Prove that $G - F$ is open and $F - G$ is closed given $F$ closed and $G$ open in $M$. 


proof:
Since  $G$ is open, $\exists B_{r}(x) \subset G$ for any $x$ and for any $r > 0$, $B_{r}(x) \cap F = \emptyset.$ So $B_{r}(x) = B_{r}(x) - F \subset G - F.$ (I am not sure how to justify $B_{r}(x) - F \subset G - F$…the venn diagram provided the intuition, is it necessary to justify this ?)
Since $F$ is closed, $\forall x \in F$, $B_{\epsilon }(x) \cap F = \emptyset$ and $B_{\epsilon}(x) \not\subset G$ for any $\epsilon >0.$ So $B_{\epsilon}(x) \cap (F - G) =B_{\epsilon}(x) \cap F - B_{\epsilon}(x) \cap G = B_{\epsilon}(x) \cap F \neq \emptyset.$
I am sure the last part proving $F-G$ is right.
 A: You have to be more careful with your quantifiers.

Statement: $G-F$ is open.
We want to prove that, for every $x\in G-F$, there exists $r>0$ such that $B_r(x)\subset G-F$.
So, let $x\in G-F$. In particular $x\in G$ and $x\notin F$.
Since $G$ is open, there is $s>0$ such that $B_s(x)\subset G$.
Since $F$ is closed, there is $t>0$ such that $B_t(x)\cap F=\emptyset$.
Now, take $r=\min\{s,t\}$. Then $B_r(x)\subset B_s(x)\subset G$ and $B_r(x)\subset B_t(x)$ so $B_r(x)\cap F=\emptyset$. Thus $B_{r}(x)\subset G-F$.

Statement: $F-G$ is closed.
We want to prove that, for every $x\notin F-G$, there exists $r>0$ such that $B_r(x)\cap (F-G)=\emptyset$.
So, let $x\notin F-G$.
If $x\notin F$, take $r>0$ such that $B_r(x)\cap F=\emptyset$ (it exists since $F$ is closed); then, easily, $B_r(x)\cap (F-G)=\emptyset$.
If $x\in F$, then we conclude that $x\in G$, because otherwise $x\in F-G$.
Thus there is $r>0$ such that $B_r(x)\subset G$ (because $G$ is open). Now it's easy to see that $B_r(x)\cap (F-G)=\emptyset$.
A: As you saw in the comments an easier way to prove this is noting that
$$
A \setminus B = A \cap B^c
$$
so taking $G$ open and $F$ closed we have $F^c$ is open and thus $G \cap F^c$ is open, then noting that $F \setminus G = (G \setminus F)^c$ we get that $F \setminus G$ is closed.
However for your argument there are some issues:


*

*"$\exists B_r(x) \subset G$ for any $x$" isn't true, consider any $x \not\in G$. Of course one may infer that you meant any $x \in G$, but this isn't very precise

*The above issue shows up again when you say "for any $r > 0, B_r(x) \cap F = \emptyset$," for which $x$ are you talking about here? The $x$ that you we inferred you were talking about above? This is not necessarily true (for example if $G \cap F \neq \emptyset$ take any $x \in G \cap F$)

*Now it isn't simply good enough to consider venn diagrams to prove this statement ($B_r(x) - F \subset G - F$, for instance how do you map an infinite dimensional metric space to a venn diagram?)


You have similar issues in your second proof, I think the issue doesn't lie in your intuition (which it appears you see what's going on) but you need to be more precise with which your quantifiers and the such (e.g. $\forall x \in G \setminus F$). If you want help starting off the proof I would gladly put a few sentences on here to help you out - just comment below. 

Spoiler:

 Let $x \in G-F$ then $x \not\in F$ so $\exists \, r > 0 : B_r(x) \cap F = \emptyset$ and since $x \in G$ we have $\exists r' > 0 : B_{r'}(x) \subset G$. Now put $R = \min\{r, r'\}$ then $B_R(x) \subset B_r(x), B_R(x) \subset B_{r'}(x)$ thus $B_R(x) \cap F = \emptyset, B_R(x) \subset G$ so that $\forall y \in B_R(x), y \in G, y \not\in F$ thus $B_R(x) \subset G-F$.

Of course we can do a similar process for showing $F-G$ is closed or consider the fact that $(G-F)^c = F-G$.
