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\section{Limits and Convergence [b]}

Review from Numbers and Sets: sequences $a_n$,
$(a_n)_{n = 1}^\infty$, $a_n \in \RR$.
\begin{definition*}
    We say that $a_n \to a$ as $n \to \infty$ if
    given $\eps > 0$, $\exists \,\, N$ such that $|a_n
    - a| < \eps$ for all $n \ge N$.
    Note $N = N(\eps)$.
\end{definition*}

\begin{definition*}[Monotonic sequence]
    A sequence is \emph{increasing} if $a_n \le
    a_{n + 1}$ for all $n$. Similarly, a sequence
    is \emph{decreasing} if $a_n \ge a_{n + 1}$
    for all $n$. The sequence is \emph{strictly
    increasing / decreasing} if equality never
    occurs. A sequence is \emph{monotonic} if it
    is either increasing or decreasing.
\end{definition*}

\begin{axiom*}[Fundamental Axiom of the Real Numbers]
    Given an increasing sequence $(a_n)_{n = 1}^\infty$
    and some $A \in \RR$ such that $a_n \le A$ for
    all $n$, there exists $a \in \RR$ such that
    $a_n \to a$ as $n \to \infty$. So an
    increasing sequence of real numbers bounded
    above \emph{converges}. Equivalently a
    decreasing sequence of real numbers bounded
    below converges. Equivalent also to: ``Every
    non-empty of real numbers bounded above has a
    \emph{supremum}''. (LUBA = Least Upper Bound
    Axiom).
\end{axiom*}

\begin{definition*}[supremum]
    Given $S \subset \RR$, $S \neq \emptyset$ we
    say that $\sup S = K$ if
    \begin{enumerate}[(i)]
        \item $x \le K \,\, \forall \,\, x \in S$
        \item given $\epsilon > 0$, $\exists x \in
            S$ such that $x > K - \eps$.
    \end{enumerate}
    \begin{note*}
        Supremum is unique. We also can define a
        similar notion of infimum.
    \end{note*}
\end{definition*}

\begin{lemma}
    \begin{enumerate}[(i)]
        \item The limit is unique. That is, if
            $a_n \to a$ and $a_n \to b$, then $a =
            b$.
        \item If $a_n \to a$ as $n \to \infty$ and
            $n_1 < n_2 < n_3 < \cdots$, then
            $a_{n_j} \to a$ as $j \to \infty$
            (subsequences converge to the same
            limit).
        \item If $a_n = c \,\,\forall\,\, n$, then
            $a_n \to c$ as $n \to \infty$.
        \item If $a_n \to a$ and $b_n \to b$, then
            $a_n + b_n \to a + b$.
        \item If $a_n \to a$ and $b_n \to b$, then
            $a_nb_n \to ab$.
        \item If $a_n \to a$, $a_n \neq 0
            \,\,\forall n$ and $a \neq 0$, then
            $\frac{1}{a_n} \to \frac{1}{a}$.
        \item If $a_n \le A \,\,\forall n$ and
            $a_n \to a$, then $a \le A$.
    \end{enumerate}
\end{lemma}

\begin{proof}
    We only do (i), (ii) and (v) and leave the
    others as exercise.
    \begin{enumerate}[(i)]
        \item[(i)] given $\eps > 0$, $\exists n_1$
            such that $|a_n - a| < \eps
            \,\,\forall\,\, n \ge n_1$, and
            $\exists n_2$ such that $|a_n - b| <
            \eps \,\,\forall\,\,n \ge n_2$. Then
            let $N = \max\{n_1, n_2\}$. Then if $n
            \ge N$,
            \[ |a - b| \le |a_n - a| + |a_n - b| <
            2\eps .\]
            If $a \neq b$, take $\eps = \frac{|a -
            b|}{3}$, then by triangle inequality
            \[ |a - b| < \frac{2}{3}|a - b| \]
            which is a contradiction if $a \neq
            b$, hence $a = b$.
        \item[(ii)] given $\eps > 0$, $\exists N$
            such that $|a_n - a| < \eps,
            \,\,\forall\,\, n \ge N$ since $n_j
            \ge j$ by induction, we have $|a_{n_j}
            - a| < \eps \,\,\forall\,\, j \ge N$,
            i.e. $a_{n_j} \to a$ as $j \to
            \infty$.
        \item[(v)] $|a_nb_n - ab| \le |a_nb_n -
            a_nb| + |a_nb - ab| = |a_n||b_n - b| +
            |b||a_n - a|$. Since $a_n \to a$,
            given $\eps > 0$, $\exists n_1$ such
            that $|a_n - a| < \eps \,\,\forall n
            \ge n_1$, and similarly since $b_n \to
            b$ $\exists n_2$ such that $|b_n - b|
            < \eps \,\,\forall\,\, n \ge n_2$. If
            $n \ge n_1(1)$, $|a_n - a| < 1$, so
            $|a_n| \le |a| + 1$. Hence
            \[ |a_nb_n - ab| \le \epsilon(|a| + 1
            + |b|) \]
            for all $n \ge n_3(\epsilon) =
            \max\{n_1(1), n_1(\epsilon),
            n_2(\epsilon)\}$.
    \end{enumerate}
\end{proof}

\begin{lemma}
    $\frac{1}{n} \to 0$ as $n \to \infty$.
\end{lemma}

\begin{proof}
    $\frac{1}{n}$ is a decreasing sequence bounded
    by below, so by the Fundamental Axiom it has a
    limit $a$. We claim that $a = 0$. Note that
    \[ \frac{1}{2n} = \half \times \frac{1}{n} \to
    \frac{a}{2} \]
    by Lemma 1.1(v). But $\frac{1}{2}$ is a
    subsequence, so by Lemma 1.1(ii),
    $\frac{1}{2n} \to a$. By uniqueness of limits
    (Lemma 1.1(i)), we have $a = \frac{a}{2}
    \implies a = 0$.
\end{proof}

\begin{remark*}
    The definition of limit of a sequence makes
    perfect sense for $a_n \in \CC$.
\end{remark*}

\begin{definition*}
    $a_n \to a$ if given $\eps > 0$, $\exists N$
    such that $\forall\,\, n \ge N$, $|a_n - a| <
    \eps$.
\end{definition*}

\noindent
The first six parts of Lemma 1.1 are the same over
$\CC$. The last one does not make sense (over
$\CC$) since it uses the \emph{order} of $\RR$.
