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Finally, using the continuity of both $f$ and $g$ applied to the positive numbers ${\u03f5}_{1}=\u03f5/(4{M}_{2}{\left|g\left(c\right)\right|}^{2})$ and ${\u03f5}_{2}=\u03f5/(4{M}_{1}{\left|g\left(c\right)\right|}^{2}),$ choose $\delta >0,$ with $\delta <min({\delta}_{1},{\delta}_{2},{\delta}^{\text{'}}),$ and such that if $|y-c|<\delta $ and $y\in S$ then $|f\left(y\right)-f\left(c\right)|<\frac{\u03f5}{4{M}_{2}/{\left|g\left(c\right)\right|}^{2}}$ and $|g\left(c\right)-g\left(y\right)|<\frac{\u03f5}{4{M}_{1}/{\left|g\left(c\right)\right|}^{2}}.$ Then, if $|y-c|<\delta $ and $y\in S$ we have that
as desired.
Let $S,T,$ and $U$ be subsets of $C,$ and let $f:S\to T$ and $g:T\to U$ be functions. Suppose $f$ is continuous at a point $c\in S$ and that $g$ is continuous at the point $f\left(c\right)\in T.$ Then the composition $g\circ f$ is continuous at $c.$
Let $\u03f5>0$ be given. Because $g$ is continuous at the point $f\left(c\right),$ there exists an $\alpha >0$ such that $\left|g\right(t)-g(f\left(c\right)\left)\right|<\u03f5$ if $|t-f(c\left)\right|<\alpha .$ Now, using this positive number $\alpha ,$ and using the fact that $f$ is continuous at the point $c,$ there exists a $\delta >0$ so that $\left|f\right(s)-f(c\left)\right|<\alpha $ if $|s-c|<\delta .$ Therefore, if $|s-c|<\delta ,$ then $\left|f\right(s)-f(c\left)\right|<\alpha ,$ and hence $\left|g\right(f\left(s\right))-g(f\left(c\right)\left)\right|=|g\circ f(s)-g\circ f(c\left)\right|<\u03f5,$ which completes the proof.
Using the previous theorems and exercises, explain why the following functions $f$ are continuous on their domains. Describe the domains as well.
Let $N$ be the set of natural numbers, let $P$ be the set of positive real numbers, and define $f:N\to P$ by $f\left(n\right)=\sqrt{1+n}.$ Prove that $f$ is continuous at each point of $N.$ Show in fact that every function $f:N\to C$ is continuous on this domain $N.$
HINT: Show that for any $\u03f5>0,$ the choice of $\delta =1$ will work.
The next result establishes an equivalence between the basic $\u03f5,\delta $ definition of continuity and a sequential formulation.In many cases, maybe most, this sequential version of continuity is easier to work with than the $\u03f5,\delta $ version.
Let $f:S\to C$ be a complex-valued function on $S,$ and let $c$ be a point in $S.$ Then $f$ is continuous at $c$ if and only if the following condition holds: For every sequence $\left\{{x}_{n}\right\}$ of elements of $S$ that converges to $c,$ the sequence $\left\{f\left({x}_{n}\right)\right\}$ converges to $f\left(c\right).$ Or, said a different way, if $\left\{{x}_{n}\right\}$ converges to $c,$ then $\left\{f\left({x}_{n}\right)\right\}$ converges to $f\left(c\right).$ And, said yet a third (somewhat less precise) way, the function $f$ converts convergent sequences to convergent sequences.
Suppose first that $f$ is continuous at $c,$ and let $\left\{{x}_{n}\right\}$ be a sequence of elements of $S$ that converges to $c.$ Let $\u03f5>0$ be given. We must find a natural number $N$ such that if $n\ge N$ then $|f\left({x}_{n}\right)-f\left(c\right)|<\u03f5.$ First, choose $\delta >0$ so that $\left|f\right(y)-f(c\left)\right|<\u03f5$ whenever $y\in S$ and $|y-c|<\delta .$ Now, choose $N$ so that $|{x}_{n}-c|<\delta $ whenever $n\ge N.$ Then if $n\ge N,$ we have that $|{x}_{n}-c|<\delta ,$ whence $|f\left({x}_{n}\right)-f\left(c\right)|<\u03f5.$ This shows that the sequence $\left\{f\left({x}_{n}\right)\right\}$ converges to $f\left(c\right),$ as desired.
We prove the converse by proving the contrapositive statement; i.e., we will show that if $f$ is not continuous at $c,$ then there does exist a sequence $\left\{{x}_{n}\right\}$ that converges to $c$ but for which the sequence $\left\{f\left({x}_{n}\right)\right\}$ does not converge to $f\left(c\right).$ Thus, suppose $f$ is not continuous at $c.$ Then there exists an ${\u03f5}_{0}>0$ such that for every $\delta >0$ there is a $y\in S$ such that $|y-c|<\delta $ but $|f\left(y\right)-f\left(c\right)|\ge {\u03f5}_{0}.$ To obtain a sequence, we apply this statement to $\delta $ 's of the form $\delta =1/n.$ Hence, for every natural number $n$ there exists a point ${x}_{n}\in S$ such that $|{x}_{n}-c|<1/n$ but $|f\left({x}_{n}\right)-f\left(c\right)|\ge {\u03f5}_{0}.$ Clearly, the sequence $\left\{{x}_{n}\right\}$ converges to $c$ since $|{x}_{n}-c|<1/n.$ On the other hand, the sequence $\left\{f\left({x}_{n}\right)\right\}$ cannot be converging to $f\left(c\right),$ because $|f\left({x}_{n}\right)-f\left(c\right)|$ is always $\ge {\u03f5}_{0}.$
This completes the proof of the theorem.
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