Removing isolated points to get a perfect set The motivating question is the following: If $F$ is a closed subset of $\mathbb{R}^1$, can one find a perfect set $E\subset F$ such that $m(E)=m(F)$ (in Lebesgue measure)?
Define $F_0=F$ and $F_n=\{\text{accumulation points of }F_{n-1}\}$ for every $n\ge1$. Take $E=\bigcap_{n=0}^\infty F_n$. Each $F_n$ has only finitely countably many isolated points, hence $m(F_n)=m(F_{n-1})$. Thus $m(E)=\lim m(F_n)=m(F)$. Also note that each $F_n$ is closed, hence so is $E$. The problem is to prove that $E$ is perfect.
To summarize:

Let $F_0$ be a closed subset of $\mathbb{R}^1$. Define $F_n=\{\text{accumulation points of }F_{n-1}\}$ for every $n\ge1$, and $E=\bigcap_{n=0}^\infty F_n$. Is $E$ necessarily perfect?

 A: Two things. First, each $F_n$ has only countably many isolated points, for example $\mathbb Z$ has only (and more than finitely many) isolated points, on the other hand, there is a neighbourhood of an isolated point which does not contain any other point of $F_n$, as uncountably many open sets in $\mathbb R$ cannot be pairwise disjoint, we can have only countably many.
Second, in general $E := F_\omega = \bigcap_{n <\omega} F_n$ will not be perfect, start with $0$, add a sequence converging to $0$, add a sequence converging to one of the added points, continue this $\omega$ often to obtain $F$. Then $E = \{0\}$. What you can do is continue after the $\omega$th step.That is for $\alpha < \omega_1$ define by induction $$ F_{\alpha + 1} = \{\text{accumulation points of $F_{\alpha}$}\},\quad F_{\lambda} = \bigcap_{\beta < \lambda} F_\beta \text{ for limit ordinals $\lambda$} $$
Then $(F_\alpha)_{\alpha < \omega_1}$ is a dreasing sequence of closed sets. As such a sequence can only be of countable length, before being constant, there is an $\alpha^* < \omega_1$ such that $F_{\alpha^*} = F_{\alpha^*+1}$, hence $F_{\alpha^*}$ is perfect. Now, as before, is the usual step, we have $m(F_{\alpha + 1}) = m(F_\alpha)$ for $\alpha < \omega_1$, as only coubntably many points are removed, and continuity of the measure gives 
$$ m(F_\lambda) = \inf_{\alpha < \lambda} m(F_\alpha) = m(F_0) $$
in the limit steps, so $m(F_{\alpha^*}) = m(F_0)$ and $F_{\alpha^*}$ is perfect.
