TITLE: Why are scattered sets G-delta?
QUESTION [7 upvotes]: I am looking for an elementary proof of the following - The problem appears in a real analysis text of A. Bruckner:
A subset $S$ of $\mathbb{R}$ is called scattered if every non empty subset $X \subseteq S$ has an isolated point. Show that every scattered set is equal to a countable intersection of open sets.
I'd really appreciate a detailed proof, if possible. Thanks.

REPLY [2 votes]: I don't know if transfinite induction is elementary. But I thought I might have a writeup of this fun fact anyways.
Function $d: \mathcal{P}(\mathbb{R})\to \mathcal{P}(\mathbb{R})$ is defined recursively as follows:
$d(X)=$ limit points of $X$;
$d^{\alpha+1}(X)=d(d^\alpha(X))$;
$d^{\alpha}(X)=\bigcap_{\beta<\alpha} d^\beta(X)$ when $\alpha$ is a limit.
Note when talking about limit points of $X$ we use the subspace topology on $X$.
We prove inductively on $\alpha<\omega_1$ the following:
$(\cdot)_\alpha: \forall X\subset \mathbb{R} \ \  X-d^\alpha(X)$ is $G_\delta$, and further more $ X-d^\alpha(X)$ can be written as $\bigcap_n U_n$ (nested/decreasing) such that $\bar{U}_n\cap d^\alpha(X)=\emptyset$.
It immediately follows that scattered sets are $G_\delta$.
Two cases.
Let $\alpha$ be a limit. Fix $\alpha_n, n\in \omega$ strictly increasing and converging to $\alpha$. By inductive hypothesis we could write $d^{\alpha_n}(X)-d^{\alpha_{n+1}}(X)=\bigcap_m U_m^n$ such that $\bar{U}_m^n \cap d^{\alpha_{n+1}}(X)=\emptyset$ (we apply $(\cdot)_{\alpha_{n+1}-\alpha_n}$ to $d^{\alpha_{n}}(X)$). Also we might assume that $U_m^j \cap U_m^k =\emptyset$ whenever $j<k$. The reason why we can do this is that given $j<k$, we know $\bigcap_m U_m^k = \bigcap_m U_m^k \cap d^{\alpha_k}(X)\subset \bigcap_m U_m^k \cap (\bigcup_{j<k} \bar{U}_m^j)^c \subset \bigcap_m U_m^k$, hence we might replace the set with the one inside if necessary.
Now we claim that $\bigcap_m \bigcup_n U_m^n=X-d^{\alpha}(X)$. One direction is easy to verify. We verify the other direction. Given $x\in \bigcap_m \bigcup_n U_m^n$, we know for $m=0$, there exists a least $k$ such that $x\in U_0^k$. We claim that $x\in \bigcap_m U_m^k$. Suppose not, let $m$ be the least such that $x\not \in  U_m^k$. If $j<k$, by assumption, we know $x\not \in U_0^j$ hence $x\not \in U_m^j$. If $j>k$, as $U_0^j\cap U_0^k=\emptyset$, we know $x\not \in U_0^j$ so $x\not \in U_m^j$. But then this contradicts with the assumption.
Now successor, say $\alpha=\beta+1$.
Let $\bigcap_m U_m$ be the result of applying $(\cdot)_\beta$. As $d^\beta(X)-d^{\beta+1}(X)$ consisting of isolated points in $d^\beta(X)$. Find $V_m$ open such that $\bar{V}_m \cap d^\beta(X)$ at at most one point, $\bar{V}_m \cap d^{\beta+1}(X)=\emptyset$ and $V_m\cap U_m =\emptyset$. The reason why this can be achieved is similar to the reason given above. Argue as before that the desired set is $\bigcap_m (V_m \cup U_m)$. Note that the inductive hypothesis is satisfied.