TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-09-09 (gpt-6-astra)

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Alexandrov's theorem, under Dependent Choice: a subspace of a complete metric space is completely metrizable exactly when it is $G_{δ}$

Statement

Assume Dependent Choice. For a subspace $Y$ of a complete metric space $X$ , $Y$ is completely metrizable if and only if $Y$ is a $G_{δ}$ subset of $X$ .

Facts & Assumptions

Given: The objects, hypotheses, and choice principles stated above.

[F1]

Under Countable Choice, if $(X, d)$ is complete and $Y \subseteq X$ is $G_{δ}$ in $X$ , then $Y$ is completely metrizable. (Under the Axiom of Countable Choice, every $G_{δ}$ subspace of a complete metric space is completely metrizable)

[F2]

Assume Dependent Choice. If $Y$ is a completely metrizable subspace of a metric space $X$ , then $Y$ is a $G_{δ}$ subset of $X$ . (Under Dependent Choice, every completely metrizable subspace of a metric space is $G_{δ}$ ).

[F3]

DC supplies a sequence beginning at any specified point of an entire relation; Countable Choice selects from any given sequence of nonempty sets. (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain, The Axiom of Countable Choice ( ${A C}_{ω}$ ))

Proof

technique · direct

1.1givenF1F2

The empty subspace satisfies both conditions.

1.2F1givenF3

The assumed DC supplies the Countable Choice needed by [F1]. [given, F3] For any sequence of nonempty sets, take the set of finite lists choosing from its first finitely many members. This set contains the empty list, and the one-term-extension relation is entire. DC starting at the empty list gives nested lists of every finite length; their union is the required choice function. No implication theorem from a later choice page is used.

2.1F1step 1.2F2

Apply [F1] with the given complete ambient metric and the Countable [step 1.2, F1, F2] Choice just derived. For the converse apply [F2] under the given DC; it needs only a compatible complete metric on $Y$ , not completeness of its inherited metric. Thus both implications have their stated hypotheses.

Alexandrov's theorem, under Dependent Choice: a subspace of a complete metric space is completely metrizable exactly when it is $G_{δ}$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Alexandrov's theorem, under Dependent Choice: a subspace of a complete metric space is completely metrizable exactly when it is GδGδ​

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Alexandrov's theorem, under Dependent Choice: a subspace of a complete metric space is completely metrizable exactly when it is $G_{δ}$