nLab derived set

This entry is about a notion in general topology, unrelated to derived geometry.

Idea

In the context of general topology, given a subset of a topological space, its derived set consists of all elements of the space that are accumulation points of the subset.

In other words, if we consider every point of the closure of the set as either isolated or non-isolated, then the derived set consists of the non-isolated points in the closure.

Another name for the derived set is the Cantor-Bendixson derivative.

Properties

Derived sets are closed, since the set of isolated points in the closure is open in the closure under the subspace topology.

Given a topological space $X$, the assignment that takes a closed subset $A$ of $X$ to its derived set $A'$ defines an operator

$$\delta: Cl(X) \to Cl(X)$$

on the set of closed subsets of $X$. It has the following easily established properties:

• $\delta(A) \subseteq A$;

• If $A \subseteq B$, then $\delta(A) \subseteq \delta(B)$;

• $\delta$ preserves finite unions.

Recall that a subset $A$ of a space $X$ is perfect if $A' = A$. For closed sets $C$, define a closed set $C^\alpha$ for each ordinal $\alpha$ by transfinite recursion:

• $C^0 \coloneqq C$;

• $C^{\alpha + 1} \coloneqq \delta(C^\alpha)$;

• For limit ordinals $\beta$, $C^\beta \coloneqq \bigcap_{\alpha \lt \beta} C^\alpha$.

The decreasing chain $C^\alpha$ eventually terminates in a perfect set. The least ordinal for which $C^{\alpha + 1} = C^\alpha$ is called the Cantor-Bendixson rank of $C$.

Proposition

In a second-countable space, every closed set has countable Cantor-Bendixson rank, and is the union of a perfect set and a countable set.

Proof

Let $U_i$ be a countable basis of $X$. Let $\alpha$ be the Cantor-Bendixson rank of a closed set $C$. For each $\beta \lt \alpha$, each point $x \in C^\beta \setminus C^{\beta + 1}$ is an isolated point, so we can find an $U_{i(x)}$ in the basis such that $U_{i(x)} \cap (C^\beta \setminus C^{\beta + 1}) = \{x\}$. It is then clear that $x \mapsto i(x)$ is injective, so each $C^\beta \setminus C^{\beta + 1}$ is countable. Similarly, whenever $C^\beta \setminus C^{\beta + 1}$ is nonempty, we can find a basis element $U_{j(\beta)}$ that isolates one of its points (say $x$), and this same $U_j$ cannot isolate any point of an earlier $C^\gamma \setminus C^{\gamma + 1}$ since $x$ is a limit point of $C^\gamma$. It follows that $\beta \mapsto j(\beta)$ is an injective map, so that $\alpha$ must be a countable ordinal, and the collection $F \coloneqq C \setminus C^\alpha = \bigcup_{\beta \lt \alpha} C^\beta \setminus C^{\beta + 1}$ is (at most) countable.

Proposition

In a Polish space, perfect sets have continuum cardinality $c = 2^{\aleph_0}$.

The proof is given there. It follows that infinite closed sets in a Polish space have cardinality either $\aleph_0$ or $2^{\aleph_0}$ (compare the continuum hypothesis).

Related concepts

• Polish space

• limit point

References

• Alexander S. Kechris, Classical descriptive set theory, Springer (1994) [doi:10.1007/978-1-4612-4190-4]

See also:

• Wikipedia: Derived set (mathematics)