LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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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.

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Baire category inside a closed bounded interval: if $[a, b]$ with $a < b$ is covered by a sequence of closed sets, then one of them contains a nondegenerate closed subinterval of $[a, b]$ ; no choice principle is used

Statement

Let $a, b \in R$ with $a < b$ and let $(F_{n})_{n \in N}$ be a sequence of closed subsets of $R$ (Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen) with

$[a, b] \subseteq ⋃_{n \in N} F_{n}$

(Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length). Then there are $n \in N$ and reals $u < v$ with

$[u, v] \subseteq F_{n} \cap [a, b] .$

No choice principle is used. The only category input is Baire category in $R$ , by nested intervals with canonically chosen rational endpoints: a countable intersection of dense open sets is dense, so $R$ is not a countable union of nowhere dense sets, whose own proof selects nothing: it fixes one enumeration of the rationals and takes least indices. Nothing further is chosen below, the argument being a direct application of that theorem to the complements of the $F_{n}$ .

Facts & Assumptions

Given: Reals $a < b$ and a sequence $(F_{n})_{n \in N}$ of closed subsets of $R$ with $[a, b] \subseteq ⋃_{n} F_{n}$ .

[L1]

A countable intersection of dense open subsets of $R$ is dense in $R$ (Baire category in $R$ , by nested intervals with canonically chosen rational endpoints: a countable intersection of dense open sets is dense, so $R$ is not a countable union of nowhere dense sets); dense means that the closure is $R$ (Limit point, isolated point, adherent point, derived set, and dense subset of $R$ ).

[L2]

$\overline{S} = R$ if and only if $N_{ε} (y) \cap S \neq \emptyset$ for every real $y$ and every real $ε > 0$ (The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points, claim 1, Interior, closure, boundary and exterior of a subset of $R$ , The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ ).

[L3]

An intersection of two closed sets is closed, and the complement of a closed set is open (Arbitrary unions and finite intersections of open subsets of $R$ are open, and dually for closed sets, claim 3, Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen).

[L4]

$[a, b]$ is closed: its complement ${x : x < a} \cup {x : x > b}$ is open, since $x < a$ gives $N_{a - x} (x) \subseteq {z : z < a}$ and $x > b$ gives $N_{x - b} (x) \subseteq {z : z > b}$ (Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen, The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L5]

$N_{ε} (y) = (y - ε, y + ε)$ , and for $a < b$ the midpoint $y : = (a + b) / 2$ and radius $ε : = (b - a) / 2 > 0$ give $N_{ε} (y) = (a, b) \subseteq [a, b]$ (The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

Proof

technique · contradiction

1.1

Put $G_{n} : = F_{n} \cap [a, b]$ for $n \in N$ . Each $G_{n}$ is closed, being an intersection of two closed sets, and $[a, b] = ⋃_{n} G_{n}$ , since $[a, b]$ is contained in the union of the $F_{n}$ and each $G_{n}$ is contained in $[a, b]$ .

L3L4

1.2

Suppose, for contradiction, that no $G_{n}$ contains a nondegenerate closed interval, that is, that there are no $n$ and no reals $u < v$ with $[u, v] \subseteq G_{n}$ .

assume-contra

2.1

Each $V_{n} : = R ∖ G_{n}$ is open, and it is dense. Openness is the complement of a closed set. For density, let $y$ be real and $ε > 0$ real; if $N_{ε} (y) \cap V_{n}$ were empty then $N_{ε} (y) \subseteq G_{n}$ , and then $[y - ε / 2, y + ε / 2]$ would be a nondegenerate closed interval inside $G_{n}$ , contrary to step 1.2.

step 1.1step 1.2L2L3L5

3.1

By the Baire category theorem the intersection $⋂_{n} V_{n}$ is dense in $R$ , so it meets the neighbourhood $N_{(b - a) / 2} ((a + b) / 2) = (a, b)$ : there is $x \in (a, b)$ with $x \notin G_{n}$ for every $n$ .

step 2.1L1L2L5

4.1

But $x \in (a, b) \subseteq [a, b] = ⋃_{n} G_{n}$ , so $x \in G_{n}$ for some $n$ , contradicting step 3.1. The assumption of step 1.2 is therefore false, and some $G_{n} = F_{n} \cap [a, b]$ contains a nondegenerate closed interval $[u, v]$ .

step 1.1step 1.2step 3.1discharge-contradiction∎

Remarks

Why the statement is about $[a, b]$ and not about $R$ . Baire category in $R$ , by nested intervals with canonically chosen rational endpoints: a countable intersection of dense open sets is dense, so $R$ is not a countable union of nowhere dense sets says that $R$ is not a countable union of nowhere dense sets. What is needed for Baire's theorem on functions of the first class (Baire's theorem: a Baire class one function on a closed bounded interval $[a, b]$ is continuous at the points of a dense subset of $[a, b]$ that is the trace of a $G_{δ}$ set, so its set of discontinuities is meager) is the same statement localised to a closed bounded interval, and the localisation is not formal: a closed set may be nowhere dense in $R$ and yet fill an interval, so the conclusion has to be stated as "contains a nondegenerate closed subinterval" rather than "has nonempty interior in $R$ ". The two are in fact the same condition here, which is what step 2.1 uses.
The hypothesis $a < b$ is not decoration. For $a = b$ the set $[a, b]$ is a single point, it is covered by the constant sequence $F_{n} = {a}$ , and no $F_{n}$ contains a nondegenerate closed interval; the conclusion fails, and the proof breaks at step 3.1, where $(a, b)$ is empty.

Depends on

Used by

Dependency tree · two levels

26 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

Baire category theorem (Wikipedia) (standard reference, not scraped)

Baire category inside a closed bounded interval: if [a,b][a,b] with a<ba<b is covered by a sequence of closed sets, then one of them contains a nondegenerate closed subinterval of [a,b][a,b]; no choice principle is used