In a topological space, a dense-in-itself set is a set without isolated point. A perfect set is a closed dense-in-itself set. A scattered set is a set that contains no nonempty dense-in-itself subset.
Given a space $X$, the union of a family of dense-in-itself sets is dense-in-itself, and the closure of a dense-in-itself set is dense-in-itself, and hence perfect. So by taking the union of all dense-in-itself subsets of $X$, one gets a perfect set $P$. And the complement of $P$ does not contain any nonempty dense-in-itself subset; so it is scattered. This proves:
Theorem: Every topological space can be decomposed into the disjoint union of a perfect set and a scattered set.
How can you show this decomposition is unique?
Best Answer
Consider any decomposition of $X$ into a perfect set $Q$ and a scattered set $U=X\setminus Q$; we wish to show that $Q$ is equal to your $P$. Trivially $Q\subseteq P$. To show $P\subseteq Q$, it suffices to show that if $A\subseteq X$ is dense-in-itself, then $A\subseteq Q$. Now note that since $U$ is open, if $A$ is dense-in-itself then so is $A\cap U$. But since $U$ is scattered, this means $A\cap U=\emptyset$, i.e. $A\subseteq Q$, as desired.