Least Upper Bound

Definition. A set $A$ of real numbers is bounded above if there is a number $x$ such that
MATH
Such a number $x$ is called an upper bound for $A$.

Definition. A number $x$ is a least upper bound of $A$ if

(1) $x$ is an upper bound of $A$.

and

(2) if $y$ is an upper bound of $A$, then $x\le y.$

Least Upper Bound Axiom

If $A$ is a set of real numbers, $A\ne \phi ,$ and if $A$ is bounded above, then $A$ has a least upper bound.

Intermediate Value Theorem

Theorem If $f$ is continuous on $[a,b]$ and $f(a)<0<f(b),$ then there is some number $x$ in $[a,b]$ such that $f(x)=0$.

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Proof. Define


MATH

Clearly $A\ne \phi $, since $a$ is in $A$; in fact,.there is some $\delta >0$ such that $A$ contains all points $x$ satisfying $a\le x<a+\delta $; this is so since $f$ is continuous on $[a,b]$ and $f(a)<0.$ Similarly, $b$ is an upper bound for $A$ and, in fact, there is a $\delta >0$ such that all points $x$ satisfying $b-\delta <x\le b$ are upper bounds for $A$ since $f(b)>0.$ From the least upper bound axiom it follows that $A$ has a least upper bound $\alpha $ and that $a<\alpha <b.$ We now wish to show that $f(\alpha )=0$, by eliminating the possibilities $f(\alpha )>0$ and $f(\alpha )<0$.

Suppose first that $f(\alpha )<$$0$. Then there is a $\delta >0$ such that $f(x)<0$ for MATH. Now there is some number $x_0$ in $A$ which satisfies MATH (because otherwise $\alpha $ would not be the least upper bound of $A).$ This means that $f$ is negative on the whole interval $[a,x_0]$. But if $x_1$ is a number between $a $ and $a+\delta $ , then $f$ is also negative on the whole interval $[x_0,x_1]$. Therefore $f$ is negative on the interval $[a,x_1]$, so $x_1\;$is in $A$. But this contradicts the fact that $\alpha $ is an upper bound for $A$; our original assumption that $f(\alpha )<0$ must be false.

Suppose, on the other hand, that $f(\alpha )>0$. Then there is a number $\delta >0$ such that $f(x)>0$ for MATH. Once again we know that there is an $x_0$ in $A$ satisfying MATH; but this means that $f$ is negative on $[a,x_0]$, which is impossible, since $f(x_0)>0$. Thus the assumption $f(a)>0$ also leads to a contradiction, leaving $f(\alpha )=0$ as the only alternative.

Theorem. If $f$ is continuous on $[a,b]$ and $f(a)f(b)<0,$ then there is some number $x$ in $[a,b]$ such that $f(x)=0.$

Theorem. If $f$ is continuous on $[a,b]$ and $f(a)<c<f(b),$ then there is some number $x$ in $[a,b]$ such that $f(x)=c.$

Proof. Consider the function $g=f-c.$

Theorem. If $f$ is continuous on $[a,b]$ and $f(a)>c>f(b),$ then there is some number $x$ in $[a,b]$ such that $f(x)=c.$

Proof. Consider the function $g=-f.$

Boundedness of Continuous Function

Theorem If $f$ is continuous at $a$, then there is a number $\delta >0$ such that $f$ is bounded above on the interval MATH

Proof. Since MATH, there is, for every $\epsilon >0$ a $\delta >0$ such that, for all $x$,


MATH

It is only necessary to apply this statement to some particular $\epsilon $, for example, $\epsilon =1$. We conclude that there is a $\delta >0$ such that, for all x,
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It follows, in particular, that if MATH then $f(x)-f(a)<1.$ This completes the proof: on the interval MATH the function $f$ is bounded above by $f(a)+1.$

Theorem. If $f$ is continuous on $[a,b]$, then $f$ is bounded above on $[a,b] $.

Proof. Let


MATH

Clearly $A\ne \phi $ (since $a$ is in $A$),and $A$ is bounded above (by $b$), so $A$ has a least upper bound $\alpha $. Notice that we are here applying the term ''bounded above'' both to the set $A$, which can be visualized as lying on the horizontal axis, and to $f$, i.e., to the sets MATH which can be visualized as lying on the vertical axis. Our first step is to prove that we actually have $\alpha =b.$ Suppose, instead, that $\alpha $$<$ $b.$ By the previous theorem there is $\delta >0\; $such that $f$ is bounded on MATH Since $\alpha $ is the least upper bound of $A$ there is some $x_0$ in $A$ satisfying MATHThis means that $f$ is bounded on $[a,x_0]$. But if $x_1$ is any number with MATHthen $f$ is also bounded on $[x_0,x_1]$. Therefore $f$ is bounded on $[a,x_1],$ so is in $A\;$contradicting the fact that $\alpha $ is an upper bound for $A$. This contradiction shows that $\alpha =b$. One detail should be mentioned: this demonstration implicitly assumed that $a<\alpha $ (so that $f$ would be defined on some interval MATH; the possibility $\alpha =a$ can be ruled out similarly: using the existence of a $\delta >0$ such that $f\;$is bounded on MATH

The proof is not quite complete--we only know that $f$ is bounded on $[a,x]$ for every $x<b,$ not necessarily that $f$ is bounded on $[a,b].$ However, only one small argument needs to be added.

There is a $\delta $$>$ 0 such that $f$ is bounded on MATH There is $x_0$ in $A$ such that $b-\delta <x_0<b.$ Thus $f$ is bounded on $[a,x_0]$ and also on $[x_0,b]$, so $f$ is bounded on $[a,b].$

Attaining Extremum

Theorem If $f$ is continuous on $[a,b],$ then there is a number $y$ in $[a,b] $ such that $f(y)\ge f(x)$ for all $x$ in $[a,b].$

Proof. We already know that $f$ is bounded on $[a,b],$which means that the set
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is bounded. This set is obviously not $\phi $, so it has a least upper bound $\alpha $. Since $\alpha \ge f(x)$ for $x$ in $[a,b]$ it suffices to show that $\alpha =f(y)$ for some $y$ in $[a,b]$.

Suppose instead that $\alpha \ne f(y)$ for all $y$ in $[a,b].$ Then the function $g$ defined by
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is continuous on $[a,b]$, since the denominator of the right side is never $0 $. On the other hand, $\alpha $ is the least upper bound of MATH; this means that


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This, in turn, means that
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But this means that $g$ is not bounded on $[a,b]$, contradicting the previous theorem.