False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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FALSE: every subset of $\mathbb{R}$ is either open or closed

Statement

False claim: every subset of $\mathbb{R}$ is open or closed (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen).

The claim treats "closed" as the negation of "open". It is not: closedness of a set is openness of its complement, and both conditions can fail at once. The half-open interval $[0,1)$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length) is the standard witness, and it fails each condition at a different point, at $0$ for openness and at $1$ for closedness.

Facts & Assumptions

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

[A1]

The false claim: every subset of $\mathbb{R}$ is open or closed.

[L1]

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

[L2]

$N_\varepsilon(x) = \{\, y : |y - x| < \varepsilon \,\}$ (The $\varepsilon$ -neighbourhood and the punctured $\varepsilon$ -neighbourhood of a point of $\mathbb{R}$ ).

[L3]

Every nonempty finite set of reals has a minimum, which is one of its members and is $\le$ both entries of a two-element set (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L4]

Absolute value: $|z| = z$ for $z \ge 0$ and $|z| = -z$ for $z < 0$ ; $|z| \ge 0$ (Basic properties of the absolute value).

[L5]

Ordered-field arithmetic: $0 < 1$ , so $2 := 1+1 > 0$ and $0 < d \cdot 2^{-1} < d$ for every $d > 0$ ; adding a constant preserves an inequality; the order is total and transitive (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.

Refutation

technique · direct

1.1

$0 \in E$ and $1 \notin E$ , since $0 \le 0 < 1$ while $1 < 1$ fails; so $1 \in \mathbb{R} \setminus E$ .

givenL5

1.2

$E$ is not open: let $\varepsilon > 0$ be real and put $y := -(\varepsilon \cdot 2^{-1})$ . Then $|y - 0| = \varepsilon \cdot 2^{-1} < \varepsilon$ by [L4] and [L5], so $y \in N_\varepsilon(0)$ ; but $y < 0$ , so $y \notin E$ . Hence no neighbourhood of the point $0$ of $E$ is contained in $E$ .

L1L2L4L5

1.3

$\mathbb{R} \setminus E$ is not open: let $\varepsilon > 0$ be real, put $t := \min\{\varepsilon, 1\} \cdot 2^{-1}$ , which is positive and satisfies $t \le \varepsilon \cdot 2^{-1} < \varepsilon$ and $t \le 1 \cdot 2^{-1}$ by [L3] and [L5], and put $y := 1 - t$ . Then $y \ge 1 - 1 \cdot 2^{-1} > 0$ and $y < 1$ , so $y \in E$ ; and $|y - 1| = t < \varepsilon$ by [L4], so $y \in N_\varepsilon(1)$ . Hence no neighbourhood of the point $1$ of $\mathbb{R} \setminus E$ is contained in $\mathbb{R} \setminus E$ .

L1L2L3L4L5

2.1

By step 1.2 the set $E$ is not open, and by steps 1.1 and 1.3 its complement is not open, so $E$ is not closed either. The subset $E$ of $\mathbb{R}$ is therefore neither open nor closed, and the claim [A1] is false.

step 1.1step 1.2step 1.3A1L1∎

Remarks

The two failures are independent and happen at different points. Openness fails only at $0$ : every $x$ with $0 < x < 1$ does have a neighbourhood inside $E$ . Closedness fails only at $1$ : every $x$ outside $E$ other than $1$ does have a neighbourhood outside $E$ . So the set is one point short of open and one point short of closed, and the two repairs move the endpoint in opposite directions, as the next remark records.
The four possibilities all occur. $\mathbb{R}$ and $\varnothing$ are both open and closed, $(0,1)$ is open and not closed, $[0,1]$ is closed and not open, and $[0,1)$ is neither (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen). "Open" and "closed" are two independent properties, not two values of one property.
The named witness is $[0,1)$ is neither open nor closed in $\mathbb{R}$ ↗; the refutation itself is carried out here.

Depends on

Used by

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

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 28 results over 9 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Open set (Wikipedia) (standard reference, not scraped)
Closed set (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 2 (Remark 2.28) (standard reference, not scraped)
MIT 18.100, Test 1 solutions (standard reference, not scraped)

FALSE: every subset of R\mathbb{R}R is either open or closed