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The least upper bound of a set, its $\varepsilon$ characterisation, and why a supremum need not be an element.
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 find the set of upper bounds of a set, read off its supremum and infimum, and say whether either belongs to the set. You will also be able to state the two-part characterisation of a least upper bound and use the second part — that some element exceeds the supremum minus any positive amount — which is the form every later proof will need. Every hypothesis in this subject is doing work somewhere in the proof. The way to know a theorem is to know which example breaks when a hypothesis is dropped.
From an introduction to proof: what an ordered field is, how a quantifier is negated, and how to argue by contradiction. The real numbers satisfy every field and order axiom the rationals do. This course is about the one axiom that separates them, and this lesson is about the vocabulary that axiom is written in.
An upper bound of a set $S$ is a number $c$ with $s \le c$ for every $s \in S$; a lower bound is the mirror. The supremum $\sup S$ is the least upper bound, and the infimum $\inf S$ the greatest lower bound. A maximum is an upper bound that is also an element. Every maximum is a supremum; most suprema are not maxima, and that gap is the reason the word exists.
Fix a set $S \subseteq \mathbb{R}$. The upper bounds of $S$ are the numbers no element exceeds. If there is one, there are infinitely many: anything above an upper bound is another. The question analysis cares about is whether that collection has a smallest member.
When it does, that member is $\sup S$, and it is pinned down by two statements, both needed:
The second is the one that does work. It says the set comes arbitrarily close to its supremum, and it is how a supremum is used in a proof rather than merely named.
$\inf S$ mirrors all of this: greatest lower bound, with some element below $\inf S + \varepsilon$ for every $\varepsilon$.
Maximum is a stronger word. $\max S$ is an element of $S$ that bounds it. The open interval from $0$ to $1$ has supremum $1$ and no maximum at all, and a subject that could only speak of maxima would have nothing to say about it. Every bounded non-empty set has a supremum; only some have a maximum.
Another way: picture
Picture the upper bounds as a ray shaded off to the right. As the shading is pushed left it eventually jams against the set. The point where it jams is the supremum — and whether that point is coloured in, belonging to the set, or left hollow is a separate fact about the set that the shading cannot see.
Another way: steps
To show that $M = \sup S$:
Step 3 is where the set's own description is used, and it is the step that fails if $M$ was too big.
Three shapes cover most cases.
A closed interval. For $[a, b]$, $\sup = b$ and $\inf = a$, and both are attained: they are elements.
An open interval. For $(a, b)$, $\sup = b$ and $\inf = a$ still, and neither is attained. The bounds did not move when the ends were removed; only their membership did.
A sequence's range. For $\{1 - 1/n : n \in \mathbb{N}\}$, the elements increase and stay below $1$, so $1$ bounds them. For leastness: given $\varepsilon > 0$, the element with $n$ large enough exceeds $1 - \varepsilon$. So $\sup = 1$, unattained, while $\inf = 0$ is attained at $n = 1$. Sets like this one are where the second half of the characterisation is unavoidable, because there is no largest element to point at.
| Set | Supremum | Attained | Infimum | Attained |
|---|---|---|---|---|
| $[0, 3]$ | $3$ | yes | $0$ | yes |
| $(0, 5)$ | $5$ | no | $0$ | no |
| $\{1 - 1/n\}$ | $1$ | no | $0$ | yes |
| $\{-n : n \in \mathbb{N}\}$ | $-1$ | yes | none | — |
An upper bound must be in the set. It must not. $7$ is an upper bound of $(0, 5)$ and so is $5$, and neither is an element. An argument that rejects a candidate bound because it lies outside the set has proved nothing.
The supremum is the largest element. Only when there is one. Asking for the largest element of $(0, 5)$ is asking a question with no answer; asking for the supremum is asking one with the answer $5$.
A set bounded below has a supremum. Bounded above, bounded below and bounded are three different claims. The completeness axiom is about the first, and a set bounded below may run off to infinity upwards.
Let $S = \{1 - 1/n : n \in \mathbb{N}\}$. Every element is below $1$, so $1$ is an upper bound.
The first half: it bounds.
Given $\varepsilon > 0$, choose $n$ with $1/n < \varepsilon$; the element $1 - 1/n$ then exceeds $1 - \varepsilon$.
The second half: nothing smaller bounds.
So $\sup S = 1$. No element equals $1$, so $S$ has no maximum, and the two facts sit together without tension.
Least upper bound, not largest element.
Let $T = \{x \in \mathbb{R} : x^2 > 4, \ x > 0\}$, which is the open ray above $2$.
Rewrite the description as a shape you recognise.
Every element exceeds $2$, so $2$ is a lower bound; and for $\varepsilon > 0$ the number $2 + \varepsilon/2$ is in $T$ and below $2 + \varepsilon$.
Both halves, in the mirrored form.
So $\inf T = 2$, unattained, and $T$ has no supremum because it is unbounded above.
The two ends answer to different questions.
The even terms are $1 + 1/n$, which decrease towards $1$ from above; the largest is the one with $n = 2$, namely $3/2$.
Split by parity before anything else.
The odd terms are $-(1 + 1/n)$, which increase towards $-1$ from below; the smallest is the one with $n = 1$, namely $-2$.
The same reading, mirrored.
So the supremum is $3/2$ and the infimum is $-2$, and both are attained — this set does have a maximum and a minimum, which is exactly why it is the easy case.
Let $S$ be the half-open interval from $3$ up to but not including $6$. Give the set of all upper bounds of $S$.
This task has no paper form; do it on a device.
What is the supremum of $\{x \in \mathbb{R} : x^2 \le 4\}$?
Answer:
For each set, say whether its supremum belongs to the set.
| its supremum belongs to the set | its supremum does not belong to the set | |
|---|---|---|
| the closed interval from $0$ to $4$ | ||
| the open interval from $0$ to $4$ | ||
| $\{4 - 1/n : n \in \mathbb{N}\}$ | ||
| $\{0, 4/2, 4\}$ |
Does the set of all natural numbers have a supremum?
Let $S$ be the open interval from $0$ to $6$. Here is an argument that its supremum is $6$. Mark the one step that is not sound.
This task has no paper form; do it on a device.
$A$ has supremum $9$ and $B$ has supremum $8$. Let $C$ be the set of all sums $x + y$ with $x \in A$ and $y \in B$. What is the supremum of $C$?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Let $S$ be the half-open interval from $1$ up to but not including $6$. Give the set of all upper bounds of $S$.
This task has no paper form; do it on a device.
You can name the upper bounds of a set, find its supremum and infimum, and say whether each is attained. Say in your own words why the open interval from zero to one has a supremum but no largest element. Next: the axiom that guarantees the supremum is there at all.
9. Your turn: the supremum and infimum of $\{(-1)^n(1 + 1/n) : n \in \mathbb{N}\}$, step 3