LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it

Statement

Let $(X, d)$ be a metric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric), let $A \subseteq X$ and let $(A, d_{A})$ be the metric subspace (Isometry, isometric embedding, and the subspace metric on a subset). Then:

Relative openness is a trace. A set $V \subseteq A$ is open in $(A, d_{A})$ (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement) if and only if $V = U \cap A$ for some $U$ open in $(X, d)$ .
Compactness read in the ambient space. $A$ is a compact subset of $X$ (Open cover, subcover, compact metric space, and compact subset of a metric space), that is $(A, d_{A})$ is a compact metric space, if and only if for every family $U$ of open subsets of $X$ with $A \subseteq ⋃ U$ there are $n \in N$ and $U_{0}, \dots, U_{n} \in U$ with $A \subseteq U_{0} \cup \dots \cup U_{n}$ , or else $A = \emptyset$ .
The same in indexed form. $A$ is a compact subset of $X$ if and only if for every set $I$ and every family $(U_{i})_{i \in I}$ of open subsets of $X$ with $A \subseteq ⋃_{i \in I} U_{i}$ there are $n \in N$ and indices $i_{0}, \dots, i_{n} \in I$ with $A \subseteq U_{i_{0}} \cup \dots \cup U_{i_{n}}$ , or else $A = \emptyset$ .

Claim 3 is the form used by almost every later proof on this page, because a cover is usually produced by a rule that attaches an open set to each point or to each index, and a set of open sets forgets that rule. No choice principle is used anywhere below; the one place a selection is made is over a finite index set, and Every natural-number-indexed list of nonempty sets has a choice function on its family of values is a theorem of ZF.

Facts & Assumptions

Given: A metric space $(X, d)$ , a subset $A \subseteq X$ , and the metric subspace $(A, d_{A})$ with $d_{A}$ the restriction of $d$ to $A \times A$ .

[L1]

Balls of a subspace are traces of ambient balls: $B_{A} (a, r) = B_{X} (a, r) \cap A$ for $a \in A$ and $r > 0$ (Isometry, isometric embedding, and the subspace metric on a subset, Open ball, closed ball and sphere in a metric space).

[L2]

A subset $W$ of a metric space is open exactly when every point of $W$ has a ball around it contained in $W$ (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

[L3]

Open balls are open, and an arbitrary union of open sets is open (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed).

[L4]

$(A, d_{A})$ is compact exactly when every family of sets open in $(A, d_{A})$ whose union is $A$ has a finite subfamily whose union is $A$ ; a family is finite when it is empty or listable as ${V_{0}, \dots, V_{n}}$ (Open cover, subcover, compact metric space, and compact subset of a metric space).

[L5]

A function with domain a natural number all of whose values are nonempty sets has a choice function, and this is a theorem of ZF (Every natural-number-indexed list of nonempty sets has a choice function on its family of values).

Proof

technique · direct

1.1L1L2

Suppose $U$ is open in $(X, d)$ and put $V : = U \cap A$ ; for $a \in V$ we have $a \in U$ , so there is $r > 0$ with $B_{X} (a, r) \subseteq U$ , whence $B_{A} (a, r) = B_{X} (a, r) \cap A \subseteq U \cap A = V$ , and $V$ is open in $(A, d_{A})$ .

1.2L1construct

Conversely let $V \subseteq A$ be open in $(A, d_{A})$ , and let $W : = {B_{X} (a, r) : a \in V, r > 0, B_{X} (a, r) \cap A \subseteq V}$ , a family cut out by a property of the pair $(a, r)$ and not by any selection; put $U : = ⋃ W$ .

2.1L3step 1.2

$U$ is open in $(X, d)$ , being a union of open balls.

2.2step 1.2

$U \cap A \subseteq V$ , since every member $B_{X} (a, r)$ of $W$ satisfies $B_{X} (a, r) \cap A \subseteq V$ by the defining condition of $W$ .

2.3L1L2step 1.2

$V \subseteq U \cap A$ : given $a \in V$ , openness of $V$ in $(A, d_{A})$ gives $r > 0$ with $B_{A} (a, r) \subseteq V$ , that is $B_{X} (a, r) \cap A \subseteq V$ , so $B_{X} (a, r) \in W$ and $a \in B_{X} (a, r) \subseteq U$ , while $a \in A$ as well.

3.1step 1.1step 2.1step 2.2step 2.3

Claim 1 is proved: by steps 2.2 and 2.3 an open $V \subseteq A$ equals $U \cap A$ with $U$ open in $X$ by step 2.1, and conversely every such trace is open in $(A, d_{A})$ by step 1.1.

4.1step 3.1L4

For claim 3, suppose first that $(A, d_{A})$ is compact, let $I$ be a set and let $(U_{i})_{i \in I}$ be open subsets of $X$ with $A \subseteq ⋃_{i \in I} U_{i}$ ; then each $U_{i} \cap A$ is open in $(A, d_{A})$ and $V : = {U_{i} \cap A : i \in I}$ is a family of open subsets of $A$ whose union is $A$ .

5.1L4step 4.1

If $A = \emptyset$ the conclusion of claim 3 holds by its second alternative, so assume $A \neq \emptyset$ ; then $V$ is an open cover of $(A, d_{A})$ , and compactness yields $n \in N$ and $V_{0}, \dots, V_{n} \in V$ with $A = V_{0} \cup \dots \cup V_{n}$ .

6.1L5step 5.1

For each $j \leq n$ the set $S_{j} : = {i \in I : U_{i} \cap A = V_{j}}$ is nonempty by the definition of $V$ , and $j \mapsto S_{j}$ is a function with domain the natural number $σ (n)$ , so a choice function for its values supplies $i_{0}, \dots, i_{n} \in I$ with $U_{i_{j}} \cap A = V_{j}$ for every $j \leq n$ .

7.1step 5.1step 6.1

Hence $A = V_{0} \cup \dots \cup V_{n} = (U_{i_{0}} \cap A) \cup \dots \cup (U_{i_{n}} \cap A) \subseteq U_{i_{0}} \cup \dots \cup U_{i_{n}}$ , which is the conclusion of claim 3 for the family $(U_{i})_{i \in I}$ , so the forward implication of claim 3 holds.

8.1step 7.1step 3.1construct

The converse of claim 3 remains, the forward implication having been settled at step 7.1; so assume the displayed condition, let $G$ be a family of sets open in $(A, d_{A})$ with union $A$ , and put $U : = {U \subseteq X : U is open in (X, d) and U \cap A \in G}$ , again a family cut out by a property, indexed by itself.

9.1step 3.1step 8.1

$A \subseteq ⋃ U$ : given $a \in A$ there is $G \in G$ with $a \in G$ , and by claim 1 there is $U$ open in $X$ with $U \cap A = G$ ; that $U$ lies in $U$ and contains $a$ .

10.1step 8.1step 9.1

If $A = \emptyset$ the empty subfamily of $G$ covers $A$ ; otherwise the assumed condition applied to the family $U$ indexed by itself gives $m \in N$ and $W_{0}, \dots, W_{m} \in U$ with $A \subseteq W_{0} \cup \dots \cup W_{m}$ .

11.1L4step 10.1

Putting $G_{j} : = W_{j} \cap A$ for $j \leq m$ gives members of $G$ with $A = (W_{0} \cap A) \cup \dots \cup (W_{m} \cap A) = G_{0} \cup \dots \cup G_{m}$ , so $G$ has a finite subcover and $(A, d_{A})$ is compact.

12.1step 7.1step 11.1∎

Claim 3 is proved by steps 7.1 and 11.1, and claim 2 is the special case of claim 3 in which $I = U$ is a family of open subsets of $X$ and $U_{i} : = i$ , the conclusion of claim 3 then naming members of $U$ itself.

Remarks

Why the ambient reading needed a proof at all. A subset $A$ of $X$ carries two candidate notions of open cover: families of sets open in $(A, d_{A})$ , and families of sets open in $X$ whose union contains $A$ . Claim 1 is what turns one into the other, and it is the reason compactness of $A$ does not depend on which space $A$ is regarded as sitting inside. Every later item on this page that covers a subset by ambient balls is using claim 2 or claim 3, and says so.

The traces do not remember their sources. A single relatively open $V$ is usually the trace of many different ambient open sets, and that is exactly why step 6.1 has to recover indices at all. Recovering infinitely many at once would be a choice principle; recovering finitely many is not, and the proof is arranged so that only finitely many are ever needed.

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources