipped with the subspace topology.
