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# 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 non-empty 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. 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 $\varnothing,\mathbb{R}\in S$ 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. 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.

## Mathematics Subject Classification

54G12*no label found*

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