Completeness and the Archimedean Property

The rational numbers have every algebraic property the reals have, so what makes $\mathbb{R}$ special? In Algebra 1 the answer was "the reals fill in the gaps, like $\sqrt{2}$." Analysis makes that precise with one axiom about suprema.

Idea clave

Completeness axiom. Every nonempty set of real numbers that is bounded above has a supremum in $\mathbb{R}$. Archimedean property (a consequence): for every real $x$ there is a positive integer $n > x$; equivalently, for every $\varepsilon > 0$ there is a positive integer $n$ with $\frac{1}{n} < \varepsilon$.

Por qué funciona

Suppose the positive integers were bounded above. By completeness they would have a supremum $s$. Then $s - 1$ is not an upper bound, so some integer $n$ has $n > s - 1$. But then $n + 1 > s$, and $n + 1$ is a positive integer, contradicting that $s$ is an upper bound. So for every $x$ some integer exceeds it. Taking $x = \frac{1}{\varepsilon}$ gives $n > \frac{1}{\varepsilon}$, that is, $\frac{1}{n} < \varepsilon$.

Completeness is exactly what fails in $\mathbb{Q}$. The set $S = \{x \in \mathbb{Q} : x > 0,\ x^{2} < 2\}$ is bounded above (by $2$, say), and in $\mathbb{R}$ its supremum is $\sqrt{2}$. But $\sqrt{2}$ is not rational, and no rational number can serve as the least upper bound: any rational upper bound $r$ has $r^{2} > 2$, and a slightly smaller rational is still an upper bound. Every limit theorem in this course (monotone sequences converge, the intermediate value theorem, the extreme value theorem) uses completeness somewhere, and each one is false in $\mathbb{Q}$.

Rationals with $x^{2} < 2$ crowd up against $\sqrt{2} \approx 1.414$. The supremum exists in $\mathbb{R}$, not in $\mathbb{Q}$.

Worked examples

Ejemplo resuelto: The Archimedean property with numbers

Find the smallest positive integer $n$ with $\frac{1}{n} < 0.003$.

$\frac{1}{n} < 0.003$ means $n > \frac{1}{0.003} = 333.\overline{3}$.
The smallest whole number above $333.\overline{3}$ is $334$.

Respuesta: $n = 334$.

Ejemplo resuelto: When equality is close

Find the smallest positive integer $n$ with $\frac{1}{n} < \frac{1}{30}$.

The condition is $n > 30$.
$n = 30$ gives equality, which is not "less than."

Respuesta: $n = 31$.

Ejemplo resuelto: A supremum outside the set's number system

Let $S = \{x \in \mathbb{Q} : x > 0,\ x^{2} < 5\}$. What is $\sup S$, and is it rational?

Every element is less than $\sqrt{5}$, and rationals come arbitrarily close to $\sqrt{5}$ (density, next topic).
So $\sup S = \sqrt{5}$, which is irrational because $5$ is not a perfect square.

Respuesta: $\sup S = \sqrt{5}$, not rational.

Error común

Assuming suprema stay in the same number system. A bounded set of rationals need not have a rational supremum, as $S$ above shows. In a proof, "let $s = \sup A$" is allowed only when $A$ is a nonempty set of reals bounded above, and then $s$ is a real number. Check both hypotheses (nonempty, bounded above) every time you invoke completeness.

Worked examples

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