Let $B \subseteq A \subseteq \mathbb{R}^n$. Show that $B$ is closed relative to $A$ iff $B = A \cap C$ for some set $C$ closed in $\mathbb{R}^n$. Let $B \subseteq A \subseteq \mathbb{R}^n$. Show that $B$ is closed relative to $A$ iff $B = A \cap C$ for some set $C$ closed in $\mathbb{R}^n$.
My solution:
We know $B$ is closed relative to $A$ if $\forall x \in A \setminus B$, $\exists$ a neighborhood $N$ such that $N \cap B = \varnothing$. So, for each $x \in B \setminus A$, we have a nghbd $N_x = D^n(x,r_x) \cap A$ such that $N_x \cap B = \varnothing$. Put $O = \bigcup_{x \in A \setminus B} D^n(x,r_x)$. Obviously $O$ is open since it is a union of open discs. So $C = \mathbb{R}^n \setminus O$ must be closed. Now, since $B$ is closed in $A$, then $A \setminus B$ must be open in $A$, therefore $A \setminus B = O \cap A$. I have here a question since I cannot see how to conclude from here that $B = C \cap A$. Can someone help see it?
the other direction: Say $B = C \cap A$ for some $C$ closed in $\mathbb{R}^n$. Therefore, every $x \in A \setminus B$ must satisfy $x \notin C$. Since $C$ is closed, $\exists$ a disc $D^n(x,r)$ such that $x \in D^n(x,r)$ and $D^n(x,r) \cap C = \varnothing$. Put $N = D^n(x,r) \cap A$. Then, $ x \in N$ and $N \cap B \subseteq N \cap C$. Therefore, $B$ is closed in $A$.
Is this proof correct? any feedback? thanks
 A: You’re doing fine up through the point at which you define $C$. At that point, however, you’re not entitled to conclude that $A\setminus B=O\cap A$; the fact that $A\setminus B$ is open in $A$ tells you that it’s equal to $U\cap A$ for some open $U$ in $\Bbb R^n$, but you’ve not actually shown that we can take this $U$ to be $O$.
It’s not hard to do so, however. Recall that $O=\bigcup_{x\in A\setminus B}N_x$, where $x\in N_x$ and $N_x\cap B=\varnothing$. On the one hand for each $x\in A\setminus B$ we have $x\in N_x\subseteq O$, so $A\setminus B\subseteq O$; and since $A\setminus B\subseteq A$, we actually have $A\setminus B\subseteq O\cap A$. On the other hand, for each $x\in A\setminus B$ we have $N_x\cap A\subseteq A\setminus B$, so that $$O\cap A=\left(\bigcup_{x\in A\setminus B}N_x\right)\cap A=\bigcup_{x\in A\setminus B}(N_x\cap A)\subseteq A\setminus B\;.$$ Thus, $O\cap A$ is indeed $A\setminus B$: your conclusion was premature, but none the less correct.
From here it’s just a bit of set algebra to see that $B=C\cap A$: 
$$C\cap A=(\Bbb R^n\setminus O)\cap A=A\setminus O=A\setminus(O\cap A)=A\setminus(A\setminus B)=B\;.$$
You should verify each of those steps, since that kind of manipulation can be quite useful. However, the actual idea involved here is much simpler: $O$ and $C$ are complementary subsets of $\Bbb R^n$, so for any $X\subseteq\Bbb R^n$, $O\cap X$ and $C\cap X$ must be complementary subsets of $X$. Every point of $\Bbb R^n$ is in exactly one of $O$ and $C$, so clearly every point of $X$ is in exactly one of $O\cap X$ and $C\cap X$. Since $O\cap A$ is $A\setminus B$, $C\cap A$ must be all the rest of $A$, which is simply $B$.
A: I think your second proof is correct.
As to the first, you only need to notice that, if $x\in(C\cap A),$ then $x$ cannot belong to $O$. That is, $x\in A\backslash A\cap O=A\backslash(A\backslash B)=B,$ hence $C\cap A\subset B.$ The other direction is similar: $B\subset A\backslash(A\backslash B)=A\backslash(A\cap O)=A\backslash O=A\cap C.$
Hope this helps.  
