 Three-stage accumulation tower

A subset of the plane can have three successive strict changes under taking its set of accumulation points. For example letPut $B=\{(1/n+1/(n^{2}m),0):n,m\ge 1\}$ and $C=\{(1/n,0):n\ge 1\}$. Then $A^{\prime }=B\cup C\cup \{0\}$, $A^{\prime \prime }=C\cup \{0\}$ and $A^{\prime \prime \prime }=\{0\}$, where $0$ denotes the origin. For fixed $n,m$, increasing $k$ gives a point of $B$; increasing $m$ gives a point of $C$; increasing $n$ gives the origin. These are all possible limits: in any convergent sequence of distinct points, either $n$ is unbounded, or a subsequence fixes $n$ and has unbounded $m$, or a subsequence fixes both and has unbounded $k$. The first-coordinate cluster intervals $[1/n,1/n+1/n^2]$ are disjoint and separated away from the origin. Repeating this classification after each differentiation proves the displayed derived sets. Each derivative removes one layer of isolated points, and all three displayed changes are strict.