Supremum & Completeness

We saw that the rationals have gaps: a bounded set can fail to have a "boundary" number inside \mathbb{Q}. The completeness axiom is the precise promise that \mathbb{R} never does this. To state it we need one idea: the least upper bound, or supremum.

Fix a nonempty set S \subseteq \mathbb{R}. A number u is an upper bound of S if it is at least as large as every element:

u \text{ is an upper bound of } S \quad\Longleftrightarrow\quad x \le u \ \text{ for all } x \in S.

A set may have many upper bounds (if u works, so does anything bigger). The supremum \sup S is the smallest of them — the least upper bound. The mirror notion, the greatest lower bound, is the infimum \inf S.

The two-part test for a supremum

Saying "s is the least upper bound" packs two separate demands. Unfolding them gives the working characterisation every proof uses. We claim: s = \sup S exactly when both

\textbf{(i)}\ \ x \le s \ \text{ for all } x \in S, \qquad\qquad \textbf{(ii)}\ \ \forall\, \varepsilon > 0\ \ \exists\, x \in S \ \text{ with } x > s - \varepsilon.

Step 1 — why (i): s must be an upper bound. The supremum is by definition an upper bound (the least one), so it certainly bounds S from above. That is exactly (i).

Step 2 — why (ii): nothing smaller can bound S. Suppose, for some \varepsilon > 0, there were no element of S exceeding s - \varepsilon. Then every x \in S satisfies x \le s - \varepsilon, making s - \varepsilon an upper bound smaller than s. That contradicts s being the least upper bound. So such an \varepsilon cannot exist: for every \varepsilon > 0 some x \in S pokes above s - \varepsilon. That is (ii).

Step 3 — the converse. Conversely, if s satisfies (i) and (ii), then (i) makes it an upper bound, and (ii) says no number below it is an upper bound (anything s - \varepsilon < s is beaten by some element of S). So s is the least upper bound — s = \sup S. The two conditions together are equivalent to being the supremum.

Worked example: \sup\{1 - \tfrac{1}{n}\} = 1

Let S = \{\, 1 - \tfrac1n : n = 1, 2, 3, \dots \,\} = \{0,\ \tfrac12,\ \tfrac23,\ \tfrac34,\ \dots\}. The terms climb toward 1 without ever reaching it. We verify \sup S = 1 against the test.

Check (i): for every n \ge 1 we have \tfrac1n > 0, so 1 - \tfrac1n < 1. Thus 1 is an upper bound.

Check (ii): given any \varepsilon > 0, we must find n with 1 - \tfrac1n > 1 - \varepsilon, i.e. \tfrac1n < \varepsilon. By the Archimedean property (see the vignette) there is an integer n > \tfrac1\varepsilon, and for that n we get \tfrac1n < \varepsilon. So some element of S lands within \varepsilon of 1.

Both halves hold, so \sup S = 1. Note 1 \notin S: a supremum need not be attained. (Here \inf S = 0, and that one is attained, at n = 1.)

Completeness axiom. Every nonempty set S \subseteq \mathbb{R} that is bounded above has a least upper bound \sup S \in \mathbb{R}. This is the defining property separating \mathbb{R} from \mathbb{Q}.
Characterisation. s = \sup S iff (i) x \le s for all x \in S, and (ii) for every \varepsilon > 0 there is an x \in S with x > s - \varepsilon.
Infimum, by symmetry. Every nonempty set bounded below has a greatest lower bound \inf S = -\sup(-S), where -S = \{-x : x \in S\}.

See the least upper bound

Below, the orange band is a set S. Every point from \sup S rightward is an upper bound (the shaded region); the supremum is the left edge of that region — the smallest number still bounding S. Drag the \varepsilon slider: the window (\sup S - \varepsilon,\ \sup S] always catches part of S, no matter how small \varepsilon is — that is condition (ii) made visible.

The Archimedean property says: for every real x there is a natural number n > x — no real is "infinitely large", and \tfrac1n can be made as small as you like. It looks obvious, but it is a consequence of completeness.

Proof. Suppose not: suppose \mathbb{N} were bounded above. Being nonempty and bounded above, by completeness it would have a supremum s = \sup \mathbb{N}. By condition (ii) with \varepsilon = 1, some n \in \mathbb{N} satisfies n > s - 1. But then n + 1 > s, and n + 1 is also a natural number — contradicting that s bounds \mathbb{N}. So \mathbb{N} is not bounded above, which is exactly the Archimedean property. Archimedes assumed it; here it is earned.

We stated completeness for sets bounded above. The dual — every nonempty set bounded below has a greatest lower bound — needs no new axiom; reflect through 0. If S is bounded below by \ell, then -S = \{-x : x \in S\} is bounded above by -\ell, so \sup(-S) exists, and

\inf S = -\sup(-S).

Reflecting the line swaps "below" with "above" and least with greatest, so one axiom delivers both bounds. This symmetry recurs throughout analysis — prove a fact for suprema, then read off its mirror for infima for free.