$\overline A - \overline B \subset \overline{A - B}$ Can someone give me a hint/explain how to show this inclusion?  $A,B$ are subsets of a topological space.  If $x \in A$, showing that $x \in  \overline{A - B}$ is obvious, but I'm not sure how to show that if $x$ is in the boundary of $A$, then $x \in \overline{A - B}$
By $\overline A$, I mean closure of $A$.
 A: Let $x\in\overline A - \overline B$. You have to show $x \in  \overline{A - B}$.
For that let $U$ be an arbitrary neighborhood of $x$. Now you have to show $U\cap (A-B)\neq \emptyset$.
Since $x\notin\overline B$ the set $X- B$ is a neighbourhood of $x$ (where $X$ is the topological space).
Then $U\cap(X- B)$ is a neighbourhood of $x$, too.
Since $x\in\overline A$ you have $A\cap (U\cap(X-B))\neq\emptyset$. But $A\cap (U\cap(X- B))=U\cap (A-B)$.
A: The closure of a set $X$ is equal to the union of $X$ and its boundary $\partial X$ (those points $b$ in the topological space such that every open set containing $b$ contains at least one point in $X$ and one not in $X$).  Let $x \in \overline{A} - \overline{B}$.  There are two cases to consider:


*

*$x \in A$ (and $x \notin \overline{B}$)

*$x \notin A$ and $x \in \partial A$ (and $x \notin \overline{B}$)


You mentioned that case 1 is obvious, so I'll explain case 2.  To show that $x \in \overline{A - B}$, assume that $x \notin A - B$ (since then it would also be in $\overline{A-B}$).  We'll show that this implies $x \in\partial(A - B)$.  To that end, let $U$ be any open set containing $x$.  We must show that $U \cap (A-B) \neq \emptyset$.  Since $x \notin \overline{B}$, there is an open set $V$ containing $x$ such that $V \cap B = \emptyset$.  Now $V \cap U \neq \emptyset$ is an open set containing $x$, so using the fact that $x \in \partial A$, there exists $y \in (V\cap U) \cap A$.  This puts $y \in U \cap (A-B)$.  Thus, by definition, $x \in \overline{A - B}$, as required.
Hope this helps!
A: Let's show that
$$
  \overline{A} - \overline{B}
  \subset
  \overline{A - \overline{B}}.
$$
This implies the desired inclusion because
$\overline{A - \overline{B}} \subset \overline{A - B}$.
(We are just saying that it is sufficient to prove for
$B$ closed!)
Let $G = \left(\overline{B}\right)^c$.
Then, we want to show that
$$
  \overline{A} \cap G
  \subset
  \overline{A \cap G}.
$$
But this is clear because for $x \in \overline{A} \cap G$,
if $V$ is a neighborhood of $x$, then $V \cap G$ is also a
neighborhood of $x$, and therefore intersects $A$,
since $x \in \overline{A}$.
That is, $V \cap G \cap A \neq \emptyset$.
So, every neighborhood of $x$ intersects $A \cap G$,
and therefore, $x \in \overline{A \cap G}$.
A: Suppose that $x\in\operatorname{cl}A\setminus\operatorname{cl}B$; then $x\notin\operatorname{cl}B$, so $x$ has an open neighborhood $U$ such that $U\cap B=\varnothing$. Suppose, to get a contradiction, that $x\notin\operatorname{cl}(A\setminus B)$; then $x$ also has an open nbhd $V$ such that $V\cap(A\setminus B)=\varnothing$. Let $W=U\cap V$; then $W\cap\Big(B\cup(A\setminus B)\Big)=\varnothing$. But $B\cup(A\setminus B)=A\cup B$, so $W\cap(A\cup B)=\varnothing$. In particular, $W\cap A=\varnothing$, contradicting the hypothesis that $x\in\operatorname{cl}A$. Thus, $x\in\operatorname{cl}(A\setminus B)$, and therefore $\operatorname{cl}A\setminus\operatorname{cl}B\subseteq\operatorname{cl}(A\setminus B)$.
A: Another possible proof using characterization of closure with sequences.
Let $l\in \overline{A} - \overline{B}$. As $l\notin\overline{B}$, there exist an open set $U$ such that $l\in U$ and $U\cap B = \emptyset$. As $l\in\overline{A}$, there exist a convergent sequence $(x_n)$ of $A$ that tends to $l$. Let $N\in\mathbb{N}$ such that for all $n\geq N$, we have $x_n\in U$. We define a new sequence $(y_n)$ by $y_n = x_{N+n}$ for all $n\geq 0$. Clearly, $(y_n)$ is a convergent sequence of $A-B$ that tends to $l$ so $l\in\overline{A-B}$.
