LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-08-05 (gpt-5.6-sol-codex-subscription)

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 subspace is compact exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it

Statement

Let $(X, T)$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), let $A \subseteq X$ and let $(A, T_{A})$ be the subspace (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). Then:

Compactness read in the ambient space. $A$ is a compact subset of $X$ (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right), that is $(A, T_{A})$ is a compact space, if and only if for every family $U \subseteq T$ 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 2 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 topological space $(X, T)$ , a subset $A \subseteq X$ , and the subspace $(A, T_{A})$ with $T_{A} = {U \cap A : U \in T}$ .

[L1]

A subset of $A$ is open in $(A, T_{A})$ exactly when it is the trace $U \cap A$ of a set $U$ open in $X$ , this being the definition of the subspace topology (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).

[L2]

$(A, T_{A})$ is compact exactly when every family of sets open in $(A, T_{A})$ 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, and compact topological space; a compact subset is a subspace that is compact in its own right).

[L3]

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

1.1

Suppose $(A, T_{A})$ 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 $(A, T_{A})$ and $V : = {U_{i} \cap A : i \in I}$ is a family of open subsets of $A$ whose union is $A$ .

L1L2

2.1

If $A = \emptyset$ the conclusion of claim 2 holds by its second alternative, so assume $A \neq \emptyset$ ; then $V$ is an open cover of $(A, T_{A})$ , and compactness yields $n \in N$ and $V_{0}, \dots, V_{n} \in V$ with $A = V_{0} \cup \dots \cup V_{n}$ .

L2step 1.1

3.1

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 $j \leq n$ .

L3step 2.1

4.1

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 2 for the family $(U_{i})_{i \in I}$ , so the forward implication of claim 2 holds.

step 2.1step 3.1

5.1

The converse of claim 2 remains, the forward implication having been settled at step 4.1; so assume the displayed condition, let $G$ be a family of sets open in $(A, T_{A})$ with union $A$ , and put $W : = {U \in T : U \cap A \in G}$ , a family cut out by a property and indexed by itself.

L1step 4.1construct

6.1

$A \subseteq ⋃ W$ : given $a \in A$ there is $G \in G$ with $a \in G$ , and by [L1] there is $U$ open in $X$ with $U \cap A = G$ ; that $U$ lies in $W$ and contains $a$ .

L1step 5.1

7.1

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

step 5.1step 6.1

8.1

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 $(A, T_{A})$ is compact.

L2step 7.1

9.1

Claim 2 is proved by steps 4.1 and 8.1, and claim 1 is the special case of claim 2 in which $I = U$ is a family of open subsets of $X$ and $U_{i} : = i$ , the conclusion of claim 2 then naming members of $U$ itself.

step 4.1step 8.1∎

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 $(A, T_{A})$ , and families of sets open in $X$ whose union contains $A$ . The trace description of the subspace topology is what turns one into the other, and it shows that compactness can be checked using ambient open sets for this fixed induced topology. Another ambient is guaranteed to give the same answer when it induces the same topology on $A$ ; if the induced topology changes, the answer may change. Every later item on this page that covers a subset by ambient open sets is using claim 1 or claim 2, 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 3.1 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.

The metric statement of the same fact is 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, whose claims 2 and 3 are claims 1 and 2 above with the open subsets of a metric space in place of the members of an abstract topology. Its proof carries an extra first claim, that relative openness in a metric subspace is a trace, which here is the definition of the subspace topology and so needs no argument. Neither statement is used in the proof of the other; that the two agree is For a metric space with its metric topology, compactness in the topological sense is compactness in the metric sense, and the two notions of compact subset coincide.

Depends on

Used by

Dependency tree · two levels

16 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

Compact space (Wikipedia) (standard reference, not scraped)
Subspace topology (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §26 (standard reference, not scraped)
Stacks Project, Section 5.12: Quasi-compact spaces and maps (standard reference, not scraped)