I have difficulties with a rather trivial topological question:
A is a discrete subset of $\mathbb{C}$ (complex numbers) and B a compact subset of $\mathbb{C}$. Why is $A \cap B$ finite? I can see that it's true if $A \cap B$ is compact, i.e. closed and bounded, but is it obvious that $A \cap B$ is closed?
Best Answer
There seems to be a contradiction between the answers of Andre Nicolas and of Arthur Fischer, yet both are correct. This depends on your definition of discrete. Andre's notion of discrete is this:
A set $S$ in a topological space $X$ is discrete if it is discrete with respect to the subspace topology. This is the same as saying that every point in $S$ has a neighborhood (in $X$) that contains no other point from $S$.
Arthur's definition is this: $S\subseteq X$ is discrete if every point in $X$ has a neighborhood that contains at most one element of $S$ (I assume all spaces to be Hausdorff).
A discrete set according to Arthur's definition is automatically closed (as a point in the closure but not in the set would violate discreteness). So most likely, if you are supposed to show that the intersection of a discrete set with a compact set is finite, you are supposed to use Arthur's definition of discreteness. By the way, note that Arthur's argument does not say anything about $\mathbb C$. This works in every Hausdorff space and mainly uses the fact that compact discrete sets are finite. (And that subsets of discrete sets are again discrete and that discrete sets are closed.)
Edit: Unfortunately Arthur Fischer's answer was deleted while I was typing mine. But it seems that my answer is still understandable.