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Does $A'\cap B' = (A \cap B)'$, where X' denotes limit points of X? Prove or give counter example.

I.e. If $x$ is not an element of $A'\cap B'$ then $x$ is not an element of either $A'$ or of $B'$. This means that either $A\cap V$ or $B\cap W$ is a subset of $x$ ($V$, $W$ being open sets). So this means $(A\cap V)\cap(B\cap W)$ subset of $x$. This means $(A\cap B)\cap(V\cap W)$ is a subset of $x$. This means that $x$ is not an element of $(A\cap B)'$ since $V\cap W$ is an open set. Is this correct in showing that one set is a subset of the other?

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    Try finding a counterexample in $[0,1]$ with $A$ and $B$ disjoint. – 2017-02-12
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    A = (0,1/2), B = (1/2,1). So intersection of A' and B' is 1/2 but (A intersection B)' is null set? – 2017-02-12
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    Yes, that's a possible example. What are the limit points of the rationals and the irrationals in the reals? – 2017-02-12
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    Null set vs R?? – 2017-02-12
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    Also it seems that one is a subset of the other. I think I proved it by contrapositive ... is that the case? – 2017-02-12
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    I.e. If x is not an element of A'∩B' then x is not an element of either A' or of B'. This means that either A∩V or B∩W is a subset of x (V, W being open sets). So this means (A∩V)∩(B∩W) subset of x. This means (A∩B)∩(V∩W) is a subset of x. This means that x is not an element of (A∩B)' since V∩W is an open set. Is this correct? – 2017-02-12
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    no, the two sets are equivalent if and only if $(1)$ $x\in (A'\cap B')\rightarrow x\in (A \cap B)'$ and $(2)$ $x \in (A\cap B)' \rightarrow x \in A'\cap B'$. If this equivalence fails for any sets A, B, and you suspect the equivalence is not true, then you need to find a counterexample to show that the statement is **not always** true. – 2017-02-12
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    No I'm not suggesting the 2 sets are equivalent. I'm suggesting that one is a subset of the other – 2017-02-12

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In the space $\mathbb R$ let $A=\mathbb Q$ and $B=\mathbb R$ \ $A.$ Then $A'=B'=\mathbb R.$ But $A\cap B,$ and hence $(A\cap B)',$ are both empty.

In general a point $p$ may be a limit point of two disjoint sets $A,B,$ so $p\in A'\cap B'$ but $(A\cap B)'$ will be empty.

A simpler example, in $\mathbb R,$ would be intervals $A=(0,1)$ and $B=(1,2),$ with $p=1.$

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    Thanks. But going the other way is one a subset of the other? Is my proof above showing this correct? – 2017-02-13
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    You should not say "subset of $x$". What you should say is that if $x\not \in A'\cap B'$ then there are open sets $V,$ and $W$, each containing $x$, with $A\cap V=B\cap W=\phi.$ So $V\cap W$ is an open set containing $x$ and disjoint from $A\cap B$ (why?). so $x\not \in (A\cap B)'.$ – 2017-02-14
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    Thanks. Since W disjoint from B clearly W∩V is disjoint from B (and W∩V clearly contains x since V and W each do....and open since intersection of open sets. A disjoint from W∩V for same reason. And since A∩B is subset of A, it's disjoint from V∩W as well ? – 2017-02-14
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    Yes that's correct. – 2017-02-14