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

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Every open subset of $R$ is a countable disjoint union of open intervals, namely its order components

Statement

Let $U \subseteq R$ be open (Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen). For $x, y \in R$ write

$H (x, y) : = {z \in R : x \leq z \leq y or y \leq z \leq x}$

for the order-convex hull of the pair, and define a relation on $U$ by

$x \sim y : ⟺ H (x, y) \subseteq U .$

Then $\sim$ is an equivalence relation on $U$ . Its equivalence classes, called the order components of $U$ , form a family $C$ with the following properties:

the members of $C$ are nonempty and pairwise disjoint, and $U = ⋃ C$ ;
every member of $C$ is an interval of one of the four open forms $(a, b)$ , $(a, \infty)$ , $(- \infty, b)$ , $(- \infty, \infty)$ of Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, and is an open set;
$C$ is at most countable (Finite, countably infinite, countable, uncountable).

So every open subset of $R$ is the union of an at most countable family of pairwise disjoint nonempty open intervals. For $U = \emptyset$ the family $C$ is empty and the union of the empty family is $\emptyset$ , so the statement holds in that case too.

No choice principle is used. The components are defined by an explicit equivalence relation, and the enumeration in claim 3 is obtained by sending a component to the least index of a rational lying in it, which is canonical by The well-ordering principle.

Facts & Assumptions

Given: An open set $U \subseteq R$ , the hull $H (x, y)$ and the relation $\sim$ as displayed in the Statement. Write $Q_{R}$ for the image of $Q$ in $R$ under the canonical embedding $q \mapsto \hat{q}$ .

[L1]

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

[L2]

$N_{ε} (u) = (u - ε, u + ε)$ , and $u \in N_{ε} (u)$ (The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ ).

[L3]

Order-convexity and the nine interval forms; each of the nine is order-convex, and $(a, b)$ , $(a, \infty)$ , $(- \infty, b)$ , $(- \infty, \infty)$ are the open forms; trichotomy and transitivity of the order (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Ordered field, Complete ordered field (least-upper-bound property)).

[L4]

Least-upper-bound property: a nonempty subset of $R$ bounded above has a least upper bound, unique, and dually a nonempty subset bounded below has a greatest lower bound, unique (Complete ordered field (least-upper-bound property), Every nonempty set bounded below has an infimum, Greatest lower bound (infimum), Suprema and infima are unique).

[L5]

Epsilon characterisations: for nonempty $S$ bounded above and $b = \sup S$ , every $ε > 0$ admits $s \in S$ with $b - ε < s$ ; for nonempty $S$ bounded below and $a = \inf S$ , every $ε > 0$ admits $s \in S$ with $s < a + ε$ (Epsilon characterisation of the supremum, Epsilon characterisation of the infimum).

[L6]

Bounded above, bounded below, and their negations: $S$ fails to be bounded above exactly when for every $w \in R$ there is $v \in S$ with $w < v$ , and fails to be bounded below exactly when for every $w$ there is $t \in S$ with $t < w$ (Lower bound, bounded below, bounded set, Complete ordered field (least-upper-bound property)).

[L7]

Strictly between any two reals lies an element of $Q_{R}$ , and $q \mapsto \hat{q}$ is injective (The rationals embed densely in the reals).

[L8]

$Q \approx N$ ( $Q$ is countably infinite); a composition of bijections is a bijection and an injection is a bijection onto its image (Injection, surjection, bijection, Equinumerous sets, $A \approx B$ and $A ⪯ B$ ); every subset of an at most countable set is at most countable (Every subset of an at most countable set is at most countable, Finite, countably infinite, countable, uncountable).

[L9]

Every nonempty subset of $N$ has a least element (The well-ordering principle).

Proof

technique · constructive

1.1

The hull satisfies $x, y \in H (x, y)$ , $H (x, y) = H (y, x)$ and $H (x, x) = {x}$ , and for all $x, y, z$ one has $H (x, z) \subseteq H (x, y) \cup H (y, z)$ : given $w \in H (x, z)$ , either $x \leq w \leq z$ , in which case $w \leq y$ puts $w$ in $H (x, y)$ and $y < w$ puts $w$ in $H (y, z)$ , or $z \leq w \leq x$ , in which case $w \leq y$ puts $w$ in $H (y, z)$ and $y < w$ puts $w$ in $H (x, y)$ . Hence $\sim$ is reflexive on $U$ (as $H (x, x) = {x} \subseteq U$ ), symmetric, and transitive.

givenL3

1.2

Let $C \subseteq R$ be nonempty, open and order-convex, and let $u \in C$ ; fix $ε > 0$ with $N_{ε} (u) \subseteq C$ . Then $u - ε / 2$ and $u + ε / 2$ lie in $N_{ε} (u) \subseteq C$ , so $u$ is neither an upper bound nor a lower bound of $C$ .

L1L2L3choose

1.3

For $x \in U$ put $C_{x} : = {y \in U : H (x, y) \subseteq U}$ , the equivalence class of $x$ , and let $C : = {C_{x} : x \in U}$ .

construct

2.1

Each $C_{x}$ is nonempty because $x \in C_{x}$ ; two classes of an equivalence relation are equal or disjoint; and every $x \in U$ lies in $C_{x}$ , so $U = ⋃ C$ . This is claim 1.

step 1.1step 1.3

2.2

Each $C_{x}$ is order-convex: let $u, v \in C_{x}$ and $u \leq w \leq v$ . From $u \sim x$ and $x \sim v$ we get $u \sim v$ , so $H (u, v) \subseteq U$ ; since $w \in H (u, v)$ we get $w \in U$ , and $H (u, w) \subseteq H (u, v) \subseteq U$ because every $t$ with $u \leq t \leq w$ satisfies $u \leq t \leq v$ , so $u \sim w$ and hence $w \in C_{u} = C_{x}$ .

step 1.1step 1.3L3

2.3

Each $C_{x}$ is open: let $u \in C_{x} \subseteq U$ and fix $ε > 0$ with $N_{ε} (u) \subseteq U$ . For $y \in N_{ε} (u)$ the hull $H (u, y)$ is contained in the order-convex set $N_{ε} (u)$ , hence in $U$ , so $u \sim y$ and $y \in C_{u} = C_{x}$ ; therefore $N_{ε} (u) \subseteq C_{x}$ .

step 1.1step 1.3L1L2L3choose

2.4

Let $C$ be nonempty, open and order-convex and bounded both above and below; then $a : = \inf C$ and $b : = \sup C$ exist by [L4]. Every $u \in C$ satisfies $a \leq u \leq b$ , and $u$ is neither an upper nor a lower bound of $C$ , so $u \neq a$ and $u \neq b$ , giving $a < u < b$ ; in particular $a < b$ and $C \subseteq (a, b)$ . Conversely let $a < w < b$ : by [L5] with $ε = b - w$ there is $v \in C$ with $w < v$ , and with $ε = w - a$ there is $t \in C$ with $t < w$ , so $t \leq w \leq v$ and order-convexity gives $w \in C$ . Hence $C = (a, b)$ .

step 1.2L3L4L5

2.5

Let $C$ be nonempty, open and order-convex. If $C$ is bounded below and not above, put $a : = \inf C$ ; as in the bounded case every $u \in C$ satisfies $a < u$ , and for $w > a$ the fact [L5] supplies $t \in C$ with $t < w$ while [L6] supplies $v \in C$ with $w < v$ , so $w \in C$ by order-convexity; hence $C = (a, \infty)$ . Symmetrically, if $C$ is bounded above and not below then $C = (- \infty, b)$ with $b : = \sup C$ . If $C$ is bounded neither above nor below then for every $w$ the fact [L6] supplies $t, v \in C$ with $t < w < v$ , so $w \in C$ and $C = R$ .

step 1.2L3L4L5L6

3.1

Every member of $C$ is nonempty, open and order-convex by steps 2.1, 2.2 and 2.3, and it is bounded above or not and bounded below or not, so steps 2.4 and 2.5 exhibit it as an interval of one of the four open forms; this is claim 2.

step 2.1step 2.2step 2.3step 2.4step 2.5L3

3.2

Every member $C$ of $C$ contains an element of $Q_{R}$ : pick $u \in C$ and, by openness, $ε > 0$ with $N_{ε} (u) \subseteq C$ ; since $u - ε < u + ε$ , the fact [L7] supplies $\hat{q}$ with $u - ε < \hat{q} < u + ε$ , and $N_{ε} (u) = (u - ε, u + ε)$ by [L2], so $\hat{q} \in C$ .

step 2.1step 2.3L2L7choose

4.1

By [L8] fix a bijection $β : N \to Q$ ; then $e : = ι \circ β$ , where $ι (q) = \hat{q}$ , is a bijection from $N$ onto $Q_{R}$ by [L7] and [L8]. For $C \in C$ the set ${n \in N : e (n) \in C}$ is nonempty by step 3.2, so $Φ (C) : = \min {n \in N : e (n) \in C}$ is defined by [L9] and no selection is made; and $Φ$ is injective, since $e (Φ (C)) \in C$ and distinct members of $C$ are disjoint by step 2.1.

step 2.1step 3.2L7L8L9construct

5.1

Hence $C$ is in bijection with $Φ [C] \subseteq N$ , and a subset of $N$ is at most countable, so $C$ is at most countable; this is claim 3.

step 4.1L8

6.1

The family $C$ constructed in step 1.3 therefore consists of pairwise disjoint nonempty open intervals whose union is $U$ , and it is at most countable, which is exactly the assertion.

step 2.1step 3.1step 5.1discharge-construct∎

Remarks

The components are forced, not chosen. A component is an equivalence class of an explicitly written relation, so the family $C$ is determined by $U$ alone, with no selection anywhere. One half of the usual uniqueness statement is immediate from that: if $U$ is written as a union of nonempty open intervals, each of those intervals is order-convex and contained in $U$ , so any two of its points are equivalent and the whole interval lies inside a single component. That the intervals must then be the components is the other half, and it is neither needed below nor proved here.
Where completeness is spent. Only in steps 2.4 and 2.5, which produce $\inf C$ and $\sup C$ from the least-upper-bound property. Everything else uses the order alone. The argument therefore does not transpose to an arbitrary ordered field, where the two bounds it asks for need not exist; the standard obstruction is the set of positive rationals whose square is below $2$ , which is bounded above in $Q$ and has no supremum there ( $\sup {q \in Q : q > 0, q^{2} < 2} = \sqrt{2}$ in $R$ , and no supremum in $Q$ ).
The two sizes in the statement pull in opposite directions. Each single component is an uncountable set, being a nonempty open set (Both $Q$ and $R ∖ Q$ are dense in $R$ , and every nonempty open subset of $R$ is uncountable), while the family of components is at most countable. There is no tension: the count in claim 3 is a count of components, not of points, and the injection of step 4.1 is into $N$ through the rationals, which are countable and dense at once.
This is one of the results whose statement is order vocabulary throughout, and Which results on this page use the order of $R$ and therefore have no general-topological analogue collects them: interval, disjoint union of intervals, and the components themselves are all defined from the order, so there is nothing here to restate where no order is present.

Depends on

Used by

Dependency tree · two levels

62 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)
J. Lebl, Basic Analysis I, §7.2 (standard reference, not scraped)
J. K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)

Every open subset of RR is a countable disjoint union of open intervals, namely its order components