TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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Arbitrary unions and finite intersections of open subsets of $R$ are open, and dually for closed sets

Statement

Let open and closed subsets of $R$ be as in Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen.

Arbitrary unions of open sets are open. If $U$ is any family of open subsets of $R$ , then $⋃ U$ is open.
Finite intersections of open sets are open. If $n \in N$ and $U_{0}, \dots, U_{n}$ are open, then $U_{0} \cap \dots \cap U_{n}$ is open.
Arbitrary intersections of closed sets are closed. If $F$ is a nonempty family of closed subsets of $R$ , then $⋂ F$ is closed.
Finite unions of closed sets are closed. If $n \in N$ and $F_{0}, \dots, F_{n}$ are closed, then $F_{0} \cup \dots \cup F_{n}$ is closed.

The word finite in claims 2 and 4 is not decoration: an arbitrary intersection of open sets need not be open, and dually an arbitrary union of closed sets need not be closed; the remarks below say where that is settled. Claim 3 asks $F$ to be nonempty only so that $⋂ F$ is a subset of $R$ without appeal to a convention about the empty intersection.

Facts & Assumptions

Given: A family $U$ of open subsets of $R$ , with $⋃ U = {x \in R : x \in U for some U \in U}$ ; a natural number $n$ and open sets $U_{0}, \dots, U_{n}$ ; a nonempty family $F$ of closed subsets of $R$ , with $⋂ F = {x \in R : x \in F for every F \in F}$ ; and closed sets $F_{0}, \dots, F_{n}$ .

[A1]

De Morgan's laws in the ambient set theory: for a nonempty family $G$ of subsets of $R$ , $R ∖ ⋂ G = ⋃ {R ∖ G : G \in G}$ , and $R ∖ (G_{0} \cup \dots \cup G_{n}) = (R ∖ G_{0}) \cap \dots \cap (R ∖ G_{n})$ . Also $R ∖ (R ∖ G) = G$ .

[L1]

$U$ is open when every $x \in U$ admits a real $ε > 0$ with $N_{ε} (x) \subseteq U$ ; $F$ is closed when $R ∖ F$ is open (Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen).

[L2]

$N_{ε} (x) = {y : ∣ y - x ∣ < ε}$ , and $0 < δ \leq ε$ gives $N_{δ} (x) \subseteq N_{ε} (x)$ (The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ ).

[L3]

Every nonempty finite set of reals has a minimum, so $\min {ε, η}$ is defined and equals one of the two entries, and is $\leq$ both (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

Proof

technique · direct

1.1givenL1choose

Let $x \in ⋃ U$ . Then $x \in U$ for some $U \in U$ , and $U$ is open, so there is $ε > 0$ with $N_{ε} (x) \subseteq U \subseteq ⋃ U$ ; as $x$ was arbitrary, $⋃ U$ is open, which is claim 1.

1.2givenL1choose

Now let $U$ and $V$ be open and let $x \in U \cap V$ ; fix $ε > 0$ with $N_{ε} (x) \subseteq U$ and $η > 0$ with $N_{η} (x) \subseteq V$ .

2.1step 1.2L2L3

Put $δ : = \min {ε, η}$ , which is one of $ε, η$ and hence $> 0$ , and satisfies $δ \leq ε$ and $δ \leq η$ ; then $N_{δ} (x) \subseteq N_{ε} (x) \subseteq U$ and $N_{δ} (x) \subseteq N_{η} (x) \subseteq V$ , so $N_{δ} (x) \subseteq U \cap V$ , and as $x$ was arbitrary $U \cap V$ is open.

2.2step 1.1A1L1

The family ${R ∖ F : F \in F}$ consists of open sets by [L1], so its union is open by step 1.1; that union is $R ∖ ⋂ F$ by [A1], so $⋂ F$ is closed, which is claim 3.

3.1step 2.1given

Claim 2 now follows by induction on $n$ : for $n = 0$ the intersection is $U_{0}$ , which is open by hypothesis; and if $W : = U_{0} \cap \dots \cap U_{n}$ is open then $U_{0} \cap \dots \cap U_{n + 1} = W \cap U_{n + 1}$ is an intersection of two open sets, hence open by step 2.1.

4.1step 3.1A1L1

Each $R ∖ F_{i}$ is open by [L1], so $(R ∖ F_{0}) \cap \dots \cap (R ∖ F_{n})$ is open by step 3.1; that set is $R ∖ (F_{0} \cup \dots \cup F_{n})$ by [A1], so $F_{0} \cup \dots \cup F_{n}$ is closed, which is claim 4.

5.1step 1.1step 2.2step 3.1step 4.1∎

Claims 1, 2, 3 and 4 are steps 1.1, 3.1, 2.2 and 4.1 respectively, so arbitrary unions and finite intersections of open sets are open, and arbitrary intersections and finite unions of closed sets are closed.

Remarks

Completeness plays no part. Nothing above uses the least-upper-bound property, or even the Archimedean property: the only facts about $R$ the proof touches are the definition of a neighbourhood, its monotonicity in the radius, and the comparison of two positive radii. What needs completeness is not the algebra of open sets but the theorems about compactness that come later.
Why finiteness cannot be dropped in claim 2. The minimum taken in step 2.1 is a minimum of finitely many positive radii, and it is positive precisely because it is one of them (Every nonempty finite set of reals has a maximum and a minimum). An infinite family of positive radii has an infimum that may be $0$ , and then no positive $δ$ survives. That is exactly what happens for the shrinking intervals of FALSE: an arbitrary intersection of open subsets of $R$ is open, whose named witness is $⋂_{k} (- 1 / k, 1 / k) = {0}$ is not open ↗.
The four claims are a rewriting of two. Claims 3 and 4 are claims 1 and 2 read through complementation, and closedness is defined by complementation (Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen), so no separate argument about closed sets is possible or needed.

Arbitrary unions and finite intersections of open subsets of $R$ are open, and dually for closed sets

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Arbitrary unions and finite intersections of open subsets of RR are open, and dually for closed sets

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Arbitrary unions and finite intersections of open subsets of $R$ are open, and dually for closed sets