Analysis I Lecture Notes
10 Supremum, infimum and completeness
Some upper bounds are more interesting than others. The set $[0, 1]$ has upper bounds including 15, 1, 1.7 and infinitely many more. Of these, 1 feels special. This is the focus of our next definition.
Definition. Let $S ⊆\mathbb R$. We say that $α ∈\mathbb R$ is the supremum of $S$, written $\operatorname{sup}S$, if
(i) $s ⩽ α$ for all $s ∈ S$; ($α$ is an upper bound of $S$)
(ii) if $s ⩽ b$ for all $s ∈ S$ then $α ⩽ b$ ($α$ is the least upper bound of $S$).
Remark. If $S$ has a supremum, then $\operatorname{sup}S$ is unique. (Check you can show this!)
Now that we have defined the supremum, we can state our final key property of $\mathbb R$ (in addition to the properties that make it an ordered field).
Completeness axiom for the real numbers Let $S$ be a non-empty subset of $\mathbb R$ that is bounded above. Then $S$ has a supremum.
Remark. There are two conditions on $S$ here: non-empty, and bounded above. They are both crucial!
It is easy to forget the non-empty condition, but it has to be there: the empty set does not have a supremum, because every real number is an upper bound for the empty set — there is no least upper bound.
The condition that $S$ is bounded above is also necessary: a set with no upper bound certainly has no supremum.
Example. Let $S = [1, 2)$. Then 2 is an upper bound, and is the least upper bound: if $b < 2$ then $b$ is not an upper bound because $\max \left(1,1+\frac{b}{2}\right) \in S$ and $\max \left(1,1+\frac{b}{2}\right)>b$. Note that in this case $\sup S \notin S$.
Let $S = (1, 2]$. Then we again have $\sup S = 2$, and this time $\sup S ∈ S$.
The supremum is the least upper bound of a set. There’s an analogous definition for lower bounds.
Definition. Let $S ⊆\mathbb R$. We say that $α ∈\mathbb R$ is the infimum of $S$, written $\inf S$, if- $s \geqslant \alpha$ for all $s ∈ S$; ($α$ is a lower bound of $S$)
- if $s ⩾ b$ for all $s ∈ S$ then $α ⩾ b$ ($α$ is the greatest lower bound of $S$).
Let’s explore some useful properties of sup and inf.
Proposition 8. (i) Let $S, T$ be non-empty subsets of $\mathbb R$, with $S ⊆ T$ and with $T$ bounded above. Then $S$ is bounded above, and $\sup S ⩽ \sup T$.
(ii) Let $T ⊆\mathbb R$ be non-empty and bounded below. Let $S =\{−t : t ∈ T\}$. Then $S$ is non-empty and bounded above. Furthermore, $\inf T$ exists, and $\inf T = − \sup S$.
Remark. (ii) and a similar result with sup and inf swapped essentially tell us that we can pass between sups and infs. Any result we prove about sup will have an analogue for inf. Also, we could have phrased the Completeness Axiom in terms of inf instead of sup. Proposition 8(ii) tells us that we don’t need separate axioms for sup and inf.
Proof. (i) Since $T$ is bounded above, it has an upper bound, say b.
Then $t ⩽ b$ for all $t ∈ T$, so certainly $t ⩽ b$ for all $t ∈ S$, so $b$ is an upper bound for $S$.
Now $S, T$ are non-empty and bounded above, so by completeness each has a supremum.
Note that $\sup T$ is an upper bound for $T$ and hence also for S, so $\sup T ⩾ \sup S$ (since $\sup S$ is the least upper bound for $S$).
(ii) Since $T$ is non-empty, so is $S$. Let $b$ be a lower bound for $T$, so $t ⩾ b$ for all $t ∈ T$.
Then $−t ⩽ −b$ for all $t ∈ T$, so $s ⩽ −b$ for all $s ∈ S$, so $−b$ is an upper bound for $S$. Now $S$ is non-empty and bounded above, so by completeness it has a supremum.
Then $s⩽\sup S$ for all $s∈S$, so $t⩾−\sup S$ for all $t ∈ T$, so $−\sup S$ is a lower bound for $T$.
Also, we saw before that if $b$ is a lower bound for $T$ then $−b$ is an upper bound for $S$.
Then $−b ⩾\sup S$ (since sup S is the least upper bound), so $b⩽-\sup S$. So $−\sup S$ is the greatest lower bound. So $\inf T$ exists and $\inf T = − \sup S$.
You might be wondering how all this relates to familiar notions of maximum and minimum so let’s explore that.
Definition. Let $S ⊆\mathbb R$ be non-empty. Take $M ∈\mathbb R$. We say that $M$ is the maximum of $S$ if
(i)$M \in S$; ($M$ is an element of $S$)
(ii)$s\le M$ for all $s\in S$ ($M$ is an upper bound for $S$).
Remark. If $S$ is empty or $S$ is not bounded above then $S$ does not have a maximum. (Check this!)
Let $S\subseteq \mathbb R$ be non-empty and bounded above, so (by completeness) $\sup S$ exists.
Then $S$ has a maximum if and only if $\sup S ∈ S$. Also, if $S$ has a maximum then $\max S = \sup S$. (Check this!)
Definition. Let $S ⊆\mathbb R$ be non-empty. Take $m∈\mathbb R$. We say that $m$ is the minimum of $S$ if
(i)$m ∈ S$; ($m$ is an element of $S$)
(ii)$s ⩾ m$ for all $s ∈ S$ ($m$ is a lower bound for $S$).
Here is a key result about the supremum, which we’ll use a lot. It is a quick consequence of the definition, but it will be useful to have formulated it in this way.
Proposition 9 (Approximation Property). Let $S ⊆\mathbb R$ be non-empty and bounded above. For any $ε>0$, there is $s_ε ∈ S$ such that $\sup S − ε < s_ε ⩽
\sup S$.
Proof. Take $ε>0$. Note that by definition of the supremum we have $s⩽\sup S$ for all $s∈S$.
Suppose, for a contradiction, that $\sup S − ε ⩾ s$ for all $s ∈ S$. Then $\sup S − ε$ is an upper bound for $S$, but $\sup S − ε <\sup S$. Contradiction. So there is $s_ε ∈ S$ with $\sup S − ε < s_ε$.
11 Existence of roots
Now that we have identified the completeness property of $\mathbb R$, we are ready to prove that $\mathbb R$ contains a square root of 2.
Theorem 10. There exists a unique positive real number $α$ such that $α^2 = 2$.
Proof. Existence Let $S =\{s ∈\mathbb R : s > 0, s^2 < 2\}$.
Idea: argue that $S$ has a supremum, and show that $\sup S$ has the required properties.
Note that $S$ is non-empty (eg $1 ∈ S$) and $S$ is bounded above, because if $x > 2$ then $x^2 > 4$ (properties of ordering) so $x\notin S$, so 2 is an upper bound for $S$.
So, by completeness, $S$ has a supremum. Let $α =\sup S$.
Note that certainly $α > 0$ (since $1∈S$ so $α⩾1$).
By trichotomy, we have $α^2 < 2$ or $α^2 = 2$ or $α^2 > 2$.
Idea: show that if $α^2 < 2$ or $α^2 > 2$ then we get a contradiction.
Case 1 Suppose, for a contradiction, that $α^2 < 2$. Then $α^2 = 2 − ε$ for some $ε > 0$.
Idea: consider $α + h$ for a small $h > 0$. Later on, we’ll choose $h$ small enough that $(α+h)^2 < 2$, and that will be a contradiction because $α+h ∈ S$.
and $α + h >\sup S$. Note that $α ⩽ 2$ (we said earlier that 2 is an upper bound for $S$). For $h ∈ (0, 1)$ we have
$(\alpha+h)^{2}=\alpha^{2}+2 \alpha h+h^{2}=2-\varepsilon+2 \alpha h+h^{2}\leqslant 2-\varepsilon+4 h+h⩽ 2 − ε + 5h$
so let $h=\min \left(\frac{\varepsilon}{10}, \frac{1}{2}\right)$ and then $(\alpha+h)^{2}<2$.
Now $α + h ∈ S$ and $α + h >\sup S$. This is a contradiction. So it is not the case that $α^2 < 2$.
Case 2 Suppose, for a contradiction, that $α^2 > 2$. Then $α^2 = 2 + ε$ for some $ε > 0$.
Idea: consider $α − h$ for a small $h > 0$. Later on, we’ll choose $h$ small enough that $(α − h)^2 > 2$, and that will lead to a contradiction because $α − h <\sup S$.
For $h ∈ (0, 1)$ we have $(\alpha-h)^{2}=\alpha^{2}-2 \alpha h+h^{2}=2+\varepsilon-2 \alpha h+h^{2}⩾ 2 + ε − 4h$
so choose $h=\min \left(\frac{\varepsilon}{8}, \frac{1}{2}, \frac{\alpha}{2}\right)$ and then $(\alpha-h)^{2}>2$ (and also $α − h > 0$)
Now $α − h <\sup S$, so by the Approximation property there is $s∈S$ with $α−h<s$.
But then $2 < (α − h)^2 < s^2 < 2$, which is a contradiction.
So it is not the case that $α^2 > 2$.
Hence, by trichotomy, $α^2 = 2$.
Uniqueness Suppose that $β$ is also a positive real number such that $β^2 = 2$.
Aim: $α = β$.
Then $0 = α^2 − β^2 = (α − β)(α + β)$ and $α + β > 0$, so $α = β$.
Proposition 11. $\mathbb Q$ is not complete (with the ordering inherited from $\mathbb R$).
Proof. If $\mathbb Q$ were complete, then the proof of Theorem 10 would work just as well in $\mathbb Q$. But we know that there is not an element of $\mathbb Q$ that squares to 2. So $\mathbb Q$ is not complete.
Theorem 12. Let $n$ be an integer with $n ⩾ 2$, and take a positive real number $r$. Then $r$ has a real $n$th root.
Proof. Exercise. (See Sheet 2 for the case of the cube root of 2.)
12 More consequences of completeness
In this course, we write $\mathbb N$ for the set of positive integers, so $\mathbb N =\mathbb Z^{>0}$.
Theorem 13 (Archimedean property of $\mathbb N$). $\mathbb N$ is not bounded above.
Proof. Idea: if there’s an upper bound then we can find a natural number just less than it, and add 1.
Suppose, for a contradiction, that $\mathbb N$ is bounded above. Then $\mathbb N$ is non-empty and bounded above, so by completeness (of $\mathbb R$) $\mathbb N$ has a supremum.
By the Approximation property with $ε=\frac12$, there is a natural number $n ∈\mathbb N$ such that $\sup\mathbb N −\frac12 < n ⩽\sup\mathbb N$. Now $n + 1 ∈\mathbb N$ and $n + 1 >\sup\mathbb N$. This is a contradiction.
Corollary 14. Let $ε > 0$. Then there is $n∈\mathbb N$ such that $0 <\frac1n < ε$.
Proof. If not, then $\frac1ε$ would be an upper bound for $\mathbb N$. This would contradict Theorem 13.
Theorem 15. Let $S$ be a non-empty subset of $\mathbb Z$.
(i) If $S$ is bounded below, then $S$ has a minimum.
(ii) If $S$ is bounded above, then $S$ has a maximum.
Proof. (i) Assume that $S$ is bounded below. Then, by completeness (applied to $\{−s : s ∈ S\}$), $S$ has an infimum.
Secret aim: $\inf S ∈ S$.
By the Approximation property (with $ε = 1$), there is $n ∈ S$ such that $\inf S ⩽ n <\inf S + 1$. Aim: $\inf S = n$.
Suppose, for a contradiction, that $\inf S < n$. Write $n =\inf S + δ$, where $0 < δ < 1$.
By the Approximation property (with $ε = δ$), there is $m ∈ S$ such that $\inf S ⩽ m <\inf S + ε = n$.
Now $m < n$ so $n − m > 0$ but $n − m$ is an integer, so $n − m ⩾ 1$. Now $n ⩾ m + 1 ⩾\inf S + 1$. This is a contradiction. So $n =\inf S ∈ S$ so $\inf S =\min S$.
(ii) Similar.
Proposition 16. Take $a, b ∈\mathbb R$ with $a < b$. Then
(i) there is $x ∈\mathbb Q$ such that $a < x < b$ (the rationals are dense in the reals); and
(ii) there is $y ∈\mathbb R \setminus \mathbb Q$ such that $a < y < b$ (the irrationals are dense in the reals).
Proof. Exercise (see Sheet 2)
Summary of our work so far$\mathbb R$ is a complete ordered field.