if $A=B \cup C$ and $A,B$ are open and $C$ is closed then $C \subset B$? 
if $A=B \cup C$ and $A,B$ are open and $C$ is closed then $C \subset B$?

I thought about this statement for another question that I had.

Suppose $A,B,C$ are subsets of $\mathbb{R}^n $ such that $A=B\cup C$ and $A,B$ are open and aren't closed, $C$ is closed and isn't open, does that necessaraly mean that $C\subset B$?

thanks/
 A: First of all, if you use the usual topology on $\mathbb R^n$, then the space is connected so the only sets that are both open and closed at the same time are $\mathbb R^n$ and $\emptyset$.
For a simple counterexample, consider the intervals $A = (-2, 2), B=(-2, -1) \cup (1, 2)$ and $C = [-1, 1]$.
A: Let $D$ denote all of the boundary points of $C$.  At first glance, I guessed that the conjecture is correct, because $C$, which is closed must contain all of $D$.  Therefore $B\cup C$ must contain all of $D$.  However, since $B\cup C$ is specified as open, each point in $D$ can not be a boundary point in $B\cup C$.  This follows, because $B\cup C$, which is open, can not contain any of its boundary points.
This implies that each boundary point in $D$ must be contained in some open neighborhood of $B\cup C$.  This superficially (and wrongly) suggested that all of $C$ must be contained in some open neighborhood of $B$.
There are two things wrong with this suggested conclusion:

*

*Even if it were true that every point $D$ is contained in some open neighborhood around $B$, that does not imply that each
point in $C$ is contained in some open neighborhood around $B$.
A case in point is $B = (-2,-1) \cup (1,2)$ and $C = [(-3/2),(3/2)].$


*As has already been demonstrated by the counter example given in Numbra's response, it is feasible that every point in $D$ will also be a boundary point of $B$.  In such a case, since $B$ is open, it won't contain any point in $D$, and therefore can not be a superset to $C$.
In fact, a less extreme counter-example would be if exactly one point $d \in D$ is also a boundary point of $B$.  This (by itself) implies that $d \in C$ and $d \not\in B$, which (again) implies that $B$ is not a superset of $C$.
