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.

(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 $S$).

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).

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

The condition that $S$ is bounded above is also necessary: a set with no upper bound certainly has no supremum.

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.

- $s \geqslant \alpha$ 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 $S$).

(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.

(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 $S$).

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!)

(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.

\sup 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.

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.

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 $α^2 < 2$.

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 $α = β$.

12 More consequences of completeness

In this course, we write $\mathbb N$ for the set of positive integers, so $\mathbb N =\mathbb Z^{>0}$.

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.

(i) If $S$ is bounded below, then $S$ has a minimum.

(ii) If $S$ is bounded above, then $S$ has a maximum.

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.

(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).

Summary of our work so far

$\mathbb R$ is a complete ordered field.