The Supremum Axiom

Find the smallest ceiling over a bounded set. That ceiling, missing in the rationals, completes the reals.

The supremum is the least of all upper bounds. Ceilings are many, but the lowest one is unique. Rationals sometimes lack that lowest ceiling. The reals are the world where it always exists.

For example, measure every 🧑 classmate's height. The smallest line above them all is the supremum. It touches the tallest height exactly. Everyone fits underneath.

Examples & Exercises

Concept check

Prove $\sup S = 1$ for $S = \left\{1 - \frac{1}{n} : n \geq 1\right\}$ .

(Note $1 \notin S$ .)

\sup\left\{1-\frac{1}{n}\right\} = 1 \notin S

Candidate

1.2

—

0.2

above the true supremum

1

$1 \in S$ , and $1$ is the maximum of $S$
$1 \notin S$ , but $1$ is the least upper bound of $S$
Since $1 \notin S$ , $\sup S$ does not exist
$\sup S = 0$ , because the $n = 1$ element $0$ is the largest

Show explanation

Answer: $1 \notin S$ , but $1$ is the least upper bound of $S$

Since $1 - \frac{1}{n} < 1$ for every $n$ , $1$ is an upper bound; by the Archimedean property no $u < 1$ is an upper bound, so $1$ is the least upper bound, i.e. $\sup S = 1$ . The elements increase from $0$ toward $1$ , so $0$ is the smallest element, and $1$ itself is not an element of $S$ for any $n$ — a supremum need not belong to the set, and its existence is guaranteed by the completeness axiom.

Prove $\sup(A+B) = \sup A + \sup B$ for nonempty bounded-above $A, B \subset \mathbb{R}$ .

Tightening the candidate

1.5

down to the true supremum.

Show proof

(≤) For all $a \in A$ and $b \in B$ , $a + b \leq \sup A + \sup B$ , so $\sup A + \sup B$ is an upper bound of $A + B$ . Hence $\sup(A+B) \leq \sup A + \sup B$ .
(≥) Let $\varepsilon > 0$ . We can choose $a \in A$ with $a > \sup A - \frac{\varepsilon}{2}$ and $b \in B$ with $b > \sup B - \frac{\varepsilon}{2}$ (definition of supremum).
Then $a + b > \sup A + \sup B - \varepsilon$ , so $\sup(A+B) \geq \sup A + \sup B - \varepsilon$ .
Since $\varepsilon$ was arbitrary, $\sup(A+B) \geq \sup A + \sup B$ . Combining the two inequalities gives $\sup(A+B) = \sup A + \sup B$ .

Accept from completeness that $s = \sup\{x \in \mathbb{Q} : x \geq 0, x^2 < 2\}$ exists, and verify $s^2 = 2$ . (The existence step is idea-level.)

s^2 < 2 \Rightarrow \text{contradiction},\quad s^2 > 2 \Rightarrow \text{contradiction}

\sqrt{2} \approx 1.4142

— rational truncations approach the hole.

Show proof

(Lemma — cited) The rationals $\mathbb{Q}$ are dense in $\mathbb{R}$ : every open interval of reals contains a rational number.
$S = \{x \in \mathbb{Q} : x \geq 0, x^2 < 2\}$ is nonempty ( $1 \in S$ ) and bounded above (e.g. $2$ is an upper bound). By the completeness axiom, $s = \sup S$ exists.
Suppose $s^2 < 2$ . Since $1 \in S$ and $s = \sup S$ , we have $s \ge 1$ . Put $\varepsilon = 2 - s^2 > 0$ and pick $h > 0$ with $h < \min\left(1, \frac{\varepsilon}{2s+1}\right)$ ; by the lemma there is a rational $q = s + h'$ with $0 < h' < h$ (so $q > s \ge 0$ ).
Bound the second factor explicitly: since $h' < 1$ , $q^2 = s^2 + 2sh' + h'^2 = s^2 + h'(2s + h') < s^2 + h'(2s+1) < s^2 + \varepsilon = 2$ . Hence $q \in S$ with $q > s$ , contradicting that $s$ is an upper bound.
Suppose $s^2 > 2$ . Again $1 \in S$ gives $s \ge 1 > 0$ — so division by $2s$ is legitimate. Put $\delta = s^2 - 2 > 0$ ; since $\delta < s^2$ , we have $\frac{\delta}{2s} < \frac{s}{2} < s$ , and by the lemma there is a rational $r$ with $s - \frac{\delta}{2s} < r < s$ , in particular $r > s/2 \ge 0$ .
Now $0 < s - r < \frac{\delta}{2s}$ and $0 < s + r < 2s$ , so $s^2 - r^2 = (s-r)(s+r) < \frac{\delta}{2s} \cdot 2s = \delta$ , i.e. $r^2 > s^2 - \delta = 2$ . That $r$ is an upper bound: any $x \in S$ has $x \ge 0$ and $x^2 < 2 < r^2$ , and squaring is monotone on $[0, \infty)$ , so $x < r$ , i.e. $x \le r$ . Hence $r$ is an upper bound of $S$ smaller than $s$ , contradicting that $s$ is the least upper bound.
Hence $s^2 = 2$ . The completeness axiom fills the hole in the rationals, producing the real number that is $\sqrt{2}$ .

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Examples & Exercises