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$\inf\{1/n : n \ge 1\} = 0$ , not attained, while $\sup\{1/n : n \ge 1\} = 1$ is

Example

Write $n \cdot 1$ for the canonical natural number $\underbrace{1 + \cdots + 1}_{n}$ of $\mathbb{R}$ (Canonical naturals are positive and strictly increasing) and $1/n := (n \cdot 1)^{-1}$ . The set $H = \{\, 1/n \ : \ n \ge 1 \,\} = \{\, 1,\ 1/2,\ 1/3,\ \dots \,\}$ has $\inf H = 0$ , which is not attained, and $\sup H = 1$ , which is attained, being $\max H$ (Maximum and minimum of a set).

The interesting half is the infimum, and it is exactly the Archimedean property (Every complete ordered field is Archimedean) in disguise. That $0$ is a lower bound is immediate from positivity of the canonical naturals. That no positive number is a lower bound is the statement that the reciprocals of the naturals get below every positive $\varepsilon$ , which is what the Archimedean property asserts once applied to $\varepsilon^{-1}$ . Leastness is then read off from the epsilon characterisation Epsilon characterisation of the infimum. Nothing is asserted by inspection: for each $\varepsilon > 0$ an explicit index is produced.

Facts & Assumptions

Given: The complete ordered field $\mathbb{R}$ ; for a natural number $n \ge 1$ let $n \cdot 1$ denote the canonical natural of $\mathbb{R}$ and write $1/n := (n \cdot 1)^{-1}$ ; and let $H := \{\, 1/n : n \ge 1 \,\}$ .

[L1]

Canonical naturals: $1 \cdot 1 = 1$ ; $n \cdot 1 > 0$ for every $n \ge 1$ ; and $n \mapsto n \cdot 1$ is strictly increasing on $\{1, 2, 3, \dots\}$ , so that $1 = 1 \cdot 1 \le n \cdot 1$ for every $n \ge 1$ , with equality exactly when $n = 1$ (Canonical naturals are positive and strictly increasing).

[L2]

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

[L3]

Inverses and order: if $a > 0$ then $a^{-1} > 0$ ; if $0 < a < b$ then $0 < b^{-1} < a^{-1}$ (claims 1 and 2 of Inverses of positives are positive, and reciprocation reverses order); and, from uniqueness of the multiplicative inverse, $(a^{-1})^{-1} = a$ for every $a \ne 0$ and $1^{-1} = 1$ , since $a^{-1} a = 1$ exhibits $a$ as an inverse of $a^{-1}$ and the identity axiom $1 \cdot x = x$ at $x = 1$ exhibits $1$ as an inverse of itself (Field).

[L4]

Multiplying by a positive constant is an order equivalence: for $c > 0$ one has $x < y$ if and only if $xc < yc$ , and hence, by trichotomy, $x \le y$ if and only if $xc \le yc$ (Sign rules for products and monotonicity of multiplication, Ordered field).

[L5]

Epsilon characterisation of the infimum: for a nonempty $X \subseteq \mathbb{R}$ bounded below and a lower bound $m$ of $X$ , one has $m = \inf X$ if and only if for every $\varepsilon > 0$ there is $x \in X$ with $x < m + \varepsilon$ (Epsilon characterisation of the infimum, Greatest lower bound (infimum)).

[L6]

Maximum and attainment: $m = \max X$ means $m \in X$ and $x \le m$ for all $x \in X$ ; if a nonempty $X$ has a maximum then $\sup X$ exists and $\sup X = \max X$ (Maximum and minimum of a set, The supremum is attained exactly when a maximum exists).

[L7]

Order: trichotomy holds, so $a < a$ is impossible and the negation of $a \le b$ is $b < a$ ; the order is transitive; and $0 < 1$ (Complete ordered field (least-upper-bound property), Ordered field, The multiplicative identity is positive).

[L8]

Lower bound and bounded below: $\ell$ bounds $X$ below when $\ell \le x$ for all $x \in X$ , and $X$ is bounded below when such an $\ell$ exists (Lower bound, bounded below, bounded set).

[L9]

Minimum, and the infimum as greatest lower bound: $m = \min X$ means $m \in X$ and $m \le x$ for all $x \in X$ , so a minimum is a lower bound of $X$ belonging to $X$ (Maximum and minimum of a set); and $\inf X$ is the greatest lower bound, so $\ell \le \inf X$ for every lower bound $\ell$ of $X$ (Greatest lower bound (infimum)).

Verification

technique · direct

1.1

$H$ is nonempty: taking $n = 1$ gives the canonical natural $1 \cdot 1 = 1$ by [L1], and $1/1 = 1^{-1} = 1$ by [L3], so $1 \in H$ .

L1L3

1.2

Every element of $H$ is positive, so $0$ is a lower bound of $H$ and $H$ is bounded below: for $n \ge 1$ one has $n \cdot 1 > 0$ , hence $1/n = (n \cdot 1)^{-1} > 0$ and in particular $0 \le 1/n$ .

L1L3L8

1.3

Every element of $H$ is $\le 1$ : fix $n \ge 1$ and multiply by the positive constant $n \cdot 1$ , which by [L4] turns the inequality $(n \cdot 1)^{-1} \le 1$ into the equivalent inequality $1 = (n \cdot 1)^{-1}(n \cdot 1) \le 1 \cdot (n \cdot 1) = n \cdot 1$ , and the latter holds by [L1].

L1L3L4

1.4

Let $\varepsilon > 0$ be arbitrary; then $\varepsilon^{-1} > 0$ , and the Archimedean property applied to $\varepsilon^{-1}$ supplies a natural $n_\varepsilon \ge 1$ with $\varepsilon^{-1} < n_\varepsilon \cdot 1$ .

assume-hypL2L3choose

2.1

That index witnesses the approximation: from $0 < \varepsilon^{-1} < n_\varepsilon \cdot 1$ and [L3] one gets $0 < (n_\varepsilon \cdot 1)^{-1} < (\varepsilon^{-1})^{-1} = \varepsilon$ , that is $1/n_\varepsilon \in H$ with $1/n_\varepsilon < 0 + \varepsilon$ .

step 1.4L3L7

2.2

$0 \notin H$ : every element of $H$ is $> 0$ by 1.2, whereas $0 > 0$ is impossible by irreflexivity.

step 1.2L7

2.3

$1 = \max H$ : the element $1$ lies in $H$ by 1.1 and dominates every element of $H$ by 1.3.

step 1.1step 1.3L6

3.1

$H$ is nonempty and bounded below by $0$ , and for every $\varepsilon > 0$ some element of $H$ is $< 0 + \varepsilon$ ; the epsilon characterisation therefore gives $\inf H = 0$ .

step 1.1step 1.2step 2.1L5

3.2

Since $H$ is nonempty with maximum $1$ , the attainment criterion gives that $\sup H$ exists and $\sup H = \max H = 1 \in H$ .

step 2.3step 1.1L6

4.1

$H$ has no minimum: a minimum of $H$ would be a lower bound of $H$ lying in $H$ , hence at most the greatest lower bound $\inf H = 0$ ; but every element of $H$ is $> 0$ by 1.2, and $x \le 0$ together with $0 < x$ is impossible by trichotomy.

step 3.1step 1.2L7L9

5.1

Hence $\inf H = 0 \notin H$ and $\sup H = 1 \in H$ : the infimum of $H$ is not attained and $H$ has no minimum, while the supremum is attained and is the maximum.

step 3.1step 3.2step 2.2step 4.1L6∎

Remarks

The set $H$ is bounded, with $0 \le h \le 1$ for every $h \in H$ , and its members are pairwise distinct, since $n \mapsto 1/n$ is strictly decreasing on $\{1, 2, 3, \dots\}$ : for $n < m$ the map $n \mapsto n \cdot 1$ is strictly increasing by [L1], so $0 < n \cdot 1 < m \cdot 1$ , and inversion reverses that by [L3], giving $1/m < 1/n$ . It is the standard example showing that a bounded set with infinitely many members can attain one of its two bounds and miss the other. ("Infinite" is used here in its everyday sense: no definition of finiteness is in scope on this page, and nothing above or below depends on one.)
Positivity of every element is what makes $0$ a lower bound, and the Archimedean property is what makes it the greatest one. In a non-Archimedean ordered field the argument breaks at exactly one point, step 1.4: it is the Archimedean property (Every complete ordered field is Archimedean) that supplies, for a given $\varepsilon > 0$ , an index $n_\varepsilon$ with $\varepsilon^{-1} < n_\varepsilon \cdot 1$ , and hence with $1/n_\varepsilon < \varepsilon$ in step 2.1. And Not every ordered field is Archimedean exhibits an ordered field where no such natural exists. What that item establishes is the failure of the Archimedean property there; it says nothing about $H$ or its greatest lower bound, and this page does not compute one. So the value $\inf H = 0$ is a statement about $\mathbb{R}$ , not a formal manipulation.
Read as a sequence rather than a set, $H$ is the classical null sequence; the rational form of the same fact is The sequence $1/n$ is null.

Depends on

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Sources

Archimedean property (Wikipedia) (standard reference, not scraped)
Infimum and supremum (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 1 (standard reference, not scraped)
T. Tao, Analysis I, 3rd ed. (standard reference, not scraped)
John K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)

inf⁡{1/n:n≥1}=0\inf\{1/n : n \ge 1\} = 0inf{1/n:n≥1}=0, not attained, while sup⁡{1/n:n≥1}=1\sup\{1/n : n \ge 1\} = 1sup{1/n:n≥1}=1 is