CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

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Any two completions of a normed space are uniquely linearly isometric

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $X$ be a normed space, and let $(Y, j)$ and $(Z, k)$ be two completions of $X$ . Then there is a unique linear isometric isomorphism $U : Y \to Z$ such that $U \circ j = k .$

Facts & Assumptions

Given: The Axiom of Countable Choice; a normed space $X$ and two completions $(Y, j)$ and $(Z, k)$ of $X$ .

[A1]

Countable Choice has the meaning fixed in The Axiom of Countable Choice ( ${A C}_{ω}$ ).

[L1]

Under Countable Choice, any two metric completions are related by a unique isometry commuting with the dense embeddings (A completion is unique up to a unique isometry fixing the original space, and uniformly continuous maps into complete spaces extend through it).

[L2]

Under Countable Choice, bounded linear maps extend uniquely across a completion (Bounded linear maps extend uniquely across the completion).

[L3]

A linear isometric isomorphism is a bijective linear isometry (Linear isometries and isometric isomorphisms, Completion of a normed space).

Proof

technique · direct

1.1A1L1

By [A1] and [L1], there is a unique isometry $U : Y \to Z$ with $U \circ j = k$ .

1.2A1L2L3

On the dense subspace $j [X] \subseteq Y$ , the map $j (x) \mapsto k (x)$ is linear and norm-preserving. Applying [L2] under [A1] to this dense linear isometry extends it to a bounded linear map $\tilde{U} : Y \to Z$ with $\tilde{U} \circ j = k$ .

2.1step 1.1step 1.2L1

Both $U$ and $\tilde{U}$ are continuous maps $Y \to Z$ extending the same map on $j [X]$ , so the uniqueness in step 1.1 forces $U = \tilde{U}$ . Hence $U$ is linear.

3.1step 2.1L3∎

Since $U$ is already an isometry by step 1.1, it is a linear isometric isomorphism. Uniqueness of such a map is again the uniqueness from [L1].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

38 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Gerald Teschl, Topics in Real and Functional Analysis (standard reference, not scraped)
Theo Buhler and Dietmar A. Salamon, Functional Analysis (standard reference, not scraped)