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\newpage
\section{Continuity [3]}

Let $E \subseteq \CC$ non-empty, $f : E \to \CC$
any function, and let $a \in E$. (This includes
the case in which $f$ Is real-valued and $E
\subseteq \RR$).

\setcounter{customdefinition}{0}
\begin{definition}
    $f$ is continuous at $a \in E$ if for every
    sequence $z_n \in E$ with $z_n \to a$, we have
    $f(z_n) \to f(a)$.
\end{definition}

\begin{definition}
    $f$ is continuous at $a \in E$, if given $\eps
    > 0$, $\exists \delta > 0$ such that if $|z -
    a| < \delta$, $z \in E$, then
    \[ |f(z) - f(a)| < \eps \]
    ($\eps$-$\delta$ definition).
\end{definition}

\noindent
We will prove that these two definitions are
equivalent.

\begin{proof}
    We know that given $\eps > 0$, $\exists \delta
    > 0$ such that $|z - a| < \delta$, $z \in E$,
    then $|f(z) - f(a)| < \eps$. Let $z_n \to a$.
    Then $\exists n_0$ such that $\forall\,\,n \ge
    n_0$ we have $|z_n - a| < \delta$ hence
    $|f(z_n) - f(a)| < \eps$ so $f(z_n) \to f(a)$.
    For the other direction, assume that $f(z_n)
    \to f(a)$ whenever $z_n \to a$ ($z_n \in E$).
    Suppose $f$ is \emph{not} continuous at $a$
    according to definition 2. Then:
    \begin{center}
        $\exists \eps > 0$ such that
        $\forall\,\,\delta > 0$, there exists $z
        \in E$ such that $|z - a| < \delta$ and
        $|f(z) - f(a)| \ge \eps$.
    \end{center}
    Let $\delta = \frac{1}{n}$, from the above we
    get $z_n$ such that $|z_n - a| < \frac{1}{n}$
    and $|f(z_n) - f(a)| \ge \eps$. Clearly $z_n
    \to a$, but $f(z_n)$ does not tend to $f(a)$
    because $f(z_n) - f(a)| \ge \eps$,
    contradiction.
\end{proof}

\begin{proposition}
    $a \in E$, $g, f : E \to \CC$ continuous at
    $a$> Then so are the functions $f(z) + g(z)$,
    $f(z)g(z)$ and $\lambda f(z)$ for any constant
    $\lambda$. In addition if $f(z) \neq 0
    \,\,\forall\,\, z \in E$ then $\frac{1}{f}$ is
    continuous at $a$.
\end{proposition}

\begin{proof}
    Using definition 1, this is obvious. Using the
    analogous results for sequences (lemma 1.1),
    for example if $z_n \to a$ then $f(z_n) \to
    f(a)$ and $g(z_n) \to g(a)$ so by lemma 1.1
    $f(z_n) + g(z_n) \to f(a) + g(a)$ etc.
\end{proof}

\myskip
The function $f(z) = z$ is continuous, so by
using the proposition, we get that every
polynomial is continuous at every point in
$\CC$.

\begin{note*}
    We say that $f$ is continuous on $E$ if it is
    continuous at every $a \in E$.
\end{note*}

\begin{remark*}
    Still it is \emph{instructive} to prove
    proposition 2.1 directly from the
    $\eps$-$\delta$ definition.
\end{remark*}

\noindent
Next we look at compositions.

\begin{theorem}
    Let $f : A \to \CC$ and $g : B \to \CC$ and $g
    : B \to \CC$ be two functions such that $f(A)
    \subset B$. Suppose $f$ is continuous at $a
    \in A$ and $g$ is continuous at $f(a)$. Then
    $g \circ f : A \to \CC$ is continuous at $a$.
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/cf427b98972211ec.png}
    \end{center}
\end{theorem}

\begin{proof}
    Take any sequence $z_n \to a$. By assumption
    $f(z_n) \to f(a)$. Set $w_n = f(z_n) \in B$,
    $w_n \to f(a)$; thus $g(w_n) = g(f(z_n)) \to
    g(f(a))$.
\end{proof}

\subsubsection*{Examples}

\begin{enumerate}[(1)]
    \item $f : \RR \to \RR$
        \[ f(x) = \begin{cases}
            \sin \left( \frac{1}{x} \right) & x
            \neq 0 \\
            0 & x = 0
        \end{cases} \]
        $\sin x$ is continuous (to be proved
        later!) \\
        if $x \neq 0$, then 2.1 and 2.2 imply that
        $f(x)$ is continuous at every $x \neq 0$.
        Discontinuous at 0 because let $x_n =
        \frac{1}{\left( 2n + \half \right) \pi}$,
        then $f(x_n) = 1$, $x_n \to 0$ but $f(0) =
        0$.
    \item \eqnoskip
        \[ f(x) = \begin{cases}
            x \sin \left( \frac{1}{x} \right) & x
            \neq 0 \\
            0 & x = 0
        \end{cases} \]
        $f$ is continuous at 0, take $x_n \to 0$
        then
        \[ |f(x_n)| \le |x_n| \]
        so $f(x_n) \to 0 = f(0)$.
    \item \eqnoskip
        \[ f(x) = \begin{cases}
            1 & x \in \QQ \\
            0 & x \not\in \QQ
        \end{cases} \]
        Discontinuous at every point: if $x \in
        \QQ$, take a sequence $x_n \to x$ with
        $x_n \not\in \QQ$, then $f(x_n) = 0$ which
        doesn't tend to $f(x) = 1$. Similarly if
        $x \not\in \QQ$, take $x_n \to x$ with
        $x_n \in \QQ$. Then $f(x_n) = 1$ so
        doesn't tend to $f(x) = 0$.
\end{enumerate}
