In mathematics, a scattered space is a
topological spaceX 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
ordinal number with the
order topology is scattered. Indeed, every nonempty subset A contains a minimum element, and that element is isolated in A.
The closure of a scattered set is not necessarily scattered. For example, in the
Euclidean plane 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 . 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 with the
indiscrete topology, and are both scattered, but their union, , 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 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 . This shows that every connected component of X has a single point.)
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.