Is this right? Topology with closures I want to show that (possibly) $$cl(A-B) = cl(A) - cl(B).$$
I know that $$cl(A-B) \subset cl(A) - cl(B).$$ already, but for the other inclusion I tried this.
Let $x \in  cl(A) - cl(B)$, so that for any neighborhood $U$of $x$, we know $U \cap (A - B) =\emptyset$
However, we also get the inclusion$$U \cap (A - B)  \subset U \cap cl(A-B) \subset U \cap cl(A) - cl(B).$$
Or did I get the other way around? Because I found the counterexample $A = [0,1)$ and $B = (-1,0)$.
So that $cl(A - B) = cl(\emptyset) = \emptyset$ and as well as $$cl(A) - cl(B) = (0,1)$$
So I think I got tihs backwards
Update: As pointed out, equality does'nt really hold and my inclusion argument is incorrect as I have incorrectly applied a techinuqe to the wrong problem. Thus, mynew quesiton is, is there ever inclusion (proper)?
 A: Consider $[0,1],\{1\} \subseteq \Bbb R$, where $\Bbb R$ carries the standard topology. Then $cl([0,1] - \{1\}) = [0,1] \neq [0,1) = cl([0,1]) - cl(\{1\})$,
so the result doesn't hold true.
edit: Oh, I somehow missed  doppz's comment. Please let me know whether you want me to delete this answer.
edit: Yes, there is inclusion. In fact: $cl(A)-cl(B) \subseteq cl(A-B)$.
proof:
If $x \not \in cl(A-B)$, there is an open neighborhood $x \in U$, such that $U \cap (A-B) = \emptyset$. On the other hand, if $x \not \in cl(B)$, there is an open neighborhood $x \in V$, such that $V \cap B = \emptyset$. Now observe that 
$$U \cap (A-B) = (U \cap A)-(U \cap B)$$, so that $U \cap A \subseteq U \cap B$. This gives
$$V \cap U \cap A  \subseteq V \cap U \cap B \subseteq V \cap B = \emptyset$$, so $x \not \in cl(A)$.
Let's see what we proved:
If $x\not \in cl(A-B)$ and $x \not \in cl(B)$, then $x \not \in cl(A)$. So if $x \in cl(A)\setminus cl(B)$, then we must have that $x \in cl(A-B)$, i.e. $cl(A)-cl(B) \subseteq cl(A)-cl(B)$.
Note that this also shows $cl(cl(A)-cl(B)) \subseteq cl(A-B)$.
