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.

Examples
  • 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 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 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.

Notes
  1. Steen & Seebach, p. 33
  2. Engelking, p. 59
  3. See proposition 2.8 in Al-Hajri, Monerah; Belaid, Karim; Belaid, Lamia Jaafar (2016). "Scattered Spaces, Compactifications and an Application to Image Classification Problem". Tatra Mountains Mathematical Publications. 66: 1–12. doi:10.1515/tmmp-2016-0015. S2CID 199470332.
  4. https://math.stackexchange.com/questions/3854864
  5. https://math.stackexchange.com/questions/376116
  6. Willard, problem 30E, p. 219
  7. https://math.stackexchange.com/questions/3856152
  8. https://math.stackexchange.com/questions/742025

References
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