Completeness, sup/inf, Archimedean property

In one sentence

Completeness is the single axiom that separates $\mathbb{R}$ from $\mathbb{Q}$, and almost every question in this area is really asking which property fails once you take it away.

Why the exam asks it

Because it is the cheapest way to test whether you know what a supremum is, as opposed to how to compute one. A candidate who thinks "sup means maximum" will answer several options wrongly in a single question, and the paper is designed to find them. Expect at least one option turning on a supremum that is not attained, and one on a property that holds in $\mathbb{R}$ but not in $\mathbb{Q}$.

The idea

The completeness axiom: every non-empty subset of $\mathbb{R}$ that is bounded above has a least upper bound in $\mathbb{R}$.

That is the whole of it, and everything else in this topic is a consequence.

To see what it rules out, look at $A = \{x \in \mathbb{Q} : x^2 < 2\}$ inside $\mathbb{Q}$. It is non-empty. It is bounded above — by $2$, say. And it has no least rational upper bound: given any rational upper bound you can always find a smaller one, because the candidate limit is irrational.

$\mathbb{Q}$ is not missing a few points. It is torn at every irrational, and completeness is the axiom that repairs it.

Working with sup and inf

The characterisation you should reach for, every time:

$s = \sup A \iff s \text{ is an upper bound, and for every } \varepsilon > 0 \text{ there is } a \in A \text{ with } a > s - \varepsilon.$

The second half is the useful half. It says $s$ can be approached from inside $A$ — nothing smaller than $s$ can bound $A$, because you can always find an element of $A$ above it.

Note what is not claimed: that $s \in A$. The supremum need not be attained. $\sup(0,1) = 1$, and $1 \notin (0,1)$. For a compact set the supremum is attained, which is the whole content of "a continuous function on a closed bounded interval attains its bounds".

Two identities worth having ready, and one non-identity:

• $\sup(A + B) = \sup A + \sup B$.
• $\sup(-A) = -\inf A$.
• $\sup(AB) \ne \sup A \cdot \sup B$ in general — negatives ruin it.

The Archimedean property

For every $x \in \mathbb{R}$ there is a natural number $n > x$.

This looks too obvious to name, which is exactly why it is named: it is a theorem, proved from completeness, not a triviality about numbers. There exist ordered fields where it fails — the field of rational functions ordered by eventual growth, in which $x$ exceeds every constant, so no integer dominates it.

From it follow $1/n \to 0$, and the density of $\mathbb{Q}$ in $\mathbb{R}$: between any two reals lies a rational, and also an irrational.

Where intuition breaks

"The supremum is the maximum." Only when it happens to lie in the set. $\sup(0,1) = 1$ and the set has no maximum at all. This is the single most productive distractor in the topic, because it converts a true statement about sup into a false one about max without changing a symbol on the page.

"$\sup(A \cap B) = \min(\sup A, \sup B)$." Tempting, symmetric, and false. Take $A = (0,1)$ and $B = (2,3)$: the intersection is empty. Even with overlap it fails — the intersection can sit well below both suprema. This is finite-dimensional intuition of a particular kind: reasoning about sets as though they were intervals that always meet.

"Every ordered field is Archimedean." False, and the counterexample is the rational-function field above. The reason this matters is that it exposes what completeness is doing: Archimedes is not built into the idea of order, it is bought with the completeness axiom.

"$\mathbb{Q}$ is countable, so it is small; therefore its complement is large in every sense." Both statements are true and the inference is still worth care. $\mathbb{Q}$ has measure zero, so in that sense it is small. But it is dense — every interval contains infinitely many rationals. Small in measure and everywhere at once are perfectly compatible, and questions that pair the two are testing whether you conflate them.

"An uncountable set cannot have measure zero." The Cantor set. Uncountable, and measure zero. This one is worth memorising outright, because there is no way to see it by intuition.

The exam's angle

Questions in this area almost always take one of three shapes.

1. Four statements about sup and inf, one of which quietly assumes attainment. Check each against a set whose supremum is not in it — $(0,1)$ answers most of them in seconds.
2. "Which of these properties of $\mathbb{R}$ fails in $\mathbb{Q}$?" The answer is completeness and everything equivalent to it: monotone convergence, Bolzano–Weierstrass, Cauchy completeness, nested intervals, Heine–Borel. Order, density and countability of $\mathbb{Q}$ are not affected.
3. A cardinality or measure statement. Keep three facts to hand: $\mathbb{Q}$ countable, $\mathbb{R}$ uncountable, Cantor set uncountable with measure zero.

The night before

• Completeness: every non-empty set bounded above has a least upper bound. That is the axiom; the rest of the list is equivalent to it.
• $s = \sup A$ means $s$ bounds $A$ and for every $\varepsilon$ some element exceeds $s - \varepsilon$.
• Sup need not be attained. $(0,1)$ is the disproof of half the false options in this topic.
• Archimedes is a theorem, not an obvious fact — non-Archimedean ordered fields exist.
• $\mathbb{Q}$: countable, measure zero, dense. All three at once, and no contradiction.
• Cantor set: uncountable, measure zero.