Question about adherence of sets please 
I have that $A$ is compact and i want to prove that for a set $U$ we have $A=
\overline{A\setminus U}\cap \overline {U}$
I do this:
$x\in \overline{A\setminus U}\cap \overline {U}\Longleftrightarrow \begin{cases}\forall V\in \mathcal{V}_x, V\cap(A\setminus U)\neq \emptyset\\\text{and}\\ \forall V\in \mathcal{V}_x, V\cap U\neq \emptyset\end{cases}\Longleftrightarrow x\in \overline{A}=A $
But i don't know how to prove that $A=\overline{A}\subset\overline{A\setminus U}\cap \overline {U}$ 
Can we prove directly that $\overline{A}= \overline{A\setminus U}\cap \overline {U}$
Please help me 
Thank you.
 A: As noted in other answers even for examples on $\mathbb{R}$ with the normal topology this isn't true. However maybe a more detailed description of what the points in $X=\overline{A\setminus U}\cap \overline {U}$ look like will make it clearer why this isn't equal to $A$ (at least without conditions on $A$ or $U$).
If we have any point $x$ in the set $X=\overline{A\setminus U}\cap \overline {U}$ then you have already noted that we can concluded $x$ is in $\overline{A}$ (and then if our topological space is Hausdorff then it is also in $A$) but we can actually say much more about the point $x$.
While it may sound obvious to note, we know that $x$ is in $\overline{U}$. Therefore we must have that $x$ is either in the interior or the boundary of $U$.
We also can see that $x$ is in the closure of the complement of $U$ (specifically it is in the closure of its complement in $A$ but adding in extra stuff doesn't hurt here). Then $x$ clearly couldn't be in the interior of $U$ (otherwise it couldn't be in the closure of its complement) so it is a point on the boundary of $U$.
Therefore we can see that unless all the points of $A$ are also boundary points of $U$ it isn't possible to have that $A =X$.
A: What do you know about $U$? It is not valid for every $U$; certainly, $A:=[0,2]\subseteq\mathbb{R}$ is not equal to $\overline{[0,2]\setminus [0,1]}\cap \overline{[0,1]}=\{1\}$.
Note that if $A$ is a compact subset of a Hausdorf space $X$, then $A$ is closed in $X$ (not sure if this is your question).
A: Take $A = [0, 3] \subset \mathbb{R}, U = [1, 2].$ Then $A \setminus U = [0, 1) \cup (2, 3] \Rightarrow \overline{A \setminus U} = [0, 1] \cup [2, 3].$ So..?
