Scattered space

In mathematics, a scattered space is a topological space X that contains no nonempty dense-in-itself subset.^[1]^[2] Equivalently, every nonempty subset A of X contains a point isolated in A.

A subset of a topological space is called a scattered set if it is a scattered space with the subspace topology.

Every discrete space is scattered.
Every ordinal number with the order topology is scattered. Indeed, every nonempty subset A contains a minimum element, and that element is isolated in A.
A space X with the particular point topology, in particular the Sierpinski space, is scattered. This is an example of a scattered space that is not a T₁ space.
The closure of a scattered set is not necessarily scattered. For example, in the Euclidean plane $\mathbb {R} ^{2}$ take a countably infinite discrete set A in the unit disk, with the points getting denser and denser as one approaches the boundary. For example, take the union of the vertices of a series of n-gons centered at the origin, with radius getting closer and closer to 1. Then the closure of A will contain the whole circle of radius 1, which is dense-in-itself.

Properties

In a topological space X the closure of a dense-in-itself subset is a perfect set. So X is scattered if and only if it does not contain any nonempty perfect set.
Every subset of a scattered space is scattered. Being scattered is a hereditary property.
Every scattered space X is a T₀ space. (Proof: Given two distinct points x, y in X, at least one of them, say x, will be isolated in $\{x,y\}$ . That means there is neighborhood of x in X that does not contain y.)
In a T₀ space the union of two scattered sets is scattered.^[3]^[4] Note that the T₀ assumption is necessary here. For example, if $X=\{a,b\}$ with the indiscrete topology, $\{a\}$ and $\{b\}$ are both scattered, but their union, $X$ , is not scattered as it has no isolated point.
Every T₁ scattered space is totally disconnected.
(Proof: If C is a nonempty connected subset of X, it contains a point x isolated in C. So the singleton $\{x\}$ is both open in C (because x is isolated) and closed in C (because of the T₁ property). Because C is connected, it must be equal to $\{x\}$ . This shows that every connected component of X has a single point.)
Every second countable scattered space is countable.^[5]
Every topological space X can be written in a unique way as the disjoint union of a perfect set and a scattered set.^[6]^[7]
Every second countable space X can be written in a unique way as the disjoint union of a perfect set and a countable scattered open set.
(Proof: Use the perfect + scattered decomposition and the fact above about second countable scattered spaces, together with the fact that a subset of a second countable space is second countable.)
Furthermore, every closed subset of a second countable X can be written uniquely as the disjoint union of a perfect subset of X and a countable scattered subset of X.^[8] This holds in particular in any Polish space, which is the contents of the Cantor–Bendixson theorem.

Properties

Notes

References

Related Articles