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

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$\mathbb{Q} \cap [0,2]$ is bounded and disconnected, so being an interval of $\mathbb{Q}$ is not enough

Statement refuted

Refuted claim: the set $E := \mathbb{Q}_{\mathbb{R}} \cap [0,2]$ of rationals between $0$ and $2$ is connected (Separated sets, disconnection, and connected subset of $\mathbb{R}$ ), where $\mathbb{Q}_{\mathbb{R}}$ is the copy of $\mathbb{Q}$ inside $\mathbb{R}$ (The rationals embed densely in the reals).

$E$ is bounded, and it contains every rational lying between its two endpoints, so it is order-convex as a subset of the ordered field $\mathbb{Q}$ : it is an interval of that field. As a subset of $\mathbb{R}$ it is nevertheless disconnected, split at the irrational point $\sqrt 2$ . So the equivalence of A subset of $\mathbb{R}$ is connected if and only if it is order-convex, that is, an interval genuinely uses the completeness of $\mathbb{R}$ , and "is an interval of the order it carries from $\mathbb{Q}$ " is not enough to make a set connected.

Facts & Assumptions

Given: The copy $\mathbb{Q}_{\mathbb{R}}$ of $\mathbb{Q}$ in $\mathbb{R}$ , the set $E := \mathbb{Q}_{\mathbb{R}} \cap [0,2]$ , and the real $r$ with $r \ge 0$ and $r^2 = 2$ .

[A1]

The refuted claim: $E$ is connected.

[L1]

Separated sets, disconnection and connectedness (Separated sets, disconnection, and connected subset of $\mathbb{R}$ ).

[L2]

$\overline{A}$ is the smallest closed superset of $A$ , so $\overline{A} \subseteq F$ for every closed $F \supseteq A$ (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, Interior, closure, boundary and exterior of a subset of $\mathbb{R}$ ).

[L3]

Each of $(-\infty, c]$ and $[c,\infty)$ is a closed set and each of $(-\infty,c)$ and $(c,\infty)$ is an open set (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen, Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

[L4]

In a complete ordered field every $a \ge 0$ has a unique $s \ge 0$ with $s^2 = a$ (Square roots exist: a unique $\sqrt{a} \ge 0$ with $(\sqrt{a})^2 = a$ ; the positives are $\{x^2 : x \neq 0\}$ ).

[L5]

No rational squares to $2$ (FALSE: some rational number squares to 2); the map $q \mapsto \hat q$ is an injective embedding of ordered fields, so it preserves sums, products and the order (The rationals embed densely in the reals, The rationals as equivalence classes of pairs of integers).

[L6]

Squaring is strictly monotone on the nonnegatives: $0 \le a < b$ gives $a^2 < b^2$ (Squaring is monotone on the nonnegatives); $0 < 1$ and the order is total and transitive (The multiplicative identity is positive, Ordered field, Complete ordered field (least-upper-bound property)).

[L7]

A set is bounded when it has an upper and a lower bound (Lower bound, bounded below, bounded set).

Counterexample

technique · direct

1.1

By [L4] there is a unique real $r \ge 0$ with $r^2 = 2$ , and $0 < r < 2$ : indeed $r \ne 0$ since $0^2 = 0 \ne 2$ , while $r \ge 2$ would give $r^2 \ge 4 > 2$ by [L6].

L4L6

1.2

$r \notin \mathbb{Q}_{\mathbb{R}}$ : if $r = \hat q$ for a rational $q$ , then $\widehat{q^2} = \hat q \cdot \hat q = r^2 = 2 = \hat 2$ by [L5], and injectivity of the embedding gives $q^2 = 2$ in $\mathbb{Q}$ , contradicting [L5].

1.3

$E$ is bounded, since $0 \le y \le 2$ for every $y \in E$ by the definition of $E$ .

givenL7

2.1

Put $A := E \cap (-\infty, r)$ and $B := E \cap (r, \infty)$ . Then $A \cup B = E$ , because every $y \in E$ satisfies $y \ne r$ by step 1.2 and hence $y < r$ or $y > r$ ; and both are nonempty, since $0 \in A$ and $2 \in B$ by step 1.1, both being rationals in $[0,2]$ .

step 1.1step 1.2L3L5L6

3.1

$A$ and $B$ are separated: $(-\infty, r]$ is closed and contains $A$ , so $\overline{A} \subseteq (-\infty,r]$ by [L2] and hence $\overline{A} \cap B \subseteq (-\infty,r] \cap (r,\infty) = \varnothing$ ; symmetrically $\overline{B} \subseteq [r,\infty)$ and $A \cap \overline{B} = \varnothing$ . Hence $(A,B)$ is a disconnection of $E$ and $E$ is disconnected, so the claim [A1] is refuted.

step 2.1A1L1L2L3∎

Remarks

What the witness shows about A subset of $\mathbb{R}$ is connected if and only if it is order-convex, that is, an interval. That theorem says a subset of $\mathbb{R}$ is connected exactly when it is order-convex in $\mathbb{R}$ . $E$ is not order-convex in $\mathbb{R}$ : the point $r$ lies between $0$ and $2$ and is not in $E$ . So no contradiction arises, and the example locates precisely what "interval" has to mean in the theorem: order-convex with respect to the complete order, not with respect to the order of a dense subfield.
Where completeness is spent in the theorem, and why it is absent here. The proof of A subset of $\mathbb{R}$ is connected if and only if it is order-convex, that is, an interval produces a supremum, and that supremum is the point at which the two pieces of a would-be disconnection must meet. For $E$ the corresponding supremum is $r$ , which exists in $\mathbb{R}$ and not in $\mathbb{Q}$ ; inside $\mathbb{Q}$ there is no point at which to detect the split, which is exactly why $E$ looks like an interval there.
The same phenomenon in a different guise is $\{q \in \mathbb{Q} : q \ge 0,\ q^2 < 2\}$ is closed and bounded in $\mathbb{Q}$ and is not compact: a set that behaves well inside $\mathbb{Q}$ because the real number that would spoil it is missing from $\mathbb{Q}$ . In both cases the missing number is $\sqrt 2$ .

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 66 results over 29 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

Connected space (Wikipedia) (standard reference, not scraped)
Square root of 2 (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 2 (Example 2.44) (standard reference, not scraped)

Q∩[0,2]\mathbb{Q} \cap [0,2]Q∩[0,2] is bounded and disconnected, so being an interval of Q\mathbb{Q}Q is not enough