Closure of the difference of two sets vs difference of their closures

Question

This exercise is from Chapter 2, Section 17, number 8 (c), pag. 101, from Munkres's Topology.

Let $A$, $B$ denote subsets of a space $X$. Determine whether the following equation holds: $$\overline {A - B} = \overline A - \overline B.$$ If it fails, determine whether one of the inclusions $\supset$ or $\subset$ holds.

This exercise was very difficult for me. I tried the following way.

Let $A^\prime$ be the set of all the limit points of $A$ that not belong to $A$; that is, $A \bigcap A^\prime = \varnothing$. Let $B^\prime$ be the set of all the limit points of $B$ that not belong to $B$; that is, $B \bigcap B^\prime = \varnothing$. By last, let $C$ be the set of all the limit points of $A - B$ that not belong to $A - B$; then, $(A - B) \bigcap C = \varnothing$.

We proceed by contradiction. We have that $\overline A - \overline B = (A \bigcup A^\prime) - (B \bigcup B^\prime)$. Assume that there exists a point $x$ such that $x \in (\overline A - \overline B)$, but $x \not\in \overline {A - B}$. Then, the point $x$ does not belong neither to $B$ nor to $B^\prime$, because in such case $x$ would not belong to $\overline A - \overline B$. Besides, it can not belong to $A$ either, because in this case it would also belong to $\overline {A - B}$. Assume that $x \in A^\prime$; in such case, $x$ is one of the limit points of $A$ that not belong to $A$. Now, $x \not\in C$, because in this case it would also be in $\overline {A - B}$, which contradicts our initial hypothesis. But, to explain that $x \not\in C$, it is necessary to assume that $B$ extract from $A$ a subset which does not contain $x$, but it has $x$ as limit point. In such case, $x$ would be a limit point of $B$ as well, and this implies $x \in B$ or $x \in B^\prime$, contradicting our prior assertion. Therefore, it is not possible that $x \in (\overline A - \overline B)$ and $x \not\in \overline {A - B}$. It follows that $\overline {A - B} \supset \overline A - \overline B$.

I'm not sure of this proof. Someone can tell me if there is another way of solving it? Thanks.

Example of non-equality.

Let $(0, 2)$ and $(1, 3]$ be intervals in $\mathbb{R}$. Then, the difference between the intervals is $(0, 1]$, and the closure of the difference is $[0, 1]$. On the other hand, the closures of the intervals $(0, 2)$ and $(1, 3]$ are $[0, 2]$ and $[1, 3]$, respectively. Consequently, the difference between closures is $[0, 1) \ne [0, 1]$.

Answer 1

To see that $\overline{A} - \overline{B} \subset \overline{A-B}$: let $x \in \overline{A} - \overline{B}$, and let $O$ be any open subset that contains $x$. As $x \notin \overline{B}$, we have that $X - \overline{B}$ is also an open set that contains $x$, so $O \cap (X - \overline{B})$ is also an open neighbourhood of $x$ so it intersects $A$, as $x \in \overline{A}$. In particular, every neighbourhood $O$ of $x$ intersects $A - \overline{B} \subset A - B$. So $x \in \overline{A-B}$.

Answer 2

Once one knows which inclusion is valid, in metric spaces the proof can be made much shorter than the one suggested above. Let $x$ in $\overline A \setminus\overline B$, then $x$ is in $\overline A$ hence there exists a sequence $(x_n)$ in $A$ such that $x_n\to x$. But $x$ is not in $\overline B$ hence $x_n$ is in $B$ at most finitely many times. Thus, $x_n$ is in $A\setminus B$ infinitely many times, which proves that an infinite subsequence of $(x_n)$ included in $A\setminus B$ converges to $x$, and in particular, $x$ is in $\overline{A\setminus B}$.

This proves the inclusion $\overline A\setminus\overline B\subseteq\overline{A\setminus B}$. A counterexample to the inclusion $\overline{A\setminus B}\subseteq\overline A\setminus\overline B$ is easy to come by. Hint: try $A$ and $B$ subsets of the real line and $B$ a singleton.

Answer 3

Take an example $[1,5]-(1,5)=\{1,5\}$ it’s closure $\{1,5\}$ but if you take the closure first $[1,5]-[1,5]=\phi$

Answer 4

We have that $A\setminus B=A\cap B^c$. This reduces to exhibiting $A,B$ for which $\overline{A\cap B}\subsetneq \overline A\cap \overline B$. We always have the inclusion holds, that is $\overline{A\cap B}\subseteq \overline A\cap\overline B$, since $\overline A,\overline B$ are closed subsets that contain $A\cap B$.