= Solution
Take $A_1=\mathbb Q^2$. Every <open disk> contains infinitely many points with both coordinates rational, so every point of the plane is an <accumulation point> of $A_1$. Every disk also contains infinitely many points with an irrational first coordinate, and these lie in the <complement>. Thus
$$
\boxed{A_1'=\mathbb R^2=(\mathbb R^2\setminus A_1)'.}
$$
The infinitely-many-points observation ensures that one can avoid the <centre> of the disk itself in both cases.
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