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

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.

An explicit open subset of $R$ written as the disjoint union of its component intervals

Example

Take

$U : = (- 1, 0) \cup (0, 1) \cup (2, \infty) .$

This is an open subset of $R$ , and its order components in the sense of Every open subset of $R$ is a countable disjoint union of open intervals, namely its order components are exactly the three intervals $(- 1, 0)$ , $(0, 1)$ and $(2, \infty)$ : they are pairwise disjoint, their union is $U$ , and there are three of them, a finite and hence at most countable family. The example is chosen so that one component is unbounded and two are bounded, and so that the two bounded ones are separated by a single missing point, $0$ , rather than by a gap of positive length.

Facts & Assumptions

Given: The set $U : = (- 1, 0) \cup (0, 1) \cup (2, \infty)$ , together with the order-convex hull $H (x, y) = {z : x \leq z \leq y or y \leq z \leq x}$ and the relation $x \sim y ⟺ H (x, y) \subseteq U$ on $U$ , both as in Every open subset of $R$ is a countable disjoint union of open intervals, namely its order components. Write $A : = (- 1, 0)$ , $B : = (0, 1)$ and $C : = (2, \infty)$ .

[L1]

For an open $U$ , the relation $\sim$ is an equivalence relation on $U$ , its classes are the order components, and they are nonempty pairwise disjoint open intervals whose union is $U$ , forming an at most countable family (Every open subset of $R$ is a countable disjoint union of open intervals, namely its order components).

[L2]

Each of the forms $(a, b)$ and $(a, \infty)$ is an open set, and a union of open sets is open (Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Arbitrary unions and finite intersections of open subsets of $R$ are open, and dually for closed sets).

[L3]

Each of the nine interval forms is order-convex (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

Verification

technique · direct

1.1

$U$ is open: $A$ , $B$ and $C$ are open sets by [L2], and their union is open by [L2].

1.2

$A$ , $B$ and $C$ are pairwise disjoint with union $U$ : an element of $A$ is negative, an element of $B$ lies strictly between $0$ and $1$ , and an element of $C$ exceeds $2$ , so no two of the three share a point, and the union is $U$ by definition.

givenL3

2.1

Any two points of the same one of $A$ , $B$ , $C$ are equivalent: if $x, y \in A$ then $H (x, y) \subseteq A \subseteq U$ because $A$ is order-convex by [L3], so $x \sim y$ ; the same argument applies inside $B$ and inside $C$ .

step 1.2L1L3

2.2

No two points of different ones of $A$ , $B$ , $C$ are equivalent: for $x \in A$ and $y \in B$ , or for $x \in A$ and $y \in C$ , one has $x < 0 < y$ , so $0 \in H (x, y)$ while $0 \notin U$ ; for $x \in B$ and $y \in C$ one has $x < 2 < y$ , so $2 \in H (x, y)$ while $2 \notin U$ . In each case $H (x, y) \subseteq̸ U$ and $x \sim̸ y$ .

step 1.2L1

3.1

Every point of $U$ lies in exactly one of $A$ , $B$ , $C$ by step 1.2, and by steps 2.1 and 2.2 its equivalence class is precisely that one of the three; so the order components of $U$ are exactly $A$ , $B$ and $C$ , three pairwise disjoint nonempty open intervals with union $U$ , which is the decomposition promised by [L1].

step 1.2step 2.1step 2.2L1∎

Remarks

The count is the number of components, not the number of points. There are three components here, while each of them is an uncountable set (Both $Q$ and $R ∖ Q$ are dense in $R$ , and every nonempty open subset of $R$ is uncountable). The at most countable family of Every open subset of $R$ is a countable disjoint union of open intervals, namely its order components is a family of intervals, and a finite family is one instance of it.
What keeps two components apart may be a single missing point. $A$ and $B$ are kept apart by $0 \notin U$ alone, and no gap of positive length is required, although $B$ and $C$ happen to have one. This is why the components are defined by an equivalence relation on $U$ and not by measuring distances between the pieces.
Reading the decomposition off the formula is legitimate here only because the three pieces were checked to be the classes. A presentation of an open set as a union of open intervals is not automatically its decomposition into components: $(0, 2) = (0, 1) \cup (0, 2)$ writes an open set as a union of open intervals that are neither disjoint nor components.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

23 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

Open set (Wikipedia) (standard reference, not scraped)
Interval (mathematics) (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 2 (Exercise 2.29) (standard reference, not scraped)

An explicit open subset of RR written as the disjoint union of its component intervals