3 The Least Upper Bound Axiom
Highlights
- The definition of the supremum of a set and the Least Upper Bound axiom of \(\R\)
- The Archimedean Properties of \(\R\)
3.1 Preparation
In this section, we will lay out an axiom for \(\R\) that distinguishes it from other ordered fields. Leading up to this axiom, we make a few definitions.
Definition 3.1 Let \(S\subseteq\R\) be a subset of the real numbers.
- We say \(b\in\R\) is an upper bound of \(S\) if for all \(x\in S,\) \(b\geq x.\) If \(S\) has an upper bound, we say \(S\) is bounded above.
- We say that \(u\in\R\) the least upper bound or supremum of \(S\) if \(u\) is an upper bound and for all upper bounds \(b\) of \(S,\) \(u\leq b.\) In this case we write \[u=\sup S.\]
- If \(u\) is the least upper bound \(S\) and \(u\in S,\) we say that \(u\) is the greatest element of \(S.\)
Exercise 3.2
- Write down three upper bounds for the interval \(S=[0,3).\) Does \(S\) have a least upper bound; if so what is it? Does \(S\) have a greatest element; if so, what is it?
- For each set below, say what it’s supremum and greatest element are or say why it doesn’t have one.
- \(\left\{\frac{n}{n+1}\st n\in\N\right\}\)
- \((2,\infty)\)
- \(\left\{\frac{1}{n}\st n\in\N\right\}\)
- \((-\infty,3]\)
- \(\{x\st x^2\leq 3\}\)
- Describe in your own words which subsets of \(\R\) have upper bounds, which have least upper bounds, and which have greatest elements. This doesn’t have to be precise; just give me the vibe.
- (Exploratory) How would you change your answers to the previous problems if our universe were \(\Q\) instead of \(\R?\)
The example of \(S=[0,3)\) is illustrated below. You can play with the slider to help you think about what value(s) of \(b\) might satisfy the definitions of upper bound, supremum, or greatest element.
Exercise 3.3 Write “\(S\) is bounded above” in logical symbolic notation; see if you can do so without using the term “upper bound”.
Exercise 3.4 Suppose \(b\) is an upper bound on a set \(S.\) What does it mean for \(b\) to not be the least upper bound of \(S\)?
3.2 Least Upper Bounds
Part 2e of Exercise 3.2 is meant to draw a contrast between how least upper bounds work in \(\R\) versus how they work in \(\Q.\) In particular, the problem asks, if \[S=\{x\st x^2\leq 3\}\] does \(S\) have a least upper bound?
Whether our universe is \(\Q\) or \(\R\) or some other ordered field, \(S\) certainly has an upper bound; \(b=2,\) for example, is greater than any element of \(S\) (that is, \(2\) is greater than any number whose square is less than 3). But as you may have guessed, an ordered field has a number whose square is \(3\) if and only if the set \(S\) defined above has a least upper bound (in which case \(\sup S\) is such a number).
We have a sense that the since \(3\) has no square root in \(\Q,\) there is something missing—a “hole” in the rationals that we plug by instead thinking about \(\R.\) And as this example shows, plugging such a hole can be related to the existence of a least upper bound on a set.
This connection goes in the opposite direction as well. If we assume the supremum of \(S\) exists, then we at least have a candidate for a number whose square is \(3\) (actually proving this is more complicated than you might expect—see Theorem 3.16 below).
This is all motivation for the final axiom we’ll assume about the real numbers: a guarantee of the existence of least upper bounds that will have the effect of ensuring that any \(\R\) will not have any “holes” in this sense, the way that \(\Q\) does.
Axiom 3.5 (The Least Upper Bound Axiom) Every non-empty subset of \(\R\) that is bounded above has a least upper bound in \(\R.\)
Note that using Axiom 3.5 to conclude that a subset \(S\subseteq\R\) has a supremum requires \(S\) to satisfy two hypotheses:
- \(S\) is nonempty, and
- \(S\) is bounded above.
From now on, we’ll make a point of emphasizing which theorems require the Least Upper Bound Axiom in their proofs and how. Theorems that rely on the Least Upper Bound Axiom will be highlighted with a maroon bar.
As a first step to understanding least upper bounds, we should justify the use of the word “the” in Definition 3.1.
Generally, when define a new quantity implicitly (i.e. without giving an explicit formula), we should ask ourselves whether a unique quantity satisfies the definition. For example, a subset of \(\R\) may have more than one upper bound. That a subset has only one least upper bound (if it has one at all) is something we need to prove.
Theorem 3.6 If a subset of \(\R\) has a least upper bound, then it is unique.
Proof. Suppose \(u,v\) are both least upper boundsI am using one of the proof skeleta for proving uniqueness. See Section A.2 for some standard proof structures. for a subset \(S\subseteq\R.\) Then \(u\) is an upper bound for \(S,\) so since \(v\) is a least upper bound, \(v\leq u.\) Also, \(v\) is an upper bound for \(S,\) so since \(u\) is the least upper bound, \(u\leq v.\) Therefore \(u=v.\)
To prove a quantity is the least upper bound of a set, generally, one needs to show two things: prove that it’s an upper bound, and then prove that any other upper bound is less than or equal to it.
Example 3.7 Here we’ll do a careful proof that \(3\) is the supremum of \(S=(-\infty,3).\)
First, note that by definition \(S\) is all numbers less than 3, so for any \(x\in S,\) \(3>x.\) Therefore \(3\) is an upper bound on \(S.\)
By way of contradiction, suppose \(b<3\) and \(b\) is also an upper bound on \(S\)In Exercise 3.4 you were to show that something like this is the negation of \(3\) being the supremum. In logical symbols, saying 3 is the least upper bound is \((\forall \text{ upper bounds } b)(3\leq b).\) And thus its negation is \((\exists\text{ an upper bound } b)(b<3).\). But then consider \(x=\frac{b+3}{2}\) so that \(b<x<3.\) Since \(x<3,\) we know \(x\in S.\) But \(b<x\) which contradicts \(b\) being an upper bound on \(S.\)
Note that doing a proof by contradiction above was not really necessary. Another option would be to say “suppose \(b<3\)” and then prove \(b\) is not an upper bound.
Next we have some basic consequences of Axiom 3.5. The next theorem and its corollary are both often referred to as the “Archimedean Property” of \(\R.\)
Theorem 3.8 (Archimedean Property - Version 1) For any \(x\in\R,\) there exists \(n\in\N\) such that \(n>x.\)
Proof. By way of contradiction, suppose ___, which is the same as saying that \(x\) is an upper bound on \(\N.\) Then since \(\N\) is ___ and bounded above, \(\N\) has a least upper bound \(u.\) But then \(u-\frac{1}{2}<u,\) so \(u-\frac{1}{2}\) is not an upper bound. Therefore there exists \(n\in\N\) such that ___. But then \[n+1>u-\frac{1}{2}+1=u+\frac{1}{2}>u,\] and since \(n+1\in\N,\) we have contradicted ___.
Exercise 3.9 Fill in the blanks for the proof above. For the first blank, you’ll need to negate the original statement.
Corollary 3.10 (Archimedean Property - Version 2) For any \(\epsilon>0,\) there exists \(n\in\N\) such that \(\frac{1}{n}<\epsilon.\)
Exercise 3.11 Prove Corollary 3.10.
Remember that when we call a theorem a “Corollary”, it means that you should use the preceding theorem to prove it, and the proof will generally end up being quite short.
Exercise 3.12 Give a careful proof that \(1\) is the supremum of \(S=\left\{\frac{n}{n+1}\st n\in\N\right\}.\)
The same trick of taking an average like in Example 3.7 won’t necessarily work since that average might not be in \(S.\) Corollary 3.10 might be helpful here instead.
Exercise 3.13 Suppose \(S\subseteq\R\) is a nonempty set that is bounded above. Let \(c>0.\) We define the set \[T=\{cx\st x\in S\}.\] Prove that \(c\cdot\sup S\) is a least upper bound for \(T.\)
Exercise 3.14 Suppose \(S\) and \(T\) are nonempty subsets of \(\R\) that are bounded above. What is \(\sup(S\cup T)?\) Prove your answer.
Exercise 3.15 Prove the Archimedean Property is true when the universe is \(\Q.\) This means you’ll need to refrain from making any references to \(\R,\) and you of course can’t use the Least Upper Bound Axiom.
In Part 2e of Exercise 3.2 and the follow-up discussion, we claimed that the existence of suprema would “plug holes” in the real numbers. In particular, the supremum of \(S=\{x\st x^2\leq 3\},\) which we are assuming exists because of the Least Upper Bound Axiom, seems like it ought to be a number whose square is 3. We’ll prove a more general version of this fact below, although the proof gets somewhat technical towards the end.
Theorem 3.16 For any \(a\geq 0,\) there exists a real number \(x\geq 0\) such that \(x^2=a.\)
Proof. Since the case where \(a=0\) is trivial (since \(0^2=0\)),suppose \(a>0.\) Define \[S=\{y\st y^2\leq a\}\]
We will show that \(S\) is nonempty and bounded above so that we can claim \(\sup S\) exists.
First, we note that \(0^2=0<a,\) so \(0\in S,\) and thus \(S\) is nonempty.
Next, I claim that \(a+1\) is an upper bound on \(S.\) This is true because \[(a+1)^2=a^2+2a+1>a\]
So if \(y\in S,\) \[y^2\leq a< (a+1)^2\]
so \(y<a+1.\) Thus \(a+1\) is an upper bound on \(S,\) so \(S\) is bounded above, and \(x=\sup S\) exists, and since \(0\in S,\) \(x\geq 0.\)
By way of contradiction, suppose \(x^2\neq a.\) There are two cases: \(x^2<a\) and \(x^2>a,\) and we will finish the proof by showing each leads to a contradiction.
(Case 1: \(x^2<a\)) Let \(\epsilon=\min\left\{1,\frac{a-x^2}{2x+1}\right\}\) so that \(\epsilon>0\) and \(\epsilon^2\leq\epsilon.\) Then \[ \begin{aligned} (x+\epsilon)^2 & = x^2+2x\epsilon+\epsilon^2 \\ & \leq x^2 + 2x\epsilon + \epsilon \\ & = x^2 + \epsilon(2x+1) \\ & \leq x^2 + a - x^2 \\ & = a. \end{aligned} \]
Then \(x+\epsilon>x\) but \((x+\epsilon)^2\leq a,\) so \(x+\epsilon\in S.\) Thus \(x\) is not an upper bound on \(S,\) which is a contradiction.
(Case 2: \(x^2>a\)) Here we know \(x>0\), so let \(\epsilon=\min\left\{x,\frac{x^2-a}{2x} \right\}\) so that \(\epsilon>0\) and \(x-\epsilon>0.\) Then
\[ \begin{aligned} (x-\epsilon)^2 & = x^2-2x\epsilon+\epsilon^2 \\ & > x^2 - 2x\epsilon \\ & \geq x^2 - 2x\frac{x^2-a}{2x} \\ & = x^2 - (x^2-a) \\ & = a. \end{aligned} \]
But since \((x-\epsilon)^2>a,\) this means that for any \(y\in S,\) \(y^2<a<(x-\epsilon)^2,\) so \[y\leq|y|<|x-\epsilon|=x-\epsilon,\]
so \(x-\epsilon\) is an upper bound on \(S\) less than \(x,\) contradiction \(x\) being the least upper bound.
Thus, since \(x^2\neq a\) led to a contradiction, we must have \(x^2=a\).
3.3 Review Questions
- Give definitions of upper bound, least upper bound, and greatest element.
- What does the Least Upper Bound Axiom say? What two things do we need to know about a set \(S\) in order to apply the Least Upper Bound Axiom?
- For examples like those in Part 2 of Exercise 3.2, can you quickly identify the least upper bound when there is one?
- What do the two versions of the Archimedean Property tell us?