TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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A metric space is compact if and only if every family of closed subsets with the finite intersection property has nonempty intersection

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), with closed sets as in 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 and the finite intersection property as in Finite intersection property. For a family $\mathcal{A}$ of subsets of $X$ write

$\bigcap \mathcal{A} \;:=\; \{\, x \in X : x \in F \text{ for every } F \in \mathcal{A} \,\},$

so that $\bigcap \emptyset = X$ , matching the convention for the empty finite intersection in Finite intersection property.

Then $(X,d)$ is compact (Open cover, subcover, compact metric space, and compact subset of a metric space) if and only if every family $\mathcal{A}$ of closed subsets of $X$ with the finite intersection property satisfies $\bigcap \mathcal{A} \ne \emptyset$ .

No choice principle is used in either direction: complementation is a canonical bijection, so no member of a family ever has to be selected.

Facts & Assumptions

Given: A metric space $(X,d)$ , families of subsets of $X$ , and the two notions above.

[A1]

Elementary set algebra inside $X$ : $X \setminus (X \setminus S) = S$ for $S \subseteq X$ , and for any family $\mathcal{S}$ of subsets of $X$ one has $X \setminus \bigcup \mathcal{S} = \bigcap \{\, X \setminus S : S \in \mathcal{S} \,\}$ and $X \setminus \bigcap \mathcal{S} = \bigcup \{\, X \setminus S : S \in \mathcal{S} \,\}$ .

[L1]

$(X,d)$ is compact exactly when every family of open subsets of $X$ with union $X$ has a finite subfamily with union $X$ , the empty subfamily serving when $X = \emptyset$ (Open cover, subcover, compact metric space, and compact subset of a metric space).

[L2]

$F \subseteq X$ is closed exactly when $X \setminus F$ is open (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]

$\mathcal{A}$ has the finite intersection property exactly when $\bigcap_{i \in n} s(i) \ne \emptyset$ for every $n \in \mathbb{N}$ and every finite list $s : n \to \mathcal{A}$ , the empty list having intersection $X$ (Finite intersection property).

Proof

technique · direct

1.1

Complementation inside $X$ is its own inverse, and it exchanges the open subsets of $X$ with the closed ones: $X \setminus (X\setminus S) = S$ , and $S$ is open exactly when $X \setminus S$ is closed.

A1L2

1.2

For any family $\mathcal{S}$ of subsets of $X$ , the union of $\mathcal{S}$ is $X$ exactly when the intersection of the complements is empty, and the intersection of $\mathcal{S}$ is empty exactly when the union of the complements is $X$ .

2.1

In particular, for $F_0, \dots, F_n \subseteq X$ one has $F_0 \cap \dots \cap F_n = \emptyset$ exactly when $(X \setminus F_0) \cup \dots \cup (X\setminus F_n) = X$ .

A1step 1.2

3.1

Assume $(X,d)$ compact, let $\mathcal{A}$ be a family of closed subsets of $X$ with the finite intersection property, and suppose for contradiction that $\bigcap \mathcal{A} = \emptyset$ .

L1L3step 2.1assume-contra

4.1

Applying the finite intersection property to the empty list gives $X \ne \emptyset$ ; and $\mathcal{U} := \{\, X \setminus F : F \in \mathcal{A} \,\}$ is a family of open subsets of $X$ whose union is $X \setminus \bigcap\mathcal{A} = X$ , hence an open cover of $X$ .

L2L3step 1.1step 1.2step 3.1

5.1

Compactness applied to $\mathcal{U}$ , together with $X \ne \emptyset$ , gives $n \in \mathbb{N}$ and $U_0, \dots, U_n \in \mathcal{U}$ with $X = U_0 \cup \dots \cup U_n$ .

L1step 4.1

6.1

Each $U_i$ is $X \setminus F_i$ for the set $F_i := X \setminus U_i$ , which lies in $\mathcal{A}$ and is determined by $U_i$ alone; so $F_0 \cap \dots \cap F_n = \emptyset$ by step 2.1, and the list $i \mapsto F_i$ contradicts the finite intersection property of $\mathcal{A}$ .

L3step 1.1step 2.1step 5.1discharge-contradiction

7.1

Therefore $\bigcap \mathcal{A} \ne \emptyset$ , which is the forward implication.

step 6.1

8.1

Conversely assume that every family of closed subsets of $X$ with the finite intersection property has nonempty intersection, let $\mathcal{U}$ be a family of open subsets of $X$ with union $X$ , and suppose for contradiction that no finite subfamily of $\mathcal{U}$ has union $X$ .

L1step 7.1assume-contra

9.1

Then $X \ne \emptyset$ , since otherwise the empty subfamily would have union $X$ ; and $\mathcal{A} := \{\, X \setminus U : U \in \mathcal{U} \,\}$ is a family of closed subsets of $X$ .

L2step 1.1step 8.1

10.1

$\mathcal{A}$ has the finite intersection property: the empty list has intersection $X \ne \emptyset$ , and a list $F_0, \dots, F_n$ of members of $\mathcal{A}$ has $F_i = X \setminus U_i$ with $U_i := X \setminus F_i \in \mathcal{U}$ , so $F_0 \cap \dots \cap F_n = \emptyset$ would give $U_0 \cup \dots \cup U_n = X$ by step 2.1, a finite subfamily with union $X$ .

L3step 2.1step 9.1

11.1

By the assumed condition $\bigcap \mathcal{A} \ne \emptyset$ ; but $\bigcap \mathcal{A} = X \setminus \bigcup \mathcal{U} = \emptyset$ because $\mathcal{U}$ has union $X$ , and this contradiction is the required one.

step 1.2step 10.1discharge-contradiction

12.1

Hence some finite subfamily of $\mathcal{U}$ has union $X$ , so $(X,d)$ is compact, and with step 7.1 both implications are proved.

L1step 7.1step 11.1∎

Remarks

The empty family and the empty space. The conventions are not decoration. $\bigcap \emptyset = X$ makes the finite intersection property fail outright for every family of subsets of the empty space, so the right-hand condition is vacuously true there; and the empty space is compact, the empty subfamily covering it. The equivalence therefore holds at $X = \emptyset$ as well, with both sides true.

Why no choice is spent. The natural-looking step "the finite subcover consists of sets $X \setminus F_i$ , so pick $F_i \in \mathcal{A}$ " would be a selection if a member of the family could be the complement of several different members. It cannot: complementation inside $X$ is injective, so $F_i$ is recovered from $U_i$ by a formula. That is the whole reason this characterisation, and the completeness half of A compact metric space is complete and totally bounded, and neither implication uses any choice principle that runs through it, cost nothing.

Depends on

Used by

A compact metric space is complete and totally bounded, and neither implication uses any choice principle Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 40 results over 14 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Finite intersection property (Wikipedia) (standard reference, not scraped)
Compact space (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §26 (standard reference, not scraped)