Suprema, Infima, and the Completeness Property

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 $u-\epsilon <x_{\epsilon }$.

<img src="http://mathonline.wdfiles.com/local--files/epsilon-definition-of-the-supremum-and-infimum-of-a-bounded/Screen%20Shot%202014-10-12%20at%2010.07.13%20PM.png" alt="Screen%20Shot%202014-10-12%20at%2010.07.13%20PM.png" class="image" />

• 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 $x_{\epsilon }<w+\epsilon$.

<img src="http://mathonline.wdfiles.com/local--files/epsilon-definition-of-the-supremum-and-infimum-of-a-bounded/Screen%20Shot%202014-10-12%20at%2010.09.50%20PM.png" alt="Screen%20Shot%202014-10-12%20at%2010.09.50%20PM.png" class="image" />

• 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:

<img src="http://mathonline.wdfiles.com/local--files/the-supremum-and-infimum-of-a-function/Screen%20Shot%202014-10-12%20at%205.43.16%20PM.png" alt="Screen%20Shot%202014-10-12%20at%205.43.16%20PM.png" class="image" />

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 Failed to parse (syntax error): {\displaystyle f(x) ≤ g(x)} for all $x\in A$, then Failed to parse (syntax error): {\displaystyle \underset{A} \sup f ≤ \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 $x\in A$, Failed to parse (syntax error): {\displaystyle g(x) ≤ \underset{A} \sup g} . Now since Failed to parse (syntax error): {\displaystyle f(x) ≤ g(x)} for all $x\in A$, we have that Failed to parse (syntax error): {\displaystyle f(x) ≤ g(x) ≤ \underset{A} \sup G} .

• Furthermore we note that since Failed to parse (syntax error): {\displaystyle f(x) ≤ g(x)} for all $x\in A$, then $g$ is bounded below by $f$, and so Failed to parse (syntax error): {\displaystyle \underset{A} \sup f ≤ g(x) ≤ \underset{A} \sup G} and thus Failed to parse (syntax error): {\displaystyle \underset{A} \sup f ≤ \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 Failed to parse (syntax error): {\displaystyle f(x) ≤ g(x)} for all $x\in A$, then Failed to parse (syntax error): {\displaystyle \underset{A} \inf f ≤ \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 $x\in A$, Failed to parse (syntax error): {\displaystyle \underset{A} \inf f ≤ f(x)} . Now since Failed to parse (syntax error): {\displaystyle f(x) ≤ g(x)} for all $x\in A$ we have that Failed to parse (syntax error): {\displaystyle \underset{A} \inf f ≤ f(x) ≤ g(x)} .

• Furthermore we note that since Failed to parse (syntax error): {\displaystyle f(x) ≤ g(x)} for all $x\in A$, then $f$ is bounded above by $g$, and so Failed to parse (syntax error): {\displaystyle \underset{A} \inf f ≤ f(x) ≤ \underset{A} \inf g} and thus Failed to parse (syntax error): {\displaystyle \underset{A} \inf f ≤ \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 $S=R(f)$.