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We collect here some theorems that show some of the consequences of continuity.Some of the theorems apply to functions either of a real variable or of a complex variable,while others apply only to functions of a real variable. We begin with what may be the most famous such result, and this one is about functions of a real variable.
If $f:[a,b]\to R$ is a real-valued function that is continuous at each point of the closed interval $[a,b],$ and if $v$ is a number (value) between the numbers $f\left(a\right)$ and $f\left(b\right),$ then there exists a point $c$ between $a$ and $b$ such that $f\left(c\right)=v.$
If $v=f\left(a\right)$ or $f\left(b\right),$ we are done. Suppose then, without loss of generality, that $f\left(a\right)<v<f\left(b\right).$ Let $S$ be the set of all $x\in [a,b]$ such that $f\left(x\right)\le v,$ and note that $S$ is nonempty and bounded above. ( $a\in S,$ and $b$ is an upper bound for $S.$ ) Let $c=supS.$ Then there exists a sequence $\left\{{x}_{n}\right\}$ of elements of $S$ that converges to $c.$ (See [link] .) So, $f\left(c\right)=limf\left({x}_{n}\right)$ by [link] . Hence, $f\left(c\right)\le v.$ (Why?)
Now, arguing by contradiction, if $f\left(c\right)<v,$ let $\u03f5$ be the positive number $v-f\left(c\right).$ Because $f$ is continuous at $c,$ there must exist a $\delta >0$ such that $\left|f\right(y)-f(c\left)\right|<\u03f5$ whenever $|y-c|<\delta $ and $y\in [a,b].$ Since any smaller $\delta $ satisfies the same condition, we may also assume that $\delta <b-c.$ Consider $y=c+\delta /2.$ Then $y\in [a,b],\phantom{\rule{3.33333pt}{0ex}}|y-c|<\delta ,$ and so $\left|f\right(y)-f(c\left)\right|<\u03f5.$ Hence $f\left(y\right)<f\left(c\right)+\u03f5=v,$ which implies that $y\in S.$ But, since $c=supS,$ $c$ must satisfy $c\ge y=c+\delta /2.$ This is a contradiction, so $f\left(c\right)=v,$ and the theorem is proved.
The Intermediate Value Theorem tells us something qualitative about the range of a continuous function on an interval $[a,b].$ It tells us that the range is “connected;” i.e., if the range contains two points $c$ and $d,$ then the range contains all the points between $c$ and $d.$ It is difficult to think what the analogous assertion would be for functions of a complex variable, since “between” doesn't mean anything for complex numbers.We will eventually prove something called the Open Mapping Theorem in [link] that could be regarded as the complex analog of the Intermediate Value Theorem.
The next theorem is about functions of either a real or a complex variable.
Let $f:S\to C$ be a continuous function, and let $C$ be a compact (closed and bounded) subset of $S.$ Then the image $f\left(C\right)$ of $C$ is also compact. That is, the continuous image of a compact set is compact.
First, suppose $f\left(C\right)$ is not bounded. Thus, let $\left\{{x}_{n}\right\}$ be a sequence of elements of $C$ such that, for each $n,$ $\left|f\right({x}_{n}\left)\right|>n.$ By the Bolzano-Weierstrass Theorem, the sequence $\left\{{x}_{n}\right\}$ has a convergent subsequence $\left\{{x}_{{n}_{k}}\right\}.$ Let $x=lim{x}_{{n}_{k}}.$ Then $x\in C$ because $C$ is a closed subset of $C.$ Co, $f\left(x\right)=limf\left({x}_{{n}_{k}}\right)$ by [link] . But since $|f\left({x}_{{n}_{k}}\right)|>{n}_{k},$ the sequence $\left\{f\left({x}_{{n}_{k}}\right)\right\}$ is not bounded, so cannot be convergent. Hence, we have arrived at a contradiction, and the set $f\left(C\right)$ must be bounded.
Now, we must show that the image $f\left(C\right)$ is closed. Thus, let $y$ be a limit point of the image $f\left(C\right)$ of $C,$ and let $y=lim{y}_{n}$ where each ${y}_{n}\in f\left(C\right).$ For each $n,$ let ${x}_{n}\in C$ satisfy $f\left({x}_{n}\right)={y}_{n}.$ Again, using the Bolzano-Weierstrass Theorem, let $\left\{{x}_{{n}_{k}}\right\}$ be a convergent subsequence of the bounded sequence $\left\{{x}_{n}\right\},$ and write $x=lim{x}_{{n}_{k}}.$ Then $x\in C,$ since $C$ is closed, and from [link]
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