A closed set $F \subset \mathbb{R}$ such that $F, F', F'', F''',\dots $ are all distinct Let $F \subset \mathbb{R}$ be a closed set. Let $F'$ be the set of the limit points of $F$.
Question: Does there exists a set $F$ such that $F, F', F'', F''', ..... $ are all distinct and nonempty?
I am having hard time trying to find such a set. I was thinking maybe the Cantor set, but I am unable to prove that Cantor set has such property.
 A: This really is an addendum to Mirko Swirko's answer. I just thought a picture would be nice.
What I believe Mirko is describing is embedding the ordinal $\omega^{n+1}+1$ in the interval $[n, n+1)$ for $n=0, 1,\ldots$ with the 'top' element of the ordinal at the left-hand edge.
For example, the collection $\{\frac{1}{n}\}_{n=2}^\infty$ represents the ordinal $\omega+1$ with the top element at $0$. The next ordinal $\omega^2+1$ that would be in the interval $[1, 2)$ is represented here:
$\hskip1in$ 
The dashed line at $2$ would be included in the ordinal $\omega^3+1$ in the interval $[2,3)$. We perform this pattern in each interval $[n,n+1)$. Each of these sets are closed since they contain their limit points. Also, although it is isn't true, in general, that the union of a countable collection of closed sets is closed, when the collection of closed sets is locally finite, as it is in this case, the union is closed as well.
Now the point of this construction is that when we move from $F$ to $F'$, then $F'$ is precisely the same as $F$ just shifted over the right one unit. And, in general, $F^{(n)}$ is the same as $F$ shifted to the right $n$ units. Thus we have a closed set $F$ such that $F^{(n)}$ is distinct from $F^{(m)}$ when $n\neq m$. (Actually, to make it exactly the same by shifting, we need to add an isolated point at $-1$ to the original $F$ I described, which causes no problem)
