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

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$Z$ is closed and not compact, and $(0, 1)$ is bounded and not compact: neither hypothesis of Heine-Borel can be dropped

Statement refuted

Refuted claims: (i) every closed subset of $R$ is compact; (ii) every bounded subset of $R$ is compact (Open cover, subcover, compact subset of $R$ (every open cover has a finite subcover), and sequentially compact subset).

Both are refuted, so neither hypothesis of A subset of $R$ is compact if and only if it is closed and bounded can be dropped. The witness for (i) is the set $Z$ of integers of $R$ , which is closed and unbounded; the witness for (ii) is the interval $(0, 1)$ , which is bounded and not closed.

What $Z$ means here. Write $Z : = {n \cdot 1_{R} : n \in N} \cup {- (n \cdot 1_{R}) : n \in N}, 0 \cdot 1_{R} : = 0,$ the set of canonical integers of $R$ . By The unique embedding of ℚ into an ordered field this is exactly the image of the integers (The integers as equivalence classes of pairs of naturals) under the unique field homomorphism $Q \to R$ , which sends $n$ to $n \cdot 1_{R}$ and $- n$ to $- (n \cdot 1_{R})$ ; as is standard we write $Z$ for it. Nothing below uses that identification: every step is carried out with the displayed description.

Facts & Assumptions

Given: The set $Z$ of canonical integers of $R$ as displayed above, and the interval $(0, 1)$ .

[A1]

The refuted claims: (i) every closed subset of $R$ is compact; (ii) every bounded subset of $R$ is compact.

[L1]

A subset of $R$ is compact exactly when it is closed and bounded (A subset of $R$ is compact if and only if it is closed and bounded, Open cover, subcover, compact subset of $R$ (every open cover has a finite subcover), and sequentially compact subset).

[L2]

Canonical naturals: $n \cdot 1_{R} > 0$ for $n \geq 1$ , the map $n \mapsto n \cdot 1_{R}$ is strictly increasing on ${1, 2, \dots}$ with $1 \cdot 1_{R} = 1$ , and $(m + n) \cdot 1_{R} = m \cdot 1_{R} + n \cdot 1_{R}$ for $m, n \geq 1$ (Canonical naturals are positive and strictly increasing).

[L3]

Archimedean property: for every $x \in R$ there is a natural $n \geq 1$ with $x < n \cdot 1_{R}$ (Every complete ordered field is Archimedean).

[L4]

$U$ is open when each of its points has a neighbourhood inside it; $F$ is closed when $R ∖ F$ is open; $N_{ε} (x) = {y : ∣ y - x ∣ < ε}$ ; each interval $(a, b)$ is an open set (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 $R$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L5]

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

[L6]

Triangle inequality: $∣ p + q ∣ \leq ∣ p ∣ + ∣ q ∣$ (The triangle inequality); $∣ z ∣ = z$ for $z \geq 0$ and $∣ z ∣ = - z$ for $z < 0$ , and $∣ - z ∣ = ∣ z ∣$ (Basic properties of the absolute value).

[L7]

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).

[L8]

$0 < 1$ , 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 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.

Counterexample

technique · direct

1.1

Distinct elements of $Z$ differ by at least $1$ in absolute value. For $m, n \in N$ with $m < n$ : if $m = 0$ then $n \cdot 1_{R} - 0 = n \cdot 1_{R} \geq 1$ by [L2], since $n \geq 1$ and the map is increasing with value $1$ at $1$ ; and if $m \geq 1$ then $m + (n - m) = n$ with $n - m \geq 1$ gives $n \cdot 1_{R} - m \cdot 1_{R} = (n - m) \cdot 1_{R} \geq 1$ by [L2]. The same computation after negation covers two distinct elements of the form $- (n \cdot 1_{R})$ , using $∣ - z ∣ = ∣ z ∣$ from [L6]. Finally, for distinct $n \cdot 1_{R}$ and $- (m \cdot 1_{R})$ at least one of $n, m$ is $\geq 1$ , so their difference $n \cdot 1_{R} + m \cdot 1_{R}$ is a sum of two nonnegative terms one of which is $\geq 1$ , hence is $\geq 1$ .

givenL2L6L8

1.2

$Z$ is not bounded: given any $x \in R$ , [L3] supplies a natural $n \geq 1$ with $x < n \cdot 1_{R}$ , and $n \cdot 1_{R} \in Z$ , so no $x$ is an upper bound of $Z$ and $Z$ has no upper bound at all.

givenL3L5

1.3

$(0, 1)$ is bounded, since $0 \leq y \leq 1$ for every $y \in (0, 1)$ , and it is not closed: $0 \notin (0, 1)$ , while for every real $ε > 0$ the point $t : = \min {ε, 1} \cdot 2^{- 1}$ satisfies $0 < t < 1$ and $∣ t - 0 ∣ = t \leq ε \cdot 2^{- 1} < ε$ by [L7] and [L8], so $t \in N_{ε} (0) \cap (0, 1)$ and no neighbourhood of $0$ lies in the complement of $(0, 1)$ .

L4L5L6L7L8

2.1

$Z$ is closed: let $x \in R ∖ Z$ . The neighbourhood $N_{1 \cdot 2^{- 1}} (x)$ contains at most one element of $Z$ , since two distinct elements $z, z^{'}$ of it would satisfy $∣ z - z^{'} ∣ = ∣ (z - x) + (x - z^{'}) ∣ \leq ∣ z - x ∣ + ∣ x - z^{'} ∣ < 1 \cdot 2^{- 1} + 1 \cdot 2^{- 1} = 1$ by [L6], contradicting step 1.1. If it contains none, then $N_{1 \cdot 2^{- 1}} (x) \cap Z = \emptyset$ . If it contains exactly one element $z$ , then $z \neq x$ because $x \notin Z$ , so $∣ x - z ∣ > 0$ , and $ε : = \min {1 \cdot 2^{- 1}, ∣ x - z ∣}$ is positive by [L7]; then $N_{ε} (x) \subseteq N_{1 \cdot 2^{- 1}} (x)$ , so any element of $Z$ in $N_{ε} (x)$ must be $z$ , whereas $∣ z - x ∣ \geq ε$ excludes $z$ . In both cases some neighbourhood of $x$ misses $Z$ , so $R ∖ Z$ is open.

step 1.1L4L6L7L8

3.1

By step 2.1 the set $Z$ is closed and by step 1.2 it is not bounded, so [L1] denies that it is compact, refuting claim (i) of [A1]; and by step 1.3 the set $(0, 1)$ is bounded and not closed, so [L1] denies that it is compact, refuting claim (ii). Neither hypothesis of [L1] is therefore removable.

step 1.2step 1.3step 2.1A1L1∎

Remarks

The two failures are of opposite kinds. $Z$ has all its limit points, of which it has none, and escapes to infinity; $(0, 1)$ stays inside a bounded region and loses its two endpoints. Compactness rules out both, and A subset of $R$ is compact if and only if it is closed and bounded says these are the only two ways to fail for a subset of $R$ .
An explicit cover for each. For $Z$ the intervals $(- n, n)$ with $n \geq 1$ cover $R$ and any finite subfamily has union $(- N, N)$ , which omits the element $N \cdot 1_{R}$ of $Z$ . For $(0, 1)$ the cover ${(1 / k, 1)}$ of The cover ${(1 / k, 1)}$ of $(0, 1)$ has no finite subcover, so $(0, 1)$ is not compact does the same job. Neither cover is needed above, since A subset of $R$ is compact if and only if it is closed and bounded already converts the failure of a hypothesis into the failure of compactness.
$Z$ is closed and has no limit points at all, which is what the separation computation really shows: its points are uniformly apart. A set of that kind is closed for free, and it is the standard example of a closed set that is as far from perfect as possible, every one of its points being isolated (Perfect subset of $R$ : closed with no isolated points).

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

36 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)
Integer (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 2 (standard reference, not scraped)

ZZ is closed and not compact, and (0,1)(0,1) is bounded and not compact: neither hypothesis of Heine-Borel can be dropped