Math Infinitum

Mapping the lesson.

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Intuition

Here the question from the first lesson is answered. The rationals have gaps, and completeness says exactly what a gap is: a set that is bounded above but has no least upper bound. The real numbers are built so that this never happens, and that single guarantee is what the whole of analysis rests on.

Return to the crowd pressing towards a wall. In the rationals the wall itself can be missing, so the crowd presses towards nothing at all. Completeness is the promise that the wall is always there to be pressed against.

The completeness axiom

This is not proved. It is the property that distinguishes the real numbers from the rationals, and it is assumed. Everything of substance in analysis is eventually traced back to it.

\begin{aligned}&S \ne \varnothing \text{ and } S \text{ bounded above}\\ &\implies \sup S \text{ exists in } \mathbb{R}\end{aligned}

What it does and does not say

The set must be non-empty. Every number bounds the empty set above, so no least one exists.
It promises a supremum, not a maximum. The bound need not belong to the set.
The mirror statement for infima follows from it and need not be assumed separately.
$\mathbb{Q}$ fails it, and the theorem below shows exactly where.
Infima come free: if $S$ is non-empty and bounded below then $-S$ is bounded above, and $\inf S=-\sup(-S)$ .

Take the rationals whose square is below 2. They are bounded above — by 1.5, by 2, by any rational beyond $\sqrt2$ — and among the rationals there is no least such bound, because $\sqrt2$ is not one. That missing number is the gap, and completeness is the promise that the reals have none.

The rationals are not complete

\begin{aligned}&S = \{q \in \mathbb{Q} : q > 0,\ q^2 < 2\}\\ &\text{has no supremum in } \mathbb{Q}\end{aligned}

The set is non-empty and bounded above, so completeness would hand us a least upper bound. Every rational candidate fails: too small and it is not a bound, too large and it is not least, and equal is impossible by the first lesson.

Proof steps

S \ne \varnothing \text{ and } 2 \text{ bounds } S \text{ above}

One belongs to $S$ , and anything above two has square above four.

\text{Suppose } s \in \mathbb{Q} \text{ were the least upper bound}

Assume the rationals do supply one, and examine the three possibilities for its square.

s^2 = 2 \text{ is impossible, as no rational squares to } 2

This is the theorem from the first lesson of the chapter.

s^2 < 2 \implies \text{some larger rational still lies in } S

Then $s$ does not bound $S$ above, so it was not an upper bound at all.

s^2 > 2 \implies \text{some smaller rational still bounds } S

Then $s$ was not the least such number.

\begin{aligned}&\text{Every case fails, so no rational}\\ &\text{least upper bound exists}\end{aligned}

The set is bounded above with no supremum, which is precisely a gap.

Applications

Practice

What completeness promises

S \ne \varnothing,\ S \text{ bounded above} \implies \sup S \text{ exists}

Two conditions in, one guaranteed number out.

Try it

Which condition does completeness require of a set before promising a supremum?