CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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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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$[0, 1)$ is neither open nor closed in $R$

Statement refuted

Refuted claim: every subset of $R$ is open or closed (FALSE: every subset of $R$ is either open or closed, Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen).

The witness is the half-open interval $E : = [0, 1)$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length). It fails openness at its left endpoint $0$ , which belongs to $E$ while every neighbourhood of $0$ reaches below $0$ , and it fails closedness at $1$ , which does not belong to $E$ while every neighbourhood of $1$ reaches into $E$ . The refutation is carried out in full in FALSE: every subset of $R$ is either open or closed and is recorded here as the named counterexample.

Facts & Assumptions

Given: The interval $E : = [0, 1) = {x \in R : 0 \leq x < 1}$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[A1]

The refuted claim: every subset of $R$ is open or closed.

[L1]

$E$ is neither open nor closed (FALSE: every subset of $R$ is either open or closed).

[L2]

$U$ is open when every $x \in U$ admits $ε > 0$ with $N_{ε} (x) \subseteq U$ ; $F$ is closed when $R ∖ F$ is open; $N_{ε} (x) = {y : ∣ y - x ∣ < ε}$ (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 $R$ ).

[L3]

Every nonempty finite set of reals has a minimum, which is one of its members (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L4]

Absolute value: $∣ z ∣ = z$ for $z \geq 0$ and $∣ z ∣ = - z$ for $z < 0$ (Basic properties of the absolute value); $0 < 1$ , so $2 : = 1 + 1 > 0$ and $0 < d \cdot 2^{- 1} < d$ for $d > 0$ ; adding a constant preserves an inequality and the order is total (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field, Complete ordered field (least-upper-bound property)). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used below follow by adjoining the equality case, in which the two sides coincide.

Counterexample

technique · direct

1.1

$0 \in E$ and $1 \notin E$ , so $E$ is a legitimate instance of the claim [A1] and $1 \in R ∖ E$ .

A1L4

1.2

$E$ is not open: for a real $ε > 0$ the point $y : = - (ε \cdot 2^{- 1})$ satisfies $∣ y - 0 ∣ = ε \cdot 2^{- 1} < ε$ by [L4], so $y \in N_{ε} (0)$ , while $y < 0$ puts $y$ outside $E$ . Hence no neighbourhood of the point $0$ of $E$ is contained in $E$ .

L1L2L4

1.3

$E$ is not closed: for a real $ε > 0$ put $t : = \min {ε, 1} \cdot 2^{- 1}$ , positive by [L3] and [L4], and $y : = 1 - t$ ; then $t \leq 1 \cdot 2^{- 1}$ gives $y \geq 1 - 1 \cdot 2^{- 1} > 0$ , and $t > 0$ gives $y < 1$ , so $y \in E$ ; and $∣ y - 1 ∣ = t \leq ε \cdot 2^{- 1} < ε$ , so $y \in N_{ε} (1)$ . Hence no neighbourhood of the point $1$ of $R ∖ E$ is contained in $R ∖ E$ , so $R ∖ E$ is not open.

L1L2L3L4

2.1

By steps 1.2 and 1.3 the set $E$ is neither open nor closed, so the claim [A1] fails at $E$ and is refuted.

step 1.1step 1.2step 1.3A1L1L2∎

Remarks

The two failures are at different points and are independent. Openness fails only at $0$ , since every $x$ with $0 < x < 1$ has a neighbourhood inside $E$ ; closedness fails only at $1$ , since every point outside $E$ other than $1$ has a neighbourhood outside $E$ . Repairing either failure separately gives a set that is open or closed but not both: $(0, 1)$ is open, $[0, 1]$ is closed.
Nothing about the true results is contradicted. Arbitrary unions and finite intersections of open subsets of $R$ are open, and dually for closed sets says which combinations of open sets are open and which combinations of closed sets are closed; it says nothing about arbitrary sets, and "closed" was never the negation of "open" (Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen).
The mirror witness. $(0, 1]$ is neither open nor closed for the same two reasons with the roles of the endpoints exchanged, and $Q$ is a subset that is neither open nor closed with no endpoints at all ( $Q$ has closure $R$ , empty interior, and boundary $R$ ).

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

22 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)
Closed set (Wikipedia) (standard reference, not scraped)
Interval (mathematics) (Wikipedia) (standard reference, not scraped)
MIT 18.100, Test 1 solutions (standard reference, not scraped)

[0,1)[0,1) is neither open nor closed in RR