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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$\mathbb{Z}$ 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 $\mathbb{R}$ is compact; (ii) every bounded subset of $\mathbb{R}$ is compact (Open cover, subcover, compact subset of $\mathbb{R}$ (every open cover has a finite subcover), and sequentially compact subset).

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

What $\mathbb{Z}$ means here. Write $\mathbb{Z} \;:=\; \{\, n \cdot 1_{\mathbb{R}} : n \in \mathbb{N} \,\} \cup \{\, -(n \cdot 1_{\mathbb{R}}) : n \in \mathbb{N} \,\}, \qquad 0 \cdot 1_{\mathbb{R}} := 0,$ the set of canonical integers of $\mathbb{R}$ . 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 $\mathbb{Q} \to \mathbb{R}$ , which sends $n$ to $n \cdot 1_{\mathbb{R}}$ and $-n$ to $-(n \cdot 1_{\mathbb{R}})$ ; as is standard we write $\mathbb{Z}$ for it. Nothing below uses that identification: every step is carried out with the displayed description.

Facts & Assumptions

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

[A1]

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

[L1]

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

[L2]

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

[L3]

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

[L4]

$U$ is open when each of its points has a neighbourhood inside it; $F$ is closed when $\mathbb{R} \setminus F$ is open; $N_\varepsilon(x) = \{\, y : |y-x| < \varepsilon \,\}$ ; each interval $(a,b)$ is an open set (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen, The $\varepsilon$ -neighbourhood and the punctured $\varepsilon$ -neighbourhood of a point of $\mathbb{R}$ , Intervals of $\mathbb{R}$ : 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: $|p + q| \le |p| + |q|$ (The triangle inequality); $|z| = z$ for $z \ge 0$ 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 $\mathbb{Z}$ differ by at least $1$ in absolute value. For $m, n \in \mathbb{N}$ with $m < n$ : if $m = 0$ then $n \cdot 1_{\mathbb{R}} - 0 = n \cdot 1_{\mathbb{R}} \ge 1$ by [L2], since $n \ge 1$ and the map is increasing with value $1$ at $1$ ; and if $m \ge 1$ then $m + (n-m) = n$ with $n - m \ge 1$ gives $n \cdot 1_{\mathbb{R}} - m \cdot 1_{\mathbb{R}} = (n-m) \cdot 1_{\mathbb{R}} \ge 1$ by [L2]. The same computation after negation covers two distinct elements of the form $-(n \cdot 1_{\mathbb{R}})$ , using $|-z| = |z|$ from [L6]. Finally, for distinct $n \cdot 1_{\mathbb{R}}$ and $-(m \cdot 1_{\mathbb{R}})$ at least one of $n, m$ is $\ge 1$ , so their difference $n \cdot 1_{\mathbb{R}} + m \cdot 1_{\mathbb{R}}$ is a sum of two nonnegative terms one of which is $\ge 1$ , hence is $\ge 1$ .

givenL2L6L8

1.2

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

givenL3L5

1.3

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

L4L5L6L7L8

2.1

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

step 1.1L4L6L7L8

3.1

By step 2.1 the set $\mathbb{Z}$ 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. $\mathbb{Z}$ 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 $\mathbb{R}$ is compact if and only if it is closed and bounded says these are the only two ways to fail for a subset of $\mathbb{R}$ .
An explicit cover for each. For $\mathbb{Z}$ the intervals $(-n,n)$ with $n \ge 1$ cover $\mathbb{R}$ and any finite subfamily has union $(-N,N)$ , which omits the element $N \cdot 1_{\mathbb{R}}$ of $\mathbb{Z}$ . 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 $\mathbb{R}$ is compact if and only if it is closed and bounded already converts the failure of a hypothesis into the failure of compactness.
$\mathbb{Z}$ 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 $\mathbb{R}$ : closed with no isolated points).

Depends on

Used by

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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)

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