3.4 The supremum and infimum of a set.

Since $\mathbb{R}$ is complete, i.e., every bounded set has a least upper bound and greatest lower bound, we introduce the notion of supremum and infimum of a set.

Definition 3.12:

Let $A\subseteq \mathbb{R}$. We define the supremum of $A$, denoted $\sup(A)$ as follows:

If $A=\emptyset$, then $\sup(A) = -\infty$.
If $A$ is non-empty and is bounded above [i.e., there exists $\alpha \in \mathbb{R}$ such that for all $x\in A$, $x\leq \alpha$] then $\sup(A)$ is the least upper bound of $A$ (which we know exists by the Completeness Axiom)
If $A$ is non-empty and is not bounded above [i.e., for all $\alpha \in \mathbb{R}$, there exists $x\in A$ such that $x>\alpha$] then $\sup(A)= +\infty$.

Definition 3.13:

Let $A\subseteq \mathbb{R}$. We define the infimum of $A$, denoted $\inf(A)$ as follows:

If $A=\emptyset$, then $\inf(A) = +\infty$.
If $A$ is non-empty and is bounded below [i.e., there exists $\alpha \in \mathbb{R}$ such that for all $x\in A$, $x\geq \alpha$] then $\inf(A)$ is the greatest lower bound of $A$ (which we know exists by the Completeness Axiom)
If $A$ is non-empty and is not bounded below [i.e., for all $\alpha \in \mathbb{R}$, there exists $x\in A$ such that $x<\alpha$] then $\sup(A) = -\infty$.

Etymology:

Supremum comes from the Latin super meaning “over, above” while infimum comes from the Latin inferus meaning “below, underneath, lower” (these words gave rise to words like superior and inferior).

Example:

Let $A = \left\{\frac{1}{n}:n \in \mathbb{Z}_+\right\}$. We have already seen that $\sup(A) = 1$ (in $A$) and $\inf(A)= 0$ (not in $A$).

Proposition 3.14:

Let $a,b \in \mathbb{R}$ with $a<b$. Then

$\sup((a,b)) = \sup((a,b]) = \sup([a,b))= \sup([a,b]) = b$;
$\inf((a,b)) = \inf((a,b]) = \inf([a,b)) = \inf([a,b]) = a$;
$\sup((a,\infty)) = \sup([a,\infty)) = +\infty$;
$\inf((-\infty,a)) = \inf((-\infty,a]) = -\infty$;
$\sup((-\infty,a)) = \sup((-\infty,a]) = \inf([a,\infty)) = \inf((a,\infty)) = a$.

Proof.

We will only prove $\sup((a,b)) = b$ and $\sup((a,\infty)) = +\infty$ and leave the rest as the arguments are very similar.

Let $a,b \in \mathbb{R}$ with $a<b$, we will show $\sup((a,b)) = b$. [Note that $(a,b)$ is non-empty (as $a\neq b$) and it is bounded above.]

First we show that $b$ is an upper bound. Indeed, let $x\in (a,b)$ then by definition $a<x<b$ so $x\leq b$ as required.

Next we show that $b$ is the least upper bound [by showing any real number less than $b$ is not an upper bound]. Let $y\in \mathbb{R}$ with $y<b$.Suppose $a<y$ (i.e. $y\in (a,b)$) and let $x=\frac{b+y}{2}=y+\frac{b-y}{2}=b-\frac{b-y}{2}$. Note that $x<b$ and $x>y>a$, so $x\in (a,b)$. Since $x>y$, we have $y$ is not an upper bound. Suppose $y\leq a$ (i.e. $y\notin(a,b)$) and let $x=\frac{a+b}{2}=a+\frac{b-a}{2}=b-\frac{b-a}{2}$. Then $x<b$ and $x>a\leq y$, so $x\in(a,b)$. Since $x>y$, we have $y$ is not an upper bound. In either cases, $y$ is not an upper bound, so $b$ is the least upper bound.

Let $a \in \mathbb{R}$, we show that $\sup((a,\infty))=+\infty$. [Note that $(a,\infty)$ is non-empty, so we want to show it is not bounded above].

Suppose for contradiction that $(a,\infty)$ is bounded above [i.e., $\sup((a,\infty))\neq +\infty$]. Let $u \in \mathbb{R}$ be an upper bound for $(a,\infty)$. Set $x=|a|+|u|+1 \in \mathbb{R}$. Note that $x>|a|\geq a$, so $x\in (a,\infty)$. Hence $u$ being an upper bound means $x\leq u$, however $x>|u|\geq u$. This is a contradiction.

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Example:

Problem: Let \[A=\left\{\frac{n^2+1}{|n+1/2|}:n\in\mathbb{Z}\right\}.\] Show that $\sup(B) = +\infty$ and $\inf(B)=4/3$.

Solution: Let us first look at the supremum. [The question is asking us to show $B$ is not bounded above.] Let $x\in \mathbb{R}$. By the Archimedean Principle, choose $n\in\mathbb{Z}_+ \subseteq \mathbb{Z}$ such that $n>2x$ [Scratch work missing to work out why we choose this particular $n$]. Define $a = \frac{n^2+1}{n+1/2} \in A$. Then \[\begin{align*} a &= n\frac{n+\frac{1}{n}}{n+\frac{1}{2}} &\\ &\geq n\frac{n}{n+\frac{1}{2}} &\text{as $x+\frac{1}{n}\geq n$}\\ &\geq n\frac{n}{2n} & \text{as $n+\frac{1}{2}\leq 2n$} \\ &= \frac{n}{2} &\\ &> x&. \end{align*}\] So we have found $a\in A$ such that $a>x$, so $A$ is not bounded above. Hence $\sup(A) = +\infty$.

We now look at the infimum. We first show that $4/3$ is a lower bound. Let $a=\frac{n^2+1}{|n+1/2|}$ for some $n \in \mathbb{Z}$, so $a\in A$. Consider \[\begin{align*} n^2+1-(4/3)|n+1/2| &\geq n^2+1-(4/3)(|n|+1/2) &\text{ by the triangle inequality}\\ &=n^2-(4/3)|n|+1/3& \\ &=(|n|-2/3)^2 - 1/9 & \text{ by completing the square } \\ &\geq 0& \text{ since for all $n\in \mathbb{Z}$, $(|n|-2/3)^2\geq 1/9$}. \end{align*}\] Therefore $n^2+1\geq (4/3)|n+1/2|$, i.e. $a =\frac{n^2+1}{|n+1/2|}\geq 4/3$ as required.

We next show that $4/3$ is the greatest lower bound for $A$ [by showing any number greater than it is not a lower bound]. First note that by setting $n=1$, we have \[\frac{n^2+1}{|n+\frac{1}{2}|} = \frac{2}{3/2} = \frac{4}{3}.\] So $4/3\in A$. Thus, no value $y>4/3$ can be a lower bound for $B$. This shows that $4/3$ is the greatest lower bound. Hence $\inf(B) = 4/3$.

The next example is more theoretical.

Example:

Problem: Let $A$ and $B$ be bounded non-empty subsets of $\mathbb{R}$. Define the sum set as \[A+B = \{a+b:a\in A, b\in B\}.\] Show that $\sup(A+B) = \sup(A)+\sup(B)$.

Solution: Let $\alpha =\sup(A)$ and $\beta \sup(B)$. Note that $\alpha, \beta \in \mathbb{R}$ as both $A$ and $B$ are bounded. We show $\alpha+\beta = \sup(A+B)$.

We first show $\alpha+\beta$ is an upper bound for $A+B$. Let $c\in A+B$, by definition, there exists $a\in A$ and $b\in B$ such that $c=a+b$. Now $a\leq \alpha$ and $b\leq \beta$, so $c=a+b\leq \alpha+\beta.$

We now show $\alpha+\beta$ is the least upper bound for $A+B$. Let $\epsilon>0$, we show that $\alpha+\beta-\epsilon$ is not an upper bound. Since $\alpha$ is the least upper bound of $A$, then $\alpha-\epsilon/2$ is not an upper bound, so there exists $a\in A$ such that $a>\alpha-\epsilon/2$. Similarly, there exists $b\in B$ such that $b>\beta-\epsilon/2$. Define $c=a+b \in A+B$. Then \[c=a+b>(\alpha-\epsilon/2)+(\beta-\epsilon/2)=\alpha+\beta-\epsilon.\] This shows for any $\epsilon>0$, $\alpha+\beta-\epsilon$ is not an upper bound for $A+B$. So $\alpha+\beta$ is the least upper bound for $A+B$, hence the supremum of $A+B$.

Proof techniques:

Sometimes, instead of showing something is true for all $y>x$, it is easier to show it is true for all $x+\epsilon$ where $\epsilon>0$. We can do this because if $y>x$, then setting $epsilon = y-x>0$, we see that $y=x+\epsilon$.

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Definition 3.15:

Let $A\subseteq \mathbb{R}$. We say that $A$ has a maximum if $\sup(A)\in A$. In this case we write $\max (A)$ to stand for the element $a\in A$ with $a = \sup(A)$.

Similarly we say that $A$ has a minimum if $\inf(A)\in A$. In this case we write $\min (A)$ to stand for the element $a\in A$ with $a = \inf(A)$.

Note that a set may not have a minimum or a maximum.

Example:

Let \[A=\left\{\frac{n^2+1}{|n+1/2|}:n\in\mathbb{Z}\right\}.\] We have seen $\inf(A)=4/3\in A$, so $\min(A)=4/3$. However $A$ does not have a maximum as $\sup(A)=+\infty \not\in A$ (as $\infty \notin \mathbb{R}$ and $A\subseteq \mathbb{R}$).

Example:

Let $A = \left\{\frac{1}{n}: n \in \mathbb{Z}_+\right\}$. Then we have seen that $\sup(A)=1$ and $1\in A$, so $\max(A)=1$. However it does not have a minimum as we have seen $\inf(A)=0$ and $0\notin A$.