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

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FALSE: in every ordered field a closed bounded set is compact, so Heine-Borel needs no completeness

Statement

False claim: in every ordered field $F$ (Ordered field), a subset of $F$ that is closed in $F$ and bounded is compact in $F$ ; consequently the completeness hypothesis in A subset of $R$ is compact if and only if it is closed and bounded is unnecessary.

How the claim must be read. It speaks of an arbitrary ordered field, so the whole vocabulary has to be available there, and it is: for $x \in F$ and $ε \in F$ with $ε > 0$ put $N_{ε}^{F} (x) : = {y \in F : ∣ y - x ∣ < ε}$ , using the absolute value of Absolute value in an ordered field, which is defined in every ordered field; call $U \subseteq F$ open in $F$ when every $x \in U$ admits $ε > 0$ in $F$ with $N_{ε}^{F} (x) \subseteq U$ , call $C \subseteq F$ closed in $F$ when $F ∖ C$ is open in $F$ , call $S \subseteq F$ bounded when some $ℓ, u \in F$ satisfy $ℓ \leq s \leq u$ for all $s \in S$ , and call $S$ compact in $F$ when every family of sets open in $F$ whose union contains $S$ has a finite subfamily whose union already contains $S$ . These are the definitions of The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ , Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen, Lower bound, bounded below, bounded set and Open cover, subcover, compact subset of $R$ (every open cover has a finite subcover), and sequentially compact subset transposed word for word from $R$ to $F$ ; with $F = R$ they are literally those definitions.

The refutation takes $F = Q$ (The rationals as equivalence classes of pairs of integers, The rationals form a totally ordered field) and the set of nonnegative rationals whose square is below $2$ .

Facts & Assumptions

Given: The ordered field $Q$ and the set $S : = {q \in Q : q \geq 0 and q^{2} < 2}$ , together with the notions "open in $Q$ ", "closed in $Q$ ", "bounded" and "compact in $Q$ " as set out in the Statement. Here $2 : = 1 + 1$ and $4 : = 2 \cdot 2$ in $Q$ .

[A1]

The false claim: in every ordered field, a closed bounded subset is compact.

[L1]

$Q$ is a field and the relation of its order makes it a totally ordered field: the order is total and transitive, adding a constant preserves it, and a product of positives is positive (The rationals form a totally ordered field, The rationals form a field, The rationals as equivalence classes of pairs of integers, Ordered field).

[L2]

Absolute value in an ordered field: $∣ z ∣ \geq 0$ ; $∣ z ∣ = z$ for $z \geq 0$ and $∣ z ∣ = - z$ for $z < 0$ ; and for $c > 0$ one has $∣ z ∣ < c$ exactly when $- c < z < c$ (Absolute value in an ordered field, Basic properties of the absolute value).

[L3]

No rational number squares to $2$ (FALSE: some rational number squares to 2).

[L4]

In an ordered field, squaring is strictly monotone on the nonnegatives: $0 \leq a < b$ implies $a^{2} < b^{2}$ , and $0 \leq a \leq b$ implies $a^{2} \leq b^{2}$ (Squaring is monotone on the nonnegatives).

[L5]

Ordered-field arithmetic: $0 < 1$ , hence $0 < 2 < 4$ and $2 \neq 0$ ; a positive element has a positive inverse; adding a constant and multiplying by a positive preserve an inequality (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Inverses of positives are positive, and reciprocation reverses order, Ordered field). 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

$S$ is nonempty and bounded: $1 \in S$ because $1 \geq 0$ and $1^{2} = 1 < 2$ by [L5]; and every $q \in S$ satisfies $0 \leq q < 2$ , since $q \geq 2 \geq 0$ would give $q^{2} \geq 2^{2} = 4 > 2$ by [L4] and [L5], contradicting $q^{2} < 2$ .

L1L4L5

1.2

$S$ has no greatest element: let $q \in S$ and put $h : = \min {1, (2 - q^{2}) (2 q + 2)^{- 1}}$ , a definition by cases on the total order of $Q$ ; here $2 q + 2 > 0$ because $q \geq 0$ , and $2 - q^{2} > 0$ , so both entries are positive and $h > 0$ with $h \leq 1$ . Put $r : = q + h$ , so $r > q \geq 0$ . Then $h^{2} \leq h$ because $0 < h \leq 1$ , and $h (2 q + 1) \leq (2 - q^{2}) (2 q + 1) (2 q + 2)^{- 1} < 2 - q^{2}$ because $(2 q + 1) (2 q + 2)^{- 1} < 1$ and $2 - q^{2} > 0$ ; hence $r^{2} = q^{2} + 2 q h + h^{2} \leq q^{2} + h (2 q + 1) < q^{2} + (2 - q^{2}) = 2$ , so $r \in S$ and $q < r$ .

L1L4L5

1.3

$S$ is closed in $Q$ : let $q \in Q ∖ S$ , so $q < 0$ , or $q \geq 0$ and $q^{2} \geq 2$ , in which case $q^{2} \neq 2$ by [L3] gives $q^{2} > 2$ . If $q < 0$ , put $ε : = - q > 0$ ; every $y$ with $∣ y - q ∣ < ε$ satisfies $y < q + ε = 0$ by [L2], hence $y \notin S$ . If $q \geq 0$ and $q^{2} > 2$ , then $q \neq 0$ since $0^{2} = 0 < 2$ , so $q > 0$ ; put $ε : = \min {q, (q^{2} - 2) (2 q)^{- 1}} > 0$ , again a definition by cases. Every $y$ with $∣ y - q ∣ < ε$ satisfies $y > q - ε \geq 0$ , so $y^{2} > (q - ε)^{2}$ by [L4], and $(q - ε)^{2} = q^{2} - 2 q ε + ε^{2} \geq q^{2} - 2 q ε \geq q^{2} - (q^{2} - 2) = 2$ , whence $y^{2} > 2$ and $y \notin S$ . In both cases a neighbourhood of $q$ misses $S$ , so $Q ∖ S$ is open in $Q$ .

L1L2L3L4L5

1.4

For $r \in S$ put $B_{r} : = {y \in Q : y < r}$ ; each $B_{r}$ is open in $Q$ , since $y \in B_{r}$ and $ε : = r - y > 0$ give, for every $z$ with $∣ z - y ∣ < ε$ , the inequality $z < y + ε = r$ by [L2].

givenL1L2

2.1

The family $U : = {B_{r} : r \in S}$ is a cover of $S$ by sets open in $Q$ : given $q \in S$ , step 1.2 supplies $r \in S$ with $q < r$ , so $q \in B_{r}$ .

step 1.2step 1.4L1

2.2

$U$ has no finite subfamily covering $S$ : the empty subfamily fails because $S \neq \emptyset$ by step 1.1; and a nonempty finite subfamily is ${B_{r_{0}}, \dots, B_{r_{p}}}$ with every $r_{i} \in S$ , so an induction on $p$ using the totality of the order of $Q$ produces $R : = \max {r_{0}, \dots, r_{p}}$ , one of the $r_{i}$ and hence a member of $S$ ; for each $i$ one has $r_{i} \leq R$ , so $R < r_{i}$ fails and $R \notin B_{r_{i}}$ . Thus the element $R$ of $S$ lies in no member of the subfamily.

step 1.1step 1.4L1

3.1

The set $S$ is bounded by step 1.1 and closed in $Q$ by step 1.3, and by steps 2.1 and 2.2 it is not compact in $Q$ , while $Q$ is an ordered field by [L1]. So the claim [A1] fails at $F = Q$ and is false.

step 1.1step 1.3step 2.1step 2.2A1L1∎

Remarks

What the false claim gets wrong. A subset of $R$ is compact if and only if it is closed and bounded has two halves of very different strengths. The half that a compact set is closed and bounded (A compact subset of $R$ is closed and bounded) uses no completeness at all, only the Archimedean property and the existence of maxima of finite sets. The converse half is the one that rests on completeness, through Heine-Borel by bisection: every closed bounded interval $[a, b]$ is compact and the nested interval property, and it is exactly the half refuted above.
Where the missing point is. The cover of step 2.1 creeps up on a bound that $Q$ does not contain. In $R$ that bound exists, namely $\sqrt{2}$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ), and it is not rational (FALSE: some rational number squares to 2); the set $S$ is thus closed in $Q$ precisely because the point that would have to be adjoined to close it is absent from $Q$ . Read inside $R$ , the same set of numbers is bounded and not closed, and it is not compact there either.
This is a statement about ordered fields, and it is refuted in that generality. One counterexample field suffices to refute a claim about every ordered field, and $Q$ is the smallest one available here. Nothing above uses any ordered-field lemma outside its stated generality: [L2], [L4] and [L5] are all proved for an arbitrary ordered field, and the results of this page that are stated for $R$ only are not applied to $Q$ .
The named witness is ${q \in Q : q \geq 0, q^{2} < 2}$ is closed and bounded in $Q$ and is not compact ↗; the refutation is carried out here.

Depends on

Used by

Dependency tree · two levels

39 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

Heine-Borel theorem (Wikipedia) (standard reference, not scraped)
Rational number (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 2.21(g) and Thm 2.41) (standard reference, not scraped)
MIT 18.100, Test 1 solutions (standard reference, not scraped)