Suprema, Infima, and the Completeness Property

Definition of Supremum and Infimum

Definition: Let $S$ be a set that is bounded above. We say that the supremum of $S$ denoted $\sup S=u$ is a number $u$ that satisfies the conditions that $u$ is an upper bound of $S$ and $u$ is the least upper bound of $S$, that is for any $v$ that is also an upper bound of $S$ then $u\leq v$.

Definition: Let $S$ be a set that is bounded below. We say that the infimum of $S$ denoted $\inf S=w$ is a number $w$ that satisfies the conditions that $w$ is a lower bound of $S$ and $w$ is the greatest lower bound of $S$, that is for any $t$ that is also a lower bound of $S$ then $t\leq w$.

We will now reformulate these definitions with an equivalent statement that may be useful to apply in certain situations in showing that an upper bound $u$ is the supremum of a set, or showing that a lower bound $w$ is the infimum of a set.

Theorem 1: Let $S$ be a nonempty subset of the real numbers that is bounded above. The upper bound $u$ is said to be the supremum of $S$ if and only if $\forall \epsilon >0$ there exists an element $x_{\epsilon }\in S$ such that $u-\epsilon <x_{\epsilon }$.

• Proof: $\Rightarrow$ Let $S$ be a nonempty subset of the real numbers that is bounded above. We first want to show that if $u$ is an upper bound such that $\forall \epsilon >0$ there exists an element $x_{\epsilon }\in S$ such that $u-\epsilon <x_{\epsilon }$, then $u=\sup S$. Let $u$ be an upper bound of $S$ that satisfies the condition stated above, and suppose that $v<u$. Then choose $\epsilon =u-v$, and so $u-v=\epsilon >0$ and so there exists an element $x_{\epsilon }\in S$ such that $u-\epsilon <x_{\epsilon }$, and thus, $u=\sup S$.

• $\Leftarrow$ We now want to show that if $u=\sup S$ then $\forall \epsilon >0$ there exists an element $x_{\epsilon }\in S$ such that $u-\epsilon <x_{\epsilon }$. Let $u=\sup S$ and let $\epsilon >0$. We note that $u-\epsilon <u$ and so $u-\epsilon$ is not an upper bound of the set $S$. Therefore by the definition that $u=\sup S$, there exists some element $x_{\epsilon }\in S$ such that $u-\epsilon <x_{\epsilon }$. $\blacksquare$

Theorem 2: Let $S$ be a nonempty subset of the real numbers that is bounded below. The lower bound $w$ is said to be the infimum of $S$ if and only if $\forall \epsilon >0$ there exists an element $x_{\epsilon }\in S$ such that $x_{\epsilon }<w+\epsilon$.

• Proof: $\Rightarrow$ Let $S$ be a nonempty subset of the real numbers that is bounded below. We first want to show that if $w$ is a lower bound such that $\forall \epsilon >0$ there exists an element $x_{\epsilon }\in S$ such that $x_{\epsilon }<w+\epsilon$, then $w=\inf S$. Let $w$ be a lower bound of $S$ that satisfies the condition stated above, and suppose that $w<t$. Then choose $\epsilon =t-w$, and so $t-w=\epsilon >0$ and so there exists an element $x_{\epsilon }\in S$ such that $x_{\epsilon }<w+\epsilon$, and thus, $w=\inf S$.

• $\Leftarrow$ We now want to show that if $w=\inf S$ then $\forall \epsilon >0$ there exists an element $x_{\epsilon }\in S$ such that $x_{\epsilon }<w+\epsilon$. Let $w=\inf S$ and let $\epsilon >0$. We note that $w<w+\epsilon$ and so $w+\epsilon$ is not a lower bound of the set $S$. Therefore by the definition that $w=\inf S$, there exists some element $x_{\epsilon }\in S$ such that $x_{\epsilon }<w+\epsilon$. $\blacksquare$

The Supremum and Infimum of a Function

We will now begin to look at some applications of the definition of a supremum and infimum with regards to functions.

Definition: Let $f:A\to \mathbb {R}$ be a function. Then define the supremum of $f$ to be ${\underset {A}{\sup }}f:=\sup\{f(x):x\in A\}=\sup R(f)$, and define the infimum of $f$ to be ${\underset {A}{\inf }}f:=\inf\{f(x):x\in A\}=\inf R(f)$ where $R(f)$ is the range of $f$.

From the definition above, we acknowledge that the supremum and infimum of a function $f$ pertain to the set $R(f)$ that is the range of $f$. The diagram below illustrates the supremum and infimum of a function:

We will now look at some important theorems.

Theorem 1: Let $f:A\to \mathbb {R}$ and $g:A\to \mathbb {R}$ be functions such that $g$ is bounded above. If $f(x)\leq g(x)$ for all $x\in A$, then ${\underset {A}{\sup }}f\leq {\underset {A}{\sup }}g$.

• Proof: Let $f:A\to \mathbb {R}$. Let $g:A\to \mathbb {R}$ be a function that is bounded above. Since the range of $g$ is nonempty we have that for all $x\in A$, $g(x)\leq {\underset {A}{\sup }}g$. Now since $f(x)\leq g(x)$ for all $x\in A$, we have that $f(x)\leq g(x)\leq {\underset {A}{\sup }}G$.

• Furthermore we note that since $f(x)\leq g(x)$ for all $x\in A$, then $g$ is bounded below by $f$, and so ${\underset {A}{\sup }}f\leq g(x)\leq {\underset {A}{\sup }}G$ and thus ${\underset {A}{\sup }}f\leq {\underset {A}{\sup }}g$. $\blacksquare$

Theorem 2: Let $f:A\to \mathbb {R}$ and $g:A\to \mathbb {R}$ be functions such that $f$ is bounded below. If $f(x)\leq g(x)$ for all $x\in A$, then ${\underset {A}{\inf }}f\leq {\underset {A}{\inf }}g$.

• Proof: Let $f:A\to \mathbb {R}$ be a function that is bounded below. Let $g:A\to \mathbb {R}$. Since the range of $f$ is nonempty we have that for all $x\in A$, ${\underset {A}{\inf }}f\leq f(x)$. Now since $f(x)\leq g(x)$ for all $x\in A$ we have that ${\underset {A}{\inf }}f\leq f(x)\leq g(x)$.

• Furthermore we note that since $f(x)\leq g(x)$ for all $x\in A$, then $f$ is bounded above by $g$, and so ${\underset {A}{\inf }}f\leq f(x)\leq {\underset {A}{\inf }}g$ and thus ${\underset {A}{\inf }}f\leq {\underset {A}{\inf }}g$. $\blacksquare$

Theorem 3: Let $f:A\to \mathbb {R}$ be a bounded function and let $a\in \mathbb {R}$. If $a>0$ then ${\underset {A}{\sup(}}af)=a\cdot {\underset {A}{\sup }}f$, and ${\underset {A}{\inf(}}af)=a\cdot {\underset {A}{\inf }}f$. If $a<0$ then ${\underset {A}{\sup(}}af)=a\cdot {\underset {A}{\inf }}f$, and ${\underset {A}{\inf(}}af)=a\cdot {\underset {A}{\sup }}f$.

Theorem 3 immediately follows from the theorems we've already proven on <a href="/the-supremum-and-infimum-of-the-bounded-set-as">The Supremum and Infimum of The Bounded Set (aS)</a> page where $S=R(f)$.

The Completeness Property of The Real Numbers

We will now look at yet again another crucially important property of the real numbers which will allow us to call the set of $\mathbb {R}$ numbers under the operations of addition and multiplication a complete ordered field. This property will ensure that there is no "gaps" in the real number line, that is the real number line is continuous. The property is as follows.

The Completeness Property of The Real Numbers: Every nonempty subset $S$ of the real numbers that is bounded above has a supremum in $\mathbb {R}$.

The Completeness Property is also often called the "Least Upper Bound Property".

The completeness property above is a crucial axiom. A similar theorem regarding nonempty subsets $S$ of the real numbers that are bounded below exists and is proven below.

Theorem 1: Every nonempty subset $S$ of the real numbers that is bounded below has an infimum in $\mathbb {R}$.

• Proof: Let $S$ be a nonempty subset of real numbers that is bounded below. Then there exists a lower bound $m\in \mathbb {R}$ such that $m\leq s$ for every $s\in S$.

• Define the set $T$ by:

${\begin{aligned}\quad T=\{-s:s\in S\}\end{aligned}}$

• Then observe that $-s\leq -m$ for every $-s\in T$. So $T$ is a nonempty subset of real numbers that is bounded above. By the completeness property of the real numbers we have that $T$ has a supremum in $\mathbb {R}$. Let $x=\sup T$. Then $-s\leq x$ for every $-s\in T$ and if $y$ is any other upper bound of $T$ then $x\leq y$.

• So observe that $-x\leq s$ for every $s\in S$ and if $y$ is any other lower bound of $S$ then $y\leq -x$. So $\inf S=-x$. In other words, $S$ has an infimum in $\mathbb {R}$. $\blacksquare$

Licensing

Content obtained and/or adapted from:

• Epsilon Definition of the Supremum and Infimum under a CC BY-SA license
• The Supremum and Infimum of a Function under a CC BY-SA license
• The Completeness Property of the Real Numbers under a CC BY-SA license