Intersection of compact and discrete subsets 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?
 A: 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.  
A: The result is not correct, for the set $A$ of all $\frac{1}{n}$, where $n$ ranges over the positive integers, is discrete. Let $B$ be the unit disk.
As you observed, everything would be fine if $A\cap B$ were closed. But in this case it isn't.
A: If $A \cap B$ were infinite then by the compactness of $B$ there would be a limit point $z$.  Next note that every neighbourhood of $z$ would have infinite intersection with $A \cap B$ and hence with $A$, contradicting that $A$ was discrete.
Edit: I quickly deleted this after I started to think I was using the wrong notion of discreteness (closed discreteness, as mentioned in Stefan's answer below, which is stronger than discrete as a subspace).  In topology there is a notion of discrete family of sets for which the usual definition is the following: A family $\mathcal{A} = \{ A_i : i \in I \}$ of subsets of a topological space $X$ is called discrete if each point $x \in X$ has a neighbourhood which meets exactly one member of $\mathcal{A}$.  Seeing what I was supposed to show, instead of taking discrete to mean discrete as a subspace, I essentially took discrete to mean that the family of singletons from $A$ is a discrete family of sets.  It's now been undeleted to make Stefan's answer more intelligible, so be kind with your voting.
A: I'm assuming that by "discrete set" you mean that $A$ has no accumulation points--that is (since we're in a Hausdorff space), that every point in the plane has a neighborhood containing at most one point of $A$. Thus, $A$ is vacuously closed (as it contains all of its accumulation points), so since $B$ is closed, then so is $A\cap B$.
Compactness isn't enough for finiteness all by itself (consider the unit disk), so let's proceed a little further. For each point $z\in A$, we have by definition of discrete set that there is a neighborhood $U$ of $z$ such that $A\cap U=\{z\}$. To make this explicit, there is some least positive integer $n$ such that $A\cap\{w\in\Bbb C:|w-z|<\frac1n\}=\{z\}$, and given this $n$ (which depends on $z$), we'll let $U_z:=\{w\in\Bbb C:|w-z|<\frac1n\}$.
Now, the set $\mathcal{U}=\{U_z:z\in A\}$ is an open cover of $A$, and so an open cover of $A\cap B$, which we've already determined to be compact. Thus $\mathcal{U}$ has a finite subcover of $A\cap B$, say $\mathcal{V}=\{V_1,...,V_n\}$. We know by our construction of $\mathcal{U}$ that each open set $V_k\in\mathcal{V}\subseteq\mathcal{U}$ has the property that $A\cap V_k$ contains exactly one point, so $A\cap B\cap V_k$ contains at most one point. Hence, $$(A\cap B)\cap\bigcup_{k=1}^nV_k=\bigcup_{k=1}^n(A\cap B\cap V_k)$$ contains at most $n$ points. But $\mathcal{V}=\{V_1,...,V_n\}$ covers $A\cap B$, so $A\cap B\subseteq\bigcup_{k=1}^nV_k$, so $A\cap B=(A\cap B)\cap\bigcup_{k=1}^nV_k$. Thus, $A\cap B$ has at most $n$ points, so is finite.
