LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31

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Jones's bound: under choice, a closed discrete subspace of a normal space cannot have more subsets than a dense set has subsets

Statement

Assume the Axiom of Choice. If $D$ is a closed discrete subspace of a normal space $X$ and $E \subseteq X$ is dense, then there is an injection $P (D) \to P (E)$ . In cardinal notation, $2^{∣ D ∣} \leq 2^{∣ E ∣}$ .

Facts & Assumptions

Given: A normal space $X$ , a closed discrete $D \subseteq X$ , and a dense $E \subseteq X$ .

[A1]

The Axiom of Choice supplies a choice function for every family of nonempty sets (The Axiom of Choice).

[F1]

Every subset of a discrete subspace is closed in that subspace; because $D$ is closed in $X$ , each subset of $D$ is closed in $X$ (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

[F2]

Normality separates disjoint closed sets by disjoint open sets, and every nonempty open set meets a dense subset (Normal spaces and $T_{4}$ spaces, with the source disagreement over whether normality includes $T_{1}$ stated explicitly, Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets).

[F3]

Under choice, $2^{κ} = ∣ P (κ) ∣$ for every cardinal $κ$ (Cardinal sum $κ \oplus λ$ , product $κ \otimes λ$ and exponentiation $κ^{λ}$ , and why they are written apart from the ordinal operations, Assuming the Axiom of Choice, $2^{κ} = ∣ P (κ) ∣$ , and Cantor's theorem in cardinal form: $κ < 2^{κ}$ ).

Proof

technique · direct

1.1

For every $A \subseteq D$ , the sets $A$ and $D ∖ A$ are disjoint closed subsets of $X$ . By normality there is an open $U_{A}$ containing $A$ and an open $V_{A}$ containing $D ∖ A$ with $U_{A} \cap V_{A} = \emptyset$ .

F1F2

2.1

Apply [A1] to choose one such pair $(U_{A}, V_{A})$ for every $A \subseteq D$ , and define $Φ (A) = U_{A} \cap E \subseteq E$ .

A1step 1.1

3.1

If $A \neq B$ , take $d \in A ∖ B$ after interchanging them if necessary. Then $d \in U_{A} \cap V_{B}$ , a nonempty open set meeting $E$ ; a point of $E \cap U_{A} \cap V_{B}$ lies in $Φ (A)$ and not in $Φ (B)$ .

F2step 2.1

4.1

Thus $Φ$ is injective. By [F3], this is the asserted cardinal inequality.

F3step 3.1∎

Depends on

Used by

Dependency tree · two levels

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Sources

G. Gruenhage, General Topology Course Notes, Jones's lemma (standard reference, not scraped)
Samuel Gomes da Silva, Closed discrete subsets of separable spaces and relative versions of normality, countable paracompactness and property (a) (standard reference, not scraped)