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

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Literature-sourced — the exact statement appears in a cited source; only wording and notation differ.
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$\{1/k : k \ge 1\} \cup \{0\}$ is compact while $\{1/k : k \ge 1\}$ is not closed

Example

Put

$B \;:=\; \{\, 1/k : k \in \mathbb{N},\ k \ge 1 \,\} \qquad \text{and} \qquad A \;:=\; B \cup \{0\} .$

The index runs over $k \ge 1$ because $1/0$ is undefined. Then $A$ is compact (A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded) and $B$ is not closed (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen). The single point $0$ is the whole difference: it is a limit point of $B$ (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$ ) that $B$ omits, and adjoining it turns a non-closed bounded set into a compact one.

Facts & Assumptions

Given: The sets $B := \{\, 1/k : k \ge 1 \,\}$ and $A := B \cup \{0\}$ , where $1/k$ denotes the inverse of the canonical natural $k \cdot 1_{\mathbb{R}}$ , defined and positive for $k \ge 1$ .

[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, Lower bound, bounded below, bounded set).

[L2]

A set is closed exactly when it contains all its limit points; a point is a limit point of $S$ when every punctured neighbourhood of it meets $S$ (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, Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$ ).

[L3]

$U$ is open when each of its points has a neighbourhood inside it, and $S$ is closed when $\mathbb{R} \setminus S$ is open; $N_\varepsilon(x) = \{\, y : |y-x| < \varepsilon \,\}$ (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 $\mathbb{R}$ ).

[L4]

Reciprocal Archimedean property: for every real $\varepsilon > 0$ there is a natural $n \ge 1$ with $1/n < \varepsilon$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , Every complete ordered field is Archimedean).

[L5]

Canonical naturals are positive and increasing for $k \ge 1$ , with $1 \cdot 1_{\mathbb{R}} = 1$ ; reciprocation of positives is positive and reverses the order, so $1 \le m \le n$ gives $0 < 1/n \le 1/m \le 1$ (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order). 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.

[L6]

Every nonempty subset of $\mathbb{N}$ has a least element (The well-ordering principle).

[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]

Absolute value: $|z| = z$ for $z \ge 0$ and $|z| = -z$ for $z < 0$ ; the order is total and transitive (Basic properties of the absolute value, Ordered field, Complete ordered field (least-upper-bound property)).

Verification

technique · direct

1.1

Every element of $A$ lies in $[0,1]$ , and $0$ and $1$ belong to $A$ : indeed $1 = 1/1 \in B$ , and $0 < 1/k \le 1$ for every $k \ge 1$ by [L5]. In particular $A$ is bounded.

L1L5

1.2

$0$ is a limit point of $B$ and $0 \notin B$ : given a real $\varepsilon > 0$ , [L4] supplies $n \ge 1$ with $1/n < \varepsilon$ , and $1/n \in B$ with $1/n \ne 0$ , so $1/n$ lies in the punctured neighbourhood of $0$ of radius $\varepsilon$ and meets $B$ ; and $0 \notin B$ because every $1/k$ is positive by [L5].

L2L4L5

2.1

Let $x \in \mathbb{R} \setminus A$ with $x < 0$ or $x > 1$ . Put $\varepsilon := -x$ in the first case and $\varepsilon := x - 1$ in the second; then $\varepsilon > 0$ , and $y \in N_\varepsilon(x)$ gives $y < x + \varepsilon = 0$ in the first case and $y > x - \varepsilon = 1$ in the second, so $y \notin A$ by step 1.1. Hence $N_\varepsilon(x) \cap A = \varnothing$ .

step 1.1L3L8

2.2

Let $x \in \mathbb{R} \setminus A$ with $0 \le x \le 1$ . Then $x \ne 0$ and $x \ne 1$ by step 1.1, so $0 < x < 1$ . The set $\{\, k \in \mathbb{N} : k \ge 1 \text{ and } 1/k < x \,\}$ is nonempty by [L4], so it has a least element $k_0$ by [L6], and $k_0 \ne 1$ since $1/1 = 1 > x$ ; hence $k_0 \ge 2$ and $k_0 - 1 \ge 1$ . By minimality $1/(k_0-1) < x$ fails, so $x \le 1/(k_0-1)$ , and $x \ne 1/(k_0-1)$ because $x \notin A$ , so $x < 1/(k_0-1)$ . Put $\varepsilon := \min\{\, x - 1/k_0,\ 1/(k_0-1) - x \,\} > 0$ by [L7]. Then $N_\varepsilon(x) \cap A = \varnothing$ : an element of $A$ is $0$ or $1/m$ with $m \ge 1$ ; for $0$ one has $|0 - x| = x > \varepsilon$ since $\varepsilon \le x - 1/k_0 < x$ ; for $m \ge k_0$ one has $1/m \le 1/k_0 \le x - \varepsilon$ by [L5]; and for $1 \le m \le k_0 - 1$ one has $1/m \ge 1/(k_0-1) \ge x + \varepsilon$ by [L5]. In each case the element is at distance at least $\varepsilon$ from $x$ .

step 1.1L3L4L5L6L7L8

2.3

$B$ is not closed: by step 1.2 the point $0$ is a limit point of $B$ that does not lie in $B$ , so $B$ does not contain all its limit points and [L2] denies that it is closed.

step 1.2L2

3.1

$A$ is closed: every $x \in \mathbb{R} \setminus A$ falls under step 2.1 or step 2.2 by totality of the order, and in either case some $N_\varepsilon(x)$ misses $A$ , so $\mathbb{R} \setminus A$ is open. With the boundedness of step 1.1 and [L1], $A$ is compact.

step 1.1step 2.1step 2.2L1L3L8

4.1

So $A = \{\, 1/k : k \ge 1 \,\} \cup \{0\}$ is compact by step 3.1 while $B = \{\, 1/k : k \ge 1 \,\}$ is not closed by step 2.3, hence not compact by [L1].

step 2.3step 3.1L1∎

Remarks

$B$ is bounded and not compact, and it fails to be closed by a single point. Adjoining $0$ is what the verification above shows to be enough: $A = B \cup \{0\}$ is closed. The same computation, run at a point $x \notin A$ lying between two consecutive reciprocals, is the one that isolates each $1/k$ from the rest of $B$ .
$A$ is compact and has an isolated point. Every $1/k$ is isolated in $A$ , so $A$ is not perfect (Perfect subset of $\mathbb{R}$ : closed with no isolated points); compactness and perfectness are independent properties, and this is a compact set that is countable, which is possible exactly because it is not perfect (Every nonempty perfect subset of $\mathbb{R}$ is uncountable).
The index range matters. The set is indexed from $k = 1$ ; $1/0$ does not exist. Since $\mathbb{N}$ contains $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), a set written $\{1/k : k \in \mathbb{N}\}$ without a restriction would be ill formed, and the same care is needed at the threshold $K = n-1$ used in the convergence arguments on the parent page.

Depends on

Used by

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Dependency tree · next 3 levels

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Sources

Compact space (Wikipedia) (standard reference, not scraped)
Limit point (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 2 (Example 2.21(e)) (standard reference, not scraped)
J. K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)

{1/k:k≥1}∪{0}\{1/k : k \ge 1\} \cup \{0\}{1/k:k≥1}∪{0} is compact while {1/k:k≥1}\{1/k : k \ge 1\}{1/k:k≥1} is not closed