A set is open if its complement is closed? I am a little bit confused about the very fundamental property of sets. Is the statement "A set is open if its complement is closed" true? 
If both $A$ and $X$ are closed sets in $R^l$, and $A \subset X$. Then $A^c=X-A$, the complement of $A$, is neither open or closed in $X$. For example, $X=\{(x,y) \in R^2| x^2+y^2 \leq 2   \}$, and $A=\{(x,y) \in R^2| x^2+y^2 \leq 1   \}$. Then $X-A=\{(x,y) \in R^2| 1< x^2+y^2 \leq 2   \}$ is neither open or closed.
 A: "If both $A$ and $X$ are closed sets in $\mathbb R^l$, and $A\subset X$. Then $A^c=X \setminus A$ is neither open or closed in $X$."
No.  It is open in $X$.
"For example, $X=\{(x,y)\in \mathbb R^2 |x^2+y^2 \le 2\}$, and $A=\{(x,y)\in \mathbb R^2|x^2+y^2\le 1\}.$ Then $X\setminus A=\{(x,y)\in \mathbb R^2|1 < x^2+y^2\le 2\}$ is neither open or closed."
No.  It is not open in $\mathbb R^2$ (where $X \setminus A$ is NOT the complement of $A$).  But it IS open in $ X$ (where it is).
The points $(w,z)$ where $w^2 + z^2 = 2$ are all interior points of $X \setminus A$.  
Let $N(w,z) = \{(x,y) \in X|\sqrt{(x-w)^2 + (y-z)^2} < 1/2\}$ is an open neighborhod with respect to $X$.
And $N(w,z) \subset X \setminus A$.  So $(w,z)$ is an interior point $X \setminus A$.
There are no points in $X$ where $x^2 + y^2 > 2$.  So the fact that $X \setminus A$ has an "edge" at $x^2 + y^2 = 2$ does not make it not open.  It can not be thought of as an "edge" because there are no points "beyond" it. 
A: You ask

Is the statement "A set is open if its complement is closed" true? 

The answer is yes. In fact, given the definition of an open set, you can take as definition that a set is closed if the complement is open. That is, given a set $X$ with a topology a set $A$ is closed if $X-A$ is open.
If$A$ and $X$ are closed sets in $\mathbb{R}^l$ (for some $l$), then $A^{c} = \mathbb{R}^l - A$.
Now, if you put a topology on $X$ and you take a subset $Y$ of $X$, then this defines a topology on $Y$ by saying that the open sets in $Y$ are exactly those of the form $Y\cap A$ where $A$ is open in $X$. So your example 
$$
X-A=\{(x,y) \in R^2| 1< x^2+y^2 \leq 2   \}
$$
is indeed open in $X$ (but not in $\mathbb{R}^2$).
A: Yes of course the set $X\setminus A$ above is open in $X$.  I think your confusing that it's not open in $\mathbb{R}^2$. It's  because that you don't take the complement $\mathbb{R}^2\setminus A$ if you take the complement $\mathbb{R}^2\setminus A$. You will see that it will be open in $\mathbb{R}^2$.
A: Let M be the ambient space and d the metric on D, you can think of M and d as R, or R^2, or R^3,... with their usual distance functions. 
Let's recite our definitions:
A is open if and only if A is a subset of M such that for all $\alpha$ in A, there exists $\delta>0$ so that $N_\alpha(\delta)\subseteq A$. i.e. all of it's members are interior points, so that int (A)=A.
A is a closed subset of M if and only if A is a subset of M such that contains all it's boundary points ($Bd(A)\subseteq A$).
Here's the proof:
Let A$\subseteq M$ and $A^C$ be closed. Want to prove A is open using our definition of open above. We need to show for all $\alpha\in A$, there exists $\delta$>0 such that $N_\alpha(\delta)\subseteq A$:
Let $\alpha\in A$. We know $\alpha $ is not a boundary pointof A because if it were, it would also be a boundary point of $A^C$, which is a closed set, and contains all of it's boundary points which would include $\alpha$, which is impossible since $\alpha$ can not be in both A and it's complement simultaneously. Thus $\alpha$ is not a boundary point of A. Therefore, by the negation of the definition of a boundary point, there must exist $\delta>0$ so that no elements of $A^C$ are in $N_\alpha(\delta)$. Ask yourself this: if the elements of $N_\alpha(\delta)$ are not in $A^C$, where must they be? Right smack inside of A, of course. Thus for all $\alpha$ in A, there is $\delta$>0 so that $N_\alpha(\delta)\subseteq A$
Thus we've established A$\subseteq M$ and for all $\alpha$ in A, there is $\delta$>0 so that $N_\alpha(\delta)\subseteq A$. This means A is open. 
Q.E.D.
Adam V. Nease 
