Accumulation points of accumulation points of accumulation points Let $A'$ denote the set of accumulation points of $A$.
Find a subset $A$ of $\Bbb R^2$ such that $A, A', A'', A'''$ are all distinct.
I can find a set $A$ such that $A$ and $A'$ are distinct, but not one where $A,A',A'',A'''$ are all distinct.
 A: Here’s a schematic of what you need to get distinct $A,A'$, and $A''$; it generalizes easily.
$$\begin{array}{rr}
\bullet&\bullet&\bullet&\bullet&\bullet&\bullet&\longrightarrow&\bullet\\
\uparrow&\uparrow&\uparrow&\uparrow&\uparrow&\uparrow\\
\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet\\
\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet\\
\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet\\
\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet\\
\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet
\end{array}$$
A: What might be confusing here is that you won't, of course, get that $\overline{A} \neq \overline{\overline{A}}$ (where $\overline{A}$ denotes the closure of $A$). Thus, the accumulation points of $A'$ will also be accumulation points of $A$, but $A'$ may have fewer accumulation points than $A$ does.
In other words, you'll have that $A' \supset A'' \supset A''' \ldots$. This means that have to start with a set with "enough" accumulation points, otherwise you'll hit the empty set too early for $A',A'',A'''$ to be all different. The easiest way to avoid that is to work backwards, i.e. start with $A'''$, as Asaf suggested.
Let $$
  A''' = \{(0,0)\} \text{.}
$$ Then construct some $A'' \supset A'''$ such that $A''$ has accumulation point $(0,0)$, say $$
  A'' = \{(\frac{1}{n},0) \,:\, n \in \mathbb{N}\} \cup \{(0,0)\} \text{.}
$$
Now continue this idea to find $A'$. $A'$ will have to contain sequences which converge to $(\frac{1}{n},0))$ for every $n \in \mathbb{N}$. Finally, from $A'$, determine $A$. 
A: Hint: Work backwards. Start with some set, then add a sequence which converges to each point, then add a sequence which converges to each of the points in that set, and so on.
Do this carefully and you can make sure that $A$ is such set.
If you are familiar with the concept of ordinals, then you can also consider the ordinal $\omega^4$ embedded into the real numbers.
