Limit Points of the Empty (Null) Set I am trying to understand whether or not the statement $X′ ∩ Y′ ⊆ (X ∩ Y)′$  is true or false.
In trying to develop a counterexample I started asking myself what sets might make it false.  That led to the following question.  
If $(A ∩ B)$ = Empty Set, what is $(A ∩ B)′$ ?
 A: Let $X$ be a topological space and $S= \varnothing$. A $x\in X$ is a limit point of $S$ if for every $\epsilon >0$ it holds:
$$\big(B(x,\epsilon)\setminus \{x\}\big) \cap S \neq \varnothing.$$
But $S = \varnothing,$ thus the above intersection will always be the empty set for any $x\in X$. Hence, there are no $x \in X$ that are limit points of $S$.
A: A counter example to make the statement untrue would be X = the irrationals and Y= the rationals.
Then X' = Y' = the reals and $X' \cap Y' = \mathbb R $.  But $X \cap Y =\emptyset $.
So what is $\emptyset'$?  Well, every neighborhood of a limit point of the empty set must have a point of the empty set.  But the are no such points to be had so no limit points can exist.
.... which is why we say $\emptyset $ is closed.  Which is to say, $\emptyset' \subseteq \emptyset $.
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To find a counter example we need to find a point that is a limit point of both X and Y but is not a limit point of X $\cap $ Y.  In other words, every neighbor hood has  points in both X and points in Y but there is a neighborhood with no points in both... meaning X and Y are disjoint within that neighborhood.
Here's a another less vague counter example: let $X=\{(\frac {1}{2n+1},\frac {1}{2n})|n \in \mathbb N\} $ and $Y=\{(\frac 1 {2n},\frac 1 {2n-1})|n\in \mathbb N\} $. 
So 0 specifically is a limit point of both X and Y.  But no neighborhood of 0 contains any points of the intersection of X and Y, because there are no points in the intersection. So 0 is not a limit point of the intersection.
