The Real Numbers and Completeness

In this lesson you’ll learn what suprema and infima are, how to prove a number is a supremum, the least upper bound axiom, and the two immediate consequences that get used constantly: the Archimedean property and the density of the rationals.

A set $S \subseteq \mathbb{R}$ is bounded above if there is some $M$ with $x \le M$ for every $x \in S$. Such an $M$ is an upper bound. Bounded below and lower bounds are defined the same way. Bounded means both.

Upper bounds are never unique: if $M$ works, so does $M + 1$. The interesting question is whether there is a smallest one.

$M$ is the supremum of $S$, written $\sup S$, if

1. $M$ is an upper bound: $x \le M$ for all $x \in S$, and
2. $M$ is the least one: no $M' < M$ is an upper bound.

Condition 2 is awkward to use directly. This equivalent version is the one you actually work with:

$$M = \sup S \iff M \text{ is an upper bound, and } \forall \varepsilon>0 \; \exists x \in S \text{ with } x > M - \varepsilon$$

In words: nothing below $M$ is an upper bound, because you can always find an element of $S$ within $\varepsilon$ of $M$. That $\varepsilon$-characterization is how essentially every supremum proof goes.

The key subtlety: the supremum need not be an element of $S$. For $S = (0,1)$, $\sup S = 1 \notin S$. When the supremum does belong to $S$ it is called the maximum. Every maximum is a supremum; not every supremum is a maximum.

Least upper bound axiom. Every non-empty subset of $\mathbb{R}$ that is bounded above has a supremum in $\mathbb{R}$.

This is an axiom, not a theorem. It is part of what defines the real numbers, and it is exactly what the rationals lack.

The set $S = \{x \in \mathbb{Q} : x^2 < 2\}$ is non-empty and bounded above, but no rational is its least upper bound. Any rational with $p^2 < 2$ can be nudged up and still square below 2, so it isn’t an upper bound; any rational with $p^2 > 2$ can be nudged down and still exceed everything in $S$, so it isn’t least. In $\mathbb{R}$ the supremum exists and is $\sqrt2$.

By applying the axiom to $-S$, the matching statement for infima follows for free: every non-empty set bounded below has an infimum.

Every major theorem ahead is a consequence of completeness, and every one of them is false over $\mathbb{Q}$:

• Monotone convergence - a bounded increasing sequence converges. Over $\mathbb{Q}$, the decimal truncations of $\sqrt2$ increase and are bounded and converge to nothing rational.
• Bolzano-Weierstrass - every bounded sequence has a convergent subsequence.
• Cauchy completeness - every Cauchy sequence converges.
• Intermediate value theorem - over $\mathbb{Q}$, $f(x) = x^2 - 2$ goes from negative to positive without ever being zero.
• Extreme value theorem and Heine-Borel.

When a proof in this section feels like it appeared from nowhere, look for where it took a supremum. That step is almost always the load-bearing one.

For every real $x$ there is a natural number $n$ with $n > x$.

Equivalently, $\mathbb{N}$ is unbounded above, and equivalently, for every $\varepsilon > 0$ there is $n$ with $\frac1n < \varepsilon$.

This follows from completeness. Suppose $\mathbb{N}$ were bounded above; then it would have a supremum $M$. By the $\varepsilon$-characterization with $\varepsilon = 1$, there is $n \in \mathbb{N}$ with $n > M - 1$, hence $n + 1 > M$. But $n+1 \in \mathbb{N}$, contradicting that $M$ is an upper bound. ∎

The third form, ”$1/n$ can be made as small as you like,” is what gets used in practice: it is how you turn “for all $\varepsilon$” into “for all $n$ beyond some point.”

Between any two distinct reals there is a rational.

So the rationals are dense in $\mathbb{R}$: every real is a limit of rationals, and every interval, however short, contains infinitely many. The irrationals are dense too.

This is why decimal approximations work at all, and why numerical computation on rationals can approximate any real quantity to any accuracy - even though it can never represent most of them exactly.

Example 1: Find the supremum and infimum of $S = \{1 - \frac1n : n \in \mathbb{N}\}$.

Solution. The elements are $0, \frac12, \frac23, \frac34, \dots$, increasing toward 1.

Infimum. The smallest element is $0$ (at $n=1$), and $1 - \frac1n \ge 0$ always. So $\inf S = 0$, and since $0 \in S$ it is also the minimum.

Supremum. Every element is $< 1$, so 1 is an upper bound. Is it least? Take any $\varepsilon>0$. By the Archimedean property choose $n$ with $\frac1n < \varepsilon$. Then

$$1 - \tfrac1n > 1 - \varepsilon$$

so an element of $S$ exceeds $1 - \varepsilon$. Hence no number below 1 is an upper bound, and $\sup S = 1$.

$1 \notin S$, so there is no maximum. This is the standard illustration that a supremum can sit outside its set.

Example 2: Prove $\sup\{x \in \mathbb{R} : x^2 < 4\} = 2$.

Solution. The set is $(-2,2)$.

Upper bound. If $x^2 < 4$ then $|x| < 2$, so $x < 2$. ✓

Least. Let $\varepsilon>0$. We need an element exceeding $2-\varepsilon$. If $\varepsilon \ge 4$ then $0$ works. Otherwise take $x = 2 - \frac{\varepsilon}{2}$. Then $0 < x < 2$, so $x^2 < 4$ and $x \in S$, and $x > 2 - \varepsilon$. ✓

So $\sup S = 2$. ∎

Notice the shape: verify the bound, then given an arbitrary $\varepsilon$ construct a witness. Every supremum proof looks like this.

Example 3: A set whose supremum is a maximum.

Solution. For $S = [0, 3]$, we have $\sup S = 3$ and $3 \in S$, so $\max S = 3$. Likewise $\inf S = \min S = 0$.

Compare $T = [0,3)$: still $\sup T = 3$, but $3 \notin T$ so $T$ has no maximum. The supremum does not change when you remove the endpoint, which is precisely why “sup” is the useful notion and “max” is not.

Example 4: Suprema of unbounded and empty sets.

Solution. $\mathbb{N}$ is not bounded above, so it has no supremum in $\mathbb{R}$. (Some texts write $\sup \mathbb{N} = +\infty$ as shorthand; that is a convention, not a real number.)

The empty set is a genuine edge case: every real is an upper bound for $\emptyset$, so there is no least one. This is why the axiom says non-empty. Both hypotheses in “non-empty and bounded above” are doing work.

Example 5: Use the Archimedean property.

Show that $\inf\{\frac1n : n \in \mathbb{N}\} = 0$.

Solution. Lower bound. $\frac1n > 0$ for all $n$, so 0 is a lower bound. ✓

Greatest. Suppose some $c > 0$ were a lower bound, so $\frac1n \ge c$ for all $n$, that is $n \le \frac1c$ for all $n$. That makes $\mathbb{N}$ bounded above, contradicting the Archimedean property.

So no positive number is a lower bound, and $\inf = 0$. ∎

Note $0$ is not in the set, so there is no minimum. The set has elements arbitrarily close to 0 and never reaches it.

Example 6: Suprema and sums.

For non-empty bounded sets $A, B$ and $A + B = \{a + b\}$, show $\sup(A+B) = \sup A + \sup B$.

Solution. Write $\alpha = \sup A$, $\beta = \sup B$.

Upper bound. Any element of $A+B$ is $a + b \le \alpha + \beta$. ✓

Least. Let $\varepsilon>0$. Pick $a \in A$ with $a > \alpha - \frac{\varepsilon}{2}$ and $b \in B$ with $b > \beta - \frac{\varepsilon}{2}$, both possible by the $\varepsilon$-characterization. Then

$$a + b > \alpha + \beta - \varepsilon$$

so no smaller number is an upper bound. Hence $\sup(A+B) = \alpha+\beta$. ∎

The trick of splitting $\varepsilon$ into $\frac{\varepsilon}{2} + \frac{\varepsilon}{2}$ appears in nearly every proof from here on. Worth getting comfortable with it now.

Floating-point representation. A double stores a rational approximation, so the reals a computer manipulates are always rational. Density of $\mathbb{Q}$ is what makes that acceptable: any real can be approximated to within machine precision. Completeness is also why the limits your algorithms converge to are guaranteed to exist even though the machine never reaches them.

Optimization and economics. “The supremum of achievable profit” always exists for a bounded feasible set, but it may not be attained. That distinction is exactly sup versus max, and it is why optimization theory works so hard to establish that a maximum is achieved rather than merely bounded.

Numerical root finding. Bisection works by maintaining an interval containing a root and halving it. The root it converges to exists because of completeness; over $\mathbb{Q}$ the algorithm would run forever narrowing in on a value that is not there. Solving $x^2 = 2$ by bisection is literally constructing $\sqrt2$ as a supremum.

Measurement and tolerance. An engineering specification like “within $\varepsilon$ of nominal for some achievable $\varepsilon$” is an $\varepsilon$-argument. The Archimedean property is the formal statement that any tolerance, however tight, can be met by taking enough steps.

Auctions and pricing. The market-clearing price is often characterized as a supremum or infimum of a set of bids, and whether it is attained determines whether the auction has a well-defined winner.