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

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

Statement refuted

Refuted claim: the set $E : = Q_{R} \cap [0, 2]$ of rationals between $0$ and $2$ is connected (Separated sets, disconnection, and connected subset of $R$ ), where $Q_{R}$ is the copy of $Q$ inside $R$ (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 $Q$ : it is an interval of that field. As a subset of $R$ it is nevertheless disconnected, split at the irrational point $\sqrt{2}$ . So the equivalence of A subset of $R$ is connected if and only if it is order-convex, that is, an interval genuinely uses the completeness of $R$ , and "is an interval of the order it carries from $Q$ " is not enough to make a set connected.

Facts & Assumptions

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

[A1]

The refuted claim: $E$ is connected.

[L1]

Separated sets, disconnection and connectedness (Separated sets, disconnection, and connected subset of $R$ ).

[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 $R$ ).

[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 $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L4]

In a complete ordered field every $a \geq 0$ has a unique $s \geq 0$ with $s^{2} = a$ (Square roots exist: a unique $\sqrt{a} \geq 0$ 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: $0 \leq a < b$ 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 $r \geq 0$ with $r^{2} = 2$ , and $0 < r < 2$ : indeed $r \neq 0$ since $0^{2} = 0 \neq 2$ , while $r \geq 2$ would give $r^{2} \geq 4 > 2$ by [L6].

L4L6

1.2

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

1.3

$E$ is bounded, since $0 \leq y \leq 2$ 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 $y \neq r$ 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 $\overline{A} \cap B \subseteq (- \infty, r] \cap (r, \infty) = \emptyset$ ; symmetrically $\overline{B} \subseteq [r, \infty)$ and $A \cap \overline{B} = \emptyset$ . 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 $R$ is connected if and only if it is order-convex, that is, an interval. That theorem says a subset of $R$ is connected exactly when it is order-convex in $R$ . $E$ is not order-convex in $R$ : 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 $R$ 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 $R$ and not in $Q$ ; inside $Q$ 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 ${q \in Q : q \geq 0, q^{2} < 2}$ is closed and bounded in $Q$ and is not compact: a set that behaves well inside $Q$ because the real number that would spoil it is missing from $Q$ . In both cases the missing number is $\sqrt{2}$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

38 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

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]Q∩[0,2] is bounded and disconnected, so being an interval of QQ is not enough