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${1 / k : k \geq 1} \cup {0}$ is compact while ${1 / k : k \geq 1}$ is not closed

Example

Put

$B : = {1 / k : k \in N, k \geq 1} and A : = B \cup {0} .$

The index runs over $k \geq 1$ because $1 / 0$ is undefined. Then $A$ is compact (A subset of $R$ is compact if and only if it is closed and bounded) and $B$ is not closed (Open subset of $R$ (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 $R$ ) that $B$ omits, and adjoining it turns a non-closed bounded set into a compact one.

Facts & Assumptions

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

[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, 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 $R$ ).

[L3]

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

[L4]

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

[L5]

Canonical naturals are positive and increasing for $k \geq 1$ , with $1 \cdot 1_{R} = 1$ ; reciprocation of positives is positive and reverses the order, so $1 \leq m \leq n$ gives $0 < 1 / n \leq 1 / m \leq 1$ (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 $N$ 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 $z \geq 0$ 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 $0 < 1 / k \leq 1$ for every $k \geq 1$ by [L5]. In particular $A$ is bounded.

L1L5

1.2

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

L2L4L5

2.1

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

step 1.1L3L8

2.2

Let $x \in R ∖ A$ with $0 \leq x \leq 1$ . Then $x \neq 0$ and $x \neq 1$ by step 1.1, so $0 < x < 1$ . The set ${k \in N : k \geq 1 and 1 / k < x}$ is nonempty by [L4], so it has a least element $k_{0}$ by [L6], and $k_{0} \neq 1$ since $1 / 1 = 1 > x$ ; hence $k_{0} \geq 2$ and $k_{0} - 1 \geq 1$ . By minimality $1 / (k_{0} - 1) < x$ fails, so $x \leq 1 / (k_{0} - 1)$ , and $x \neq 1 / (k_{0} - 1)$ because $x \notin A$ , so $x < 1 / (k_{0} - 1)$ . Put $ε : = \min {x - 1 / k_{0}, 1 / (k_{0} - 1) - x} > 0$ by [L7]. Then $N_{ε} (x) \cap A = \emptyset$ : an element of $A$ is $0$ or $1 / m$ with $m \geq 1$ ; for $0$ one has $∣ 0 - x ∣ = x > ε$ since $ε \leq x - 1 / k_{0} < x$ ; for $m \geq k_{0}$ one has $1 / m \leq 1 / k_{0} \leq x - ε$ by [L5]; and for $1 \leq m \leq k_{0} - 1$ one has $1 / m \geq 1 / (k_{0} - 1) \geq x + ε$ by [L5]. In each case the element is at distance at least $ε$ 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 $x \in R ∖ A$ falls under step 2.1 or step 2.2 by totality of the order, and in either case some $N_{ε} (x)$ misses $A$ , so $R ∖ A$ is open. With the boundedness of step 1.1 and [L1], $A$ is compact.

step 1.1step 2.1step 2.2L1L3L8

4.1

So $A = {1 / k : k \geq 1} \cup {0}$ is compact by step 3.1 while $B = {1 / k : k \geq 1}$ 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 $R$ : 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 $R$ is uncountable).
The index range matters. The set is indexed from $k = 1$ ; $1 / 0$ does not exist. Since $N$ contains $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), a set written ${1 / k : k \in N}$ 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 · two 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≥1}∪{0} is compact while {1/k:k≥1}{1/k:k≥1} is not closed