What is Bounds, sup and inf? — Mathematics

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Definition 10.1 University Mathematics — Year 1 · অধ্যায় 10 — Real Numbers

Let $A \subseteq \R$ be nonempty. A real $M$ is an upper bound of $A$ when $a \leq M$ for all $a \in A$ ; $A$ is bounded above when it has an upper bound (similarly below, with lower bounds; bounded means both). A maximum of $A$ is an upper bound belonging to $A$ .

The supremum $\sup A$ is the least upper bound of $A$ , when it exists; the infimum $\inf A$ is the greatest lower bound.

The set A = \1 - 1n : n ∈ ℕ*\ on the number line: its points pile up toward 1 without reaching it. Every number ≥ 1 is an upper bound (the ray), and nothing smaller is, because an element of A enters each interval (1 - , 1): the two clauses of in one picture. The supremum is the left endpoint of the ray of upper bounds — and the completeness axiom is precisely the guarantee that this ray always has a left endpoint. — The set $A = \{1 - \frac1n : n \in \N^*\}$ on the number line: its points pile up toward $1$ without reaching it. Every number $\geq 1$ is an upper bound (the ray), and nothing smaller is, because an element of $A$ enters each interval $\intoo{1 - \varepsilon}{1}$ : the two clauses of Proposition 10.4 in one picture. The supremum is the left endpoint of the ray of upper bounds — and the completeness axiom is precisely the guarantee that this ray always has a left endpoint.

উদাহরণ

Example 10.5

$\sup \intoo{0}{1} = 1$ , not attained (no maximum); $\sup \intcc{0}{1} = 1 = \max$ . For $A = \{1 - \frac 1n : n \in \N^*\}$ : $\sup A = 1$ , not attained; $\inf A = \min A = 0$ . A maximum, when it exists, is the supremum; the whole point of $\sup$ is to have a substitute when the maximum does not exist.

Example 10.6 (Computing suprema in practice)

Two full workouts of Proposition 10.4.

The set $A = \{x + \frac1x : x > 0\}$ . For every $x > 0$ , $x + \frac1x - 2 = \frac{(\,\sqrt x - 1/\sqrt x\,)^2}{1} \geq 0$ , so $2$ is a lower bound; and $2 = 1 + \frac11 \in A$ : therefore $\inf A = \min A = 2$ , attained at $x = 1$ . Above, $A$ is unbounded ( $x + \frac1x > x$ can exceed any $M$ by Theorem 10.10): $\sup A$ does not exist in $\R$ (it is $+\infty$ in $\overline\R$ ).

The set $B = \bigl\{\frac{m}{m + n} : m, n \in \N^*\bigr\}$ . Every element lies in $\intoo{0}{1}$ , so $0$ and $1$ are bounds. Neither is attained: $\frac{m}{m+n} = 1$ would force $n = 0$ . For the supremum, freeze $n = 1$ and let $m$ grow: $\frac{m}{m+1} = 1 - \frac{1}{m+1} > 1 - \varepsilon$ as soon as $m + 1 > \frac1\varepsilon$ (Archimedes): $\sup B = 1$ . Symmetrically ( $m = 1$ , $n$ large), $\inf B = 0$ . The closing insight: to pin a supremum, one well-chosen one-parameter path inside the set suffices — here the path $n = 1$ — and the $\varepsilon$ -characterization asks for nothing more.

Example 10.7 (The infimum mirror)

The infimum has its own $\varepsilon$ -characterization, obtained from Proposition 10.4 through $\inf A = -\sup(-A)$ : $i = \inf A$ iff $i$ bounds $A$ below and, for every $\varepsilon > 0$ , some $a \in A$ has $a < i + \varepsilon$ . A workout with both bounds at once: let

A = \Bigl\{(-1)^n + \frac1n : n \in \N^*\Bigr\} = \Bigl\{0,\ \tfrac32,\ -\tfrac23,\ \tfrac54,\ -\tfrac45,\ \dots\Bigr\} .

Even indices give $1 + \frac1n \leq \frac32$ , with equality at $n = 2$ : since also the odd-index values are $\leq 0 < \frac32$ , we get $\sup A = \max A = \frac32$ . Odd indices give $-1 + \frac1n > -1$ , decreasing toward $-1$ : every element of $A$ is $> -1$ , and $-1 + \varepsilon$ is beaten by $-1 + \frac1n$ for odd $n > \frac1\varepsilon$ : $\inf A = -1$ , not attained. One set, all four behaviors on display: a supremum that is a maximum, an infimum that is not a minimum.

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