Past exam of the options II course of the University of Oxford / 2012 / E / E3 / b / ii

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. ThusThe infinitely-many-points observation ensures that one can avoid the centre of the disk itself in both cases.

TakeIts only accumulation point is $(0,0)$, which is absent from $A_2$. Every listed point is isolated, so none belongs to $A_2^{\prime }$. ConsequentlyFor completeness, away from the origin a sufficiently small disk meets only finitely many of the listed points and can be shrunk to exclude all but its centre, proving there are no further accumulation points.

Use three nested sequences:For fixed $n,m$, letting $k\to \infty$ produces the point $(1/n+1/(n^{2}m),0)$. Letting $m\to \infty$ produces $(1/n,0)$, and letting $n\to \infty$ produces $(0,0)$. Accordingly, putThe successive derived sets areTo see that no limits are missing, take any convergent sequence of distinct points from $A_3$. If $n$ is unbounded, a subsequence has $n\to \infty$ and converges to the origin. Otherwise fix $n$ on a subsequence. Unbounded $m$ then forces limit $(1/n,0)$; with $n,m$ fixed, distinctness forces unbounded $k$, giving the corresponding point of $B$. The same argument, with one index fewer at each stage, proves the next two equalities. Distinct $n$-clusters are separated away from the origin because their first coordinates lie in the disjoint intervals $[1/n,1/n+1/n^2]$.

The original set has positive second coordinates, the first derived set has isolated $B$-points on the axis, the second has infinitely many $C$-points, and the third is a singleton. All three required successive inequalities follow. This is a three-stage accumulation tower.