Difference of Limit Points a subset of the Limit Point of the Difference

Question

I'm trying to figure out if the following is true or not in any topological space involving limit points. Do there exist sets $A$ and $B$ such that $A'-B'\subseteq (A-B)'$? So far I can't find any counterexamples to show its false, nor a proof that it is valid either and I've spent days trying. Help please!

Answer 1

Suppose $x \in A'\setminus B'$. So $x \in A"$ and $x \notin B'$ and the latter means by definition that there is some open neighbourhood $O_x$ of $x$, such that $O_x \cap B \subseteq \{x\}$.

Now let $O$ be any open neighbourhood of $x$. Then $U = O \cap O_x$ is also an open neighbourhood of $x$, and so we have some point $a \in A \cap U$, with $a \neq x$. As $U \cap B \subseteq O_x \cap B =\{x\}$ we see that $a \notin B$ (if $a \in B$ the previous inclusions would force it to be equal to $x$), and so $a \in A \setminus B$. So every open neighbourhood of $x$ contains some point $a \in A\setminus B$ with $a \neq x$. This means that $x \in (A\setminus B)'$.

Hence $A'\setminus B' \subseteq (A\setminus B)'$ for all $A,B \subseteq X$.

Answer 2

Let $x\in A'-B'$. Then $x\in A'$, so $x\in \overline{A-\{x\}}$ and $x\notin \overline{B-\{x\}}$, as Wiliam Elliot says. Thus $x\in \overline{A-\{x\}}-\overline{B-\{x\}}$. However, $\bar A-\bar B\subseteq \overline{A-B}$ (see this for a proof) so $x\in \overline{(A-\{x\})-(B-\{x\})}=\overline{(A-B)-\{x\}}$. Hence, $x\in (A-B)'$.