LemmaStatement: Literature-sourcedProof: AI-adaptedverified 2026-09-26 (gpt-6-sol)

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Every open subspace of a completely metrizable space is completely metrizable

Statement

If $X$ is completely metrizable and $U \subseteq X$ is open, then $U$ is completely metrizable in its subspace topology.

Facts & Assumptions

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

[F1]

Let $(X, d)$ be a metric space (def-metric-space) and let $T_{d}$ be its metric topology (def-metric-topology). Call $T_{d}$ completely metrizable if some metric $ρ$ on $X$ is topologically equivalent to $d$ , that is $T_{ρ} = T_{d}$ (def-equivalent-metrics), and makes $(X, ρ)$ complete (def-complete-metric-space). Then: 1. Homeomorphism invariance. Let $(Y, e)$ be a metric space and let $h : X \to Y$ be a bijection (def-injection-surjection-bijection) such that $h$ and $h^{- 1}$ are continuous (def-metric-continuity). If $T_{d}$ is completely metrizable then so is $T_{e}$ . 2. Closed subspaces. If $T_{d}$ is completely metrizable and $A \subseteq X$ is closed in $(X, d)$ , then $T_{d_{A}}$ is completely metrizable, $d_{A}$ being the subspace metric (def-isometry-and-metric-embedding). 3. The property is strictly weaker than completeness. Let $P : = (0, \infty) \subseteq R$ (def-interval) carry $d (x, y) : = ∣ x - y ∣$ (lem-real-line-is-a-metric-space). Then $(P, d)$ is not complete, while $ρ_{P} (x, y) : = ∣ x - y ∣ + ∣ \frac{1}{x} - \frac{1}{y} ∣$ is a complete metric on $P$ with $T_{ρ_{P}} = T_{d}$ . So $T_{d}$ is completely metrizable although no completeness assumption holds for $d$ itself. Complete metrizability is a condition on the collection of open sets alone: the metric is quantified over and does not survive into the statement. That is exactly what completeness fails to be, and claim 3 shows the two conditions are genuinely different rather than merely stated differently. (Complete metrizability: admitting a topologically equivalent complete metric is preserved by homeomorphism and by closed subspaces, and $(0, \infty)$ has it without being complete).

[F2]

Let $(X, d)$ be a metric space (def-metric-space), let $A \subseteq X$ be nonempty and let $x, y \in X$ . Then $∣ d (x, A) - d (y, A) ∣ \leq d (x, y),$ with $d (\cdot, A)$ the distance to a nonempty set (def-metric-bounded-diameter). Thus the real-valued function $u \mapsto d (u, A)$ changes by at most $d (u, v)$ between $u$ and $v$ : it is $1$ -Lipschitz. ( $∣ d (x, A) - d (y, A) ∣ \leq d (x, y)$ , so the distance to a fixed nonempty set is $1$ -Lipschitz).

Proof

technique · direct

1.1givenF1

If $U = \emptyset$ , its unique metric is compatible and complete. Otherwise choose a complete metric $ρ$ on $X$ compatible with its given topology, as allowed by [F1]. Since $U$ is open, it is also $ρ$ -open.

2.1step 1.1F2

If $U = X$ , the restricted metric $ρ ∣_{U \times U}$ is compatible and complete. Hence assume $U$ is nonempty and proper, and put $F = X ∖ U$ . Then $F$ is nonempty and $ρ$ -closed. For $x \in U$ , let $δ (x) = \inf_{z \in F} ρ (x, z)$ . Openness of $U$ gives $δ (x) > 0$ , and [F2] gives $∣ δ (x) - δ (y) ∣ \leq ρ (x, y)$ .

3.1step 2.1algebra

Define $σ (x, y) = ρ (x, y) + ∣ 1 / δ (x) - 1 / δ (y) ∣$ on $U$ . This is a metric: it is nonnegative and symmetric, vanishes only when $x = y$ because $ρ$ is a metric, and satisfies the triangle inequality by adding those for $ρ$ and absolute value. Also $ρ (x, y) \leq σ (x, y)$ .

4.1step 2.1step 3.1F2

The metrics $σ$ and $ρ ∣_{U \times U}$ induce the same topology. Indeed, fix $x \in U$ and $ε > 0$ . If $ρ (x, y) < δ (x) / 2$ , then $δ (y) > δ (x) / 2$ by [F2], so $∣ 1 / δ (x) - 1 / δ (y) ∣ \leq 2 ρ (x, y) / δ (x)^{2}$ . Thus $ρ (x, y) < \min {δ (x) / 2, ε / (1 + 2 / δ (x)^{2})}$ implies $σ (x, y) < ε$ . Conversely $σ (x, y) < ε$ implies $ρ (x, y) < ε$ by step 3.1.

4.2step 1.1step 2.1step 3.1F2

Let $(x_{n})$ be $σ$ -Cauchy. Then $(x_{n})$ is $ρ$ -Cauchy and the real sequence $(1 / δ (x_{n}))$ is Cauchy by step 3.1. Completeness of $(X, ρ)$ gives a limit $x \in X$ , and every Cauchy real sequence is bounded, so $1 / δ (x_{n}) \leq M$ for some finite $M > 0$ and all $n$ . Hence $δ (x_{n}) \geq 1 / M$ ; [F2] and $x_{n} \to x$ imply $δ (x) \geq 1 / M > 0$ . If $x \in F$ , its distance to $F$ would be zero, so $x \in U$ .

5.1step 1.1step 2.1step 4.1step 4.2∎

Since $x \in U$ and $δ (x_{n}) \to δ (x) > 0$ , the reciprocal estimate of step 4.1 gives $1 / δ (x_{n}) \to 1 / δ (x)$ . Therefore $σ (x_{n}, x) \to 0$ , proving completeness of $σ$ . Steps 1.1–2.1 cover the empty and whole-space cases, and step 4.1 gives compatibility in the remaining case. Thus $U$ is completely metrizable.

Depends on

Used by

Dependency tree · two levels

34 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

David Marker, Descriptive Set Theory, §§1–2 (standard reference, not scraped)
Michael Kunzinger, General Topology, §§11.3–11.4 (standard reference, not scraped)
MFF General Topology course summary, §4.3 (standard reference, not scraped)
Jesse Peterson, Real Analysis, §§3.6–3.7 (standard reference, not scraped)