3.3 Continuity  (Page 2/2)

Finally, using the continuity of both $f$ and $g$ applied to the positive numbers ${ϵ}_1=ϵ/(4M_2{\left|g\left(c\right)\right|}^2)$ and ${ϵ}_2=ϵ/(4M_1{\left|g\left(c\right)\right|}^2),$ choose $\delta >0,$ with $\delta <min({\delta }_1,{\delta }_2,{\delta }^{\text{'}}),$ and such that if $|y-c|<\delta$ and $y\in S$ then $|f\left(y\right)-f\left(c\right)|<\frac{ϵ}{4M_2/{\left|g\left(c\right)\right|}^2}$ and $|g\left(c\right)-g\left(y\right)|<\frac{ϵ}{4M_1/{\left|g\left(c\right)\right|}^2}.$ Then, if $|y-c|<\delta$ and $y\in S$ we have that

as desired.

The composition of continuous functions is continuous.

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 $ϵ>0$ be given. Because $g$ is continuous at the point $f\left(c\right),$ there exists an $\alpha >0$ such that $\left|g\right(t)-g(f\left(c\right)\left)\right|<ϵ$ 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 $\left|g\right(f\left(s\right))-g(f\left(c\right)\left)\right|=|g\circ f(s)-g\circ f(c\left)\right|<ϵ,$ which completes the proof.

The next result establishes an equivalence between the basic $ϵ,\delta$ definition of continuity and a sequential formulation.In many cases, maybe most, this sequential version of continuity is easier to work with than the $ϵ,\delta$ 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 $ϵ>0$ be given. We must find a natural number $N$ such that if $n\ge N$ then $|f\left(x_n\right)-f\left(c\right)|<ϵ.$ First, choose $\delta >0$ so that $\left|f\right(y)-f(c\left)\right|<ϵ$ 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 $|f\left(x_n\right)-f\left(c\right)|<ϵ.$ 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 ${ϵ}_0>0$ such that for every $\delta >0$ there is a $y\in S$ such that $|y-c|<\delta$ but $|f\left(y\right)-f\left(c\right)|\ge {ϵ}_0.$ 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 $|f\left(x_n\right)-f\left(c\right)|\ge {ϵ}_0.$ 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 $\ge {ϵ}_0.$

This completes the proof of the theorem.