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Every non-empty set bounded above has a least upper bound; the approximation form, the nested interval theorem, and what fails in $\mathbb{Q}$.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to state the completeness axiom, use its approximation form to produce an element of a set arbitrarily close to the supremum, derive the existence of infima from it, and say exactly which line of an argument the rational numbers refuse. You will also be able to use the nested interval theorem, which is the shape completeness takes in nearly every later proof of this course.
From the last lesson: upper bound, supremum, infimum, and the two-part characterisation of a least upper bound. That lesson found suprema in sets that obviously had them. This one asks the question underneath: what entitles anyone to assume the supremum is there?
Complete means that every non-empty set bounded above has a least upper bound. The approximation property is the second half of the supremum characterisation used forwards: some element of the set exceeds the supremum minus any given positive amount. A family of intervals is nested when each one contains the next.
The completeness axiom. Every non-empty set of real numbers that is bounded above has a least upper bound in $\mathbb{R}$.
That is the whole of it. It is an axiom and not a theorem: no argument from the field and order axioms can produce it, because the rationals satisfy every one of those and fail this. The axiom is the definition of what makes $\mathbb{R}$ different, and every existence theorem in this course — a limit, a root, a maximum, an integral — comes back to it.
The approximation form. The characterisation of a least upper bound, used forwards, is the tool:
> if $M = \sup S$ and $\varepsilon > 0$, then some $s \in S$ has $s > M - \varepsilon$.
Read it as a supply: the axiom hands you an element as close to $M$ as you care to ask for. Nearly every use of completeness in this course is this sentence with a particular $\varepsilon$.
The infimum comes free. The axiom mentions only upper bounds, which looks lopsided. It is not: reflecting a set in the origin turns lower bounds into upper bounds, so the greatest lower bound of $S$ is minus the least upper bound of $-S$. One axiom, both ends.
Completeness is what lets a limit be produced rather than guessed. Boundedness plus monotonicity, or the Cauchy condition, names a number before anyone knows what it is; over the rationals the same argument runs and the number it names is not there.
Another way: picture
Think of the rationals as a line with invisible punctures. Approach the puncture at the square root of two from the left, through rationals, and the climb has no top inside $\mathbb{Q}$: every rational bound can be beaten by a smaller one. The axiom asserts that $\mathbb{R}$ has no such punctures, so every climb has a top.
Another way: steps
To use completeness:
Step 1 is the creative one; steps 2 to 4 are the same every time.
Let $S = \{x \in \mathbb{Q} : x > 0, \ x^2 < 2\}$.
Non-empty: $1 \in S$. True in $\mathbb{Q}$.
Bounded above: $2$ is an upper bound. True in $\mathbb{Q}$.
Has a least upper bound in $\mathbb{Q}$: false. Given any rational bound $M$ with $M^2 > 2$, the rational $M - (M^2 - 2)/(2M)$ is smaller and still a bound. So the bounds have no least member, and the collection of them has no bottom.
Both hypotheses hold and the conclusion fails. That is worth dwelling on, because it is the shape of every counterexample in this course: the theorem is not true by pure logic, and the field it is stated over is doing the work.
In $\mathbb{R}$ the same set has a supremum, and the same estimate shows its square can be neither above nor below $2$. So $\sqrt{2}$ exists — and the proof of that is nothing but the axiom plus two inequalities.
If $I_1 \supseteq I_2 \supseteq \cdots$ are closed bounded intervals $[a_n, b_n]$, their intersection is non-empty; and if the lengths $b_n - a_n$ shrink to zero, the intersection is a single point.
Why: the left ends $a_n$ increase and are bounded above by $b_1$, so $c = \sup\{a_n\}$ exists — that is the axiom, used once. Every $b_m$ is an upper bound of the $a_n$, so $c \le b_m$ for all $m$, and $a_n \le c$ by construction. Hence $c \in I_n$ for every $n$.
This is the form completeness takes in almost every later proof. Bisect an interval, keep the half where the property you want persists, and the point you are after is the one the nested intervals meet in. Bolzano-Weierstrass and the intermediate value theorem are both this argument with a different property kept.
Closed matters. The intervals $(0, 1/n)$ are nested with lengths shrinking to zero and meet in nothing at all.
It is not "there are no gaps" as a slogan. It is a statement about sets and bounds, and its uses are always of the form build a set, take its supremum. A slogan cannot be applied; this sentence can.
It does not follow from density. The rationals are dense in themselves and in the reals, and they are not complete. Density says there is always something in between; completeness says a climb has a top. Neither implies the other.
It is not about infinity. The axiom's hypothesis is boundedness, and its conclusion is a real number. No infinite quantity appears anywhere in it.
Let $S = \{x > 0 : x^2 < 2\}$ as a subset of $\mathbb{R}$. It is non-empty and bounded above by $2$.
Both hypotheses, checked explicitly.
So $M = \sup S$ exists. If $M^2 < 2$, a small step up stays in $S$, contradicting that $M$ bounds it; if $M^2 > 2$, a small step down still bounds $S$, contradicting leastness.
Squeeze the value from both sides.
Hence $M^2 = 2$. The number was never constructed — it was produced by the axiom and then identified.
Existence first, value second.
Suppose $M = \sup S$ and every element of $S$ is at most $L$. Claim: $M \le L$.
A comparison of two bounds.
If instead $M > L$, apply the property with $\varepsilon = M - L$: some $s \in S$ exceeds $M - \varepsilon = L$.
Choose the epsilon that makes the contradiction appear.
That contradicts $L$ bounding $S$, so $M \le L$. The supremum is the least bound, and this is what leastness is for.
The choice of epsilon is the whole argument.
They are nested, and their lengths shrink to zero, so the closed version would meet in one point.
Compare with the theorem before deviating from it.
A number in all of them would have to be positive and below every $1/n$; the Archimedean property, coming next lesson, rules that out.
Name the candidate and rule it out.
And $0$ itself is excluded from every interval. So the intersection is empty, and the closedness in the nested interval theorem is carrying real weight rather than tidiness.
Put the steps of a proof that $2$ is the supremum of the open interval from $0$ to $2$ into the order they must be made in.
Number the steps in order (write the number in the box):
$S = \{1 - 1/k : k \in \mathbb{N}\}$ has supremum $1$. Take $\varepsilon = \dfrac{1}{5}$. What is the smallest element of $S$ that exceeds $1 - \varepsilon$?
Answer:
Build the proof that a non-empty set bounded below has a greatest lower bound, using only the completeness axiom.
This task has no paper form; do it on a device.
Let $S$ be the set of positive rationals whose square is below $2$, where $2$ is not a perfect square. Regarded as a subset of $\mathbb{Q}$, what goes wrong?
Here is an argument that $\mathbb{Q}$ contains a number whose square is $3$. Mark the one step that $\mathbb{Q}$ does not permit.
This task has no paper form; do it on a device.
For each $n$, let $I_n$ be the closed interval from $2 - 1/n$ to $2 + 1/n$. These are nested and their lengths shrink to zero. Give the set of numbers lying in every $I_n$.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Put the steps of a proof that $3$ is the supremum of the open interval from $0$ to $3$ into the order they must be made in.
Number the steps in order (write the number in the box):
You can state the completeness axiom, use the approximation property, and point to the step an argument takes that the rationals do not allow. Say in your own words why the axiom cannot be proved from the field and order axioms. Next: the first thing the axiom buys, which is that no real number is larger than every whole number.
10. Your turn: show that the intervals from $0$ to $1/n$, ends excluded, meet in nothing, step 3