scattered space
A topological space^{} $X$ is said to be scattered if for every closed subset $C$ of $X$, the set of isolated points of $C$ is dense in $C$. Equivalently, $X$ is a scattered space if no nonempty closed subset of $X$ is dense in itself: for every closed subset $C$ of $X$, the closure^{} of the interior of $C$ is not $C$.
A subset of a topological space is called scattered if it is a scattered space with the subspace topology.
Every discrete space is scattered, since every singleton is open, hence isolated.
Scattered line. Let $\mathbb{R}$ be the real line equipped with the usual topology $T$ (formed by the open intervals). Let’s define a new topology^{} $S$ on $\mathbb{R}$ as follows: a subset $A$ is open under $S$ ($A\in S$) if $A=B\cup C$, where $B$ is open under $T$ ($B\in T$) and $C\subseteq \mathbb{R}\mathbb{Q}$, a subset of the irrational numbers. We make the following observations:

1.
$S$ is a topology on $\mathbb{R}$ which is finer than $T$

2.
$\mathbb{R}$ is a Hausdorff space under $S$,

3.
a singleton in $\mathbb{R}$ is clopen iff it contains an irrational number

4.
any subset of irrationals is scattered under the subspace topology of $\mathbb{R}$ under $S$
Proof.

1.
First note that every element of $T$ is an element of $S$, so $\mathrm{\varnothing},\mathbb{R}\in S$ in particular. Suppose ${A}_{1},{A}_{2}\in S$ with ${A}_{1}={B}_{1}\cup {C}_{1}$ and ${A}_{2}={B}_{2}\cup {C}_{2}$, where ${B}_{i},{C}_{i}$ are defined as in the setup above. Then ${A}_{1}\cap {A}_{2}=B\cup C$, where $B={B}_{1}\cap {B}_{2}\in T$ and $C=({C}_{1}\cap {B}_{2})\cup (({B}_{1}\cup {C}_{1})\cap {C}_{2})$ is a subset of the irrationals. So ${A}_{1}\cap {A}_{2}\in S$. If ${A}_{i}\in S$ with ${A}_{i}={B}_{i}\cup {C}_{i}$, then $\bigcup {A}_{i}=\bigcup {B}_{i}\cup \bigcup {C}_{i}\in S$. So $S$ is a topology which is finer than $T$

2.
$\mathbb{R}$ is Hausdorff under $S$ is clear, the topological property is inherited from $T$.

3.
First, any singleton is closed since $X$ is Hausdorff under $S$. If $x$ is irrational, then $\{x\}$ is open (under $S$) as well. So $\{x\}$ is clopen. If $x$ is rational and $\{x\}\in S$, then it is the union of a $T$open set $B$ and a subset $C$ of the irrationals. The only $T$open subset of $\{x\}$ is the empty set^{}, so $\{x\}$ is a subset of the irrationals, a contradiction^{}.

4.
Let $C$ is a subset of the irrational numbers. and considered the subspace topology under $S$. Then every point $r$ of $C$ is isolated, since $\{r\}$ is the open subset of $C$ separating it from the rest. The closure of the collection^{} of these points is clearly $C$ itself, so $C$ is scattered.
∎
The real line under the topology $S$ is called a scattered line.
Remark. Every topological space is a disjoint union^{} of a perfect set^{} and a scattered set.
Title  scattered space 

Canonical name  ScatteredSpace 
Date of creation  20130322 16:42:59 
Last modified on  20130322 16:42:59 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  6 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 54G12 
Related topic  DenseInItself 
Defines  scattered 
Defines  scattered set 
Defines  scattered line 