Find a set where removing its isolated points creates more isolated points, then removing those isolated points creates more isolated points and so on For a set $E_1 \subset \mathbb R$, denote the set of all its isolated points by $E_1^*$.
Let $E_2 = E_1\setminus E_1^*$. Similarly, $E_3 = E_2\setminus E_2^*$, and so on. Give an example of such a set $E_1$ that meets the two conditions for $n=1,2,3,\dots$ :


*

*None of the sets $E_n$ are empty 

*None of the sets $E_n^*$ are empty
 A: I can get sets $E_{n},E_{n}^{*}$ for $1\leq n\leq k.$ Let $X_{1}=\{0\}\cup\{1/n\}_{n\geq 2}.$ Now let $X_{2}=\{0\}\cup\bigcup_{n\geq2}\{1/n\}\cup\{1/n+(1/(n-1)-1/n)/j\}_{j\geq 2},$ and proceed in this way for $k$ steps, i.e., between any two consecutive points that were added in $X_{n-1},$ $x<y,$ add the sequence of points $\{x+(y-x)/j\}_{j\geq 2}.$ Let $E_{1}=X_{k-1}.$ Then at the first step, the isolated points are just those that were added in the last step, since every point $x$ that was added in $X_{n},n<k-1$ has a sequence of points in $E_{1}$ tending to $x$ (added at step $n+1$). So $E_{2}=X_{k-2},$ and this continues until $E_{k-1}=X_{1}.$ We may repeat one more time to get $E_{k}=\{0\},$ but repeating further clearly yields the empty set.
Unfortunately, this isn't the example you want, since it may only be repeated finitely many times. $X_{\infty}=\bigcup_{n\geq 1}X_{n}$ doesn't have isolated points, since given any candidate $x\in X_{\infty},$ $x$ must belong to $X_{n}$ for some $n\geq 1,$ but that means that there is a sequence of points in $X_{n+1}$ with limit $x$, and this sequence also clearly belongs to $X_{\infty}.$
However, using the idea of martin.koeberl in the comment below, we may consider instead the sets $X_{n}^{*}=2n+X_{n},$ and let $E_{1}=\bigcup_{n\geq 1}X_{n}^{*}.$ Now we have shown that $X_{k-1}^{*}$ will still be nonempty after $k$ steps in this procedure, for all $k\geq 1,$ which shows that $E_{k},E_{k}^{*}$ will be nonempty for all $k\geq 1.$ This gives the desired example.
The following picture describes the method for generating the $X_{n}.$ The green points in the first line are the points in $X_{1}=\{0\}\cup\{1/n\}_{n\geq 2}.$ On the next line, the points that were in $X_{1}$ are in red, and the added points are in green; the whole set of red and green points makes up $X_{2}=X_{1}\cup\bigcup_{n\geq 2}\{1/n+(1/n(n-1))/k\}_{k\geq 2}$ (note that $1/(n-1)-1/n=1/(n(n-1))$). This is repeated to generate $$X_{3}=X_{2}\cup\bigcup_{n\geq 2,k\geq 2}\{1/n+(1/n(n-1))/k+(1/n(n-1)k(k-1))/j\}_{j\geq 2},$$ and the process continues in this way.

