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Upper bounds, suprema and infima, the completeness axiom, and the hole in the rationals it fills.
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 with both of its hypotheses, find the supremum and infimum of a set given by a formula, prove a supremum claim in the two halves it requires — that your candidate bounds the set, and that nothing smaller does — and say in each case whether the bound is attained, so that a supremum and a maximum are never confused again. You will also be able to say what completeness buys: it is the axiom that separates $\mathbb{R}$ from $\mathbb{Q}$, and every existence theorem in the rest of this course rests on it.
You have used the real numbers all through school as the points of a line, and you have met intervals, inequalities and the idea that $\sqrt{2}$ is not a fraction. None of that is being replaced. What is added is a single axiom that says what the line has and the rationals lack, and the reason to add it is that every theorem in this course — the intermediate value theorem, the extreme value theorem, the mean value theorem, the integrability of a continuous function — is false without it.
The real numbers form an ordered field: they add, multiply and compare, and the arithmetic respects the order. So do the rationals. One axiom separates them.
The completeness axiom. Every nonempty set of real numbers that is bounded above has a least upper bound.
An upper bound for $S$ is a number $B$ with $x \le B$ for every $x \in S$. The supremum $\sup S$ is the least such $B$: it bounds $S$, and no smaller number does. The mirror statements define a lower bound and the infimum $\inf S$.
The word that does the work is least. Plenty of sets have upper bounds; the axiom says one of them is smallest, and that the smallest one is a real number. In $\mathbb{Q}$ this fails: $\{x \in \mathbb{Q} : x^2 < 2\}$ is bounded above by $2$, by $1.5$, by $1.42$ — and by no smallest rational, because any rational bound can be shaved. The least bound is $\sqrt{2}$, which is not there. The rationals have a hole exactly where the supremum should be, and completeness is the statement that $\mathbb{R}$ has none.
Supremum is not maximum. A maximum must be an element of the set. $\sup (2, 5) = 5$, and $5 \notin (2,5)$, so the interval has a supremum and no maximum. Every maximum is a supremum; the converse fails, and the sets where it fails are the interesting ones.
Another way: picture
A number line with a set shaded up to, but not including, a point. The supremum is that point. Slide a barrier leftwards from the far right until it first touches the shading: where it stops is the supremum, and whether it is standing on a point of the set decides whether there is a maximum.
Another way: steps
To find $\sup S$, two things must be shown and the second is the one that is usually skipped.
Showing that $1$ bounds $\{1 - 1/n\}$ takes one line and proves almost nothing: $2$ bounds it too, and so does $47$. The content of a supremum claim is that nothing smaller works, and the standard way to show it is to let an opponent name a smaller candidate $B - \varepsilon$ and then produce an element of the set above it.
For $S = \{1 - 1/n\}$: given $\varepsilon > 0$, choose $n > 1/\varepsilon$. Then $1/n < \varepsilon$, so $1 - 1/n > 1 - \varepsilon$, and the candidate $1 - \varepsilon$ is not a bound. Since $\varepsilon$ was arbitrary, no number below $1$ bounds $S$, and $\sup S = 1$.
That argument — name an $\varepsilon$, produce a witness — is the same shape as the epsilon-delta definition two lessons from now, and it is worth recognising here, where the setting is simple, rather than meeting it for the first time attached to a limit.
| Fact | Statement | Where it is used |
|---|---|---|
| Archimedean property | for any real $x$ there is a natural $n > x$ | every "take $n$ large enough" step |
| Density of $\mathbb{Q}$ | between any two reals there is a rational | approximation, decimal expansions |
| $\sup$ of a scaled set | $\sup(cS) = c \sup S$ for $c > 0$ | changing units in a bound |
| $\sup$ of a shifted set | $\sup(S + c) = \sup S + c$ | translating a problem to the origin |
The Archimedean property is not a separate axiom: it follows from completeness. If $\mathbb{N}$ were bounded above it would have a supremum $B$; then $B - 1$ is not a bound, so some $n > B - 1$, so $n + 1 > B$ — and $n+1$ is a natural number above the bound. The contradiction is what is what permits every "choose $n$ large enough" in the rest of this course.
Naming a bound and stopping. "$1$ is an upper bound, so $\sup S = 1$" is half an argument. Until nothing smaller has been ruled out, the claim is unproved.
Reading the largest term as the infimum. For $T = \{p + q/n\}$ the terms decrease towards $p$. The largest term, at $n = 1$, is the maximum; the infimum is at the other end and is never attained.
Applying completeness to an unbounded set. The axiom has two hypotheses, nonempty and bounded above, and a set failing either has no supremum at all. Writing $\sup = \infty$ is a convention some books adopt, but $\infty$ is not a real number and nothing in this course may treat it as one.
Assuming the supremum belongs to the set. It sometimes does. Deciding which case you are in is a separate question from finding the value, and the answer is found by checking membership, not by looking at the number.
This is the one to fix now, because it survives into limits and causes trouble there. $\sup S$ is the least upper bound; it has nothing to do with where the elements cluster. The set $\{3 + (-1)^n/n\}$ clusters at $3$, and its supremum is $\tfrac72$. The set $\{1 - 1/n\}$ clusters at $1$, and its supremum happens to be $1$ as well — which is why the two ideas are easy to confuse, and why a set that clusters somewhere other than its top is worth carrying in your head.
$S = \{1 - 1/n : n \in \mathbb{N}\} = \{0, \tfrac12, \tfrac23, \tfrac34, \ldots\}$: every element is below $1$, so $1$ is an upper bound.
Check the bound.
For any $\varepsilon > 0$, choose $n > 1/\varepsilon$; then $1 - 1/n > 1 - \varepsilon$, so no smaller number is a bound.
Nothing smaller works — this is the half that proves it.
$\sup S = 1$, and $1 \notin S$, so $S$ has no maximum. It does have a minimum, $0$, at $n = 1$.
Supremum without maximum; the two ends behave differently.
$A = \{x \in \mathbb{Q} : x > 0,\ x^2 < 2\}$ is nonempty ($1 \in A$) and bounded above ($2$ is a bound, since $x \ge 2$ would give $x^2 \ge 4$).
Both hypotheses hold.
Suppose $r \in \mathbb{Q}$ were the least rational bound. If $r^2 < 2$ one can nudge $r$ up and stay below $2$, so $r$ was not a bound; if $r^2 > 2$ one can nudge it down and still bound, so $r$ was not least.
Both cases fail, so no rational is the least bound.
And $r^2 = 2$ is impossible for a rational. So $A$ has no supremum in $\mathbb{Q}$ — it has one in $\mathbb{R}$, namely $\sqrt{2}$, and that difference is the whole content of completeness.
The axiom is what fills the hole.
The odd terms are $3 - 1/n$ and the even terms $3 + 1/n$, so the set sits either side of $3$ and closes in on it.
Split by parity first.
The largest term is $n = 2$: $3 + \tfrac12 = \tfrac72$, and it is attained, so $\sup = \tfrac72$ is a maximum.
The smallest is $n = 1$: $3 - 1 = 2$, also attained, so $\inf = 2$ is a minimum. Neither end is at $3$ — $3$ is what the set approaches, and approaching is not what a supremum measures.
Match each set fact to the conclusion it proves.
| Has both a maximum and minimum | Has neither endpoint as an extremum | Contains rational and irrational numbers | Is unbounded above and below | |
|---|---|---|---|---|
| $[0,1]$ | ||||
| $(0,1)$ | ||||
| A nonempty open interval | ||||
| $\mathbb Z$ |
Match each statement to the real-number property it uses.
| It has a least upper bound in $\mathbb R$ | It has no real upper bound, so completeness does not apply | There is a rational and an irrational number between them | Some natural number is greater than it | |
|---|---|---|---|---|
| A nonempty set is bounded above | ||||
| The set of all natural numbers | ||||
| Two distinct real numbers $a<b$ | ||||
| An arbitrary real number $x$ |
Match each set description to the correct statement about its bounds and attainment.
| Supremum 1, but no maximum | Supremum 1 and maximum 1 | Supremum 1, approached but not attained | No real supremum because there is no upper bound | |
|---|---|---|---|---|
| $S=(0,1)$ | ||||
| $S=(0,1]$ | ||||
| $S=\{1-1/n:n\in\mathbb N\}$ | ||||
| $S=\mathbb N$ |
Find $\sup S$ for $S = \{\, 3 - 1/n : n \in \mathbb{N} \,\}$.
Answer:
$S$ is two, approached from both sides, written $\{2 + (-1)^n/n : n \in \mathbb{N}\}$. Does $S$ have a maximum?
$S$ is bounded above with $\sup S = 3$. What is $\sup\{\, 4x : x \in S \,\}$?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Find $\inf T$ for $T = \{\, 5 + 6/n : n \in \mathbb{N} \,\}$.
Answer:
You can find suprema and infima, prove both halves of a supremum claim, and decide whether the bound belongs to the set. Say in your own words what the completeness axiom guarantees and why the rationals lack it. Next: the absolute value and the triangle inequality, which are how distance gets written down.
10. Your turn: $\sup$ and $\inf$ of $\{\, 3 + (-1)^n/n \,\}$, step 3