Math Notes

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\fancyhead[L]{\textbf{MATH 425 --- Real Analysis}}
\fancyhead[R]{\textit{Lecture Notes}}
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\begin{document}

\begin{center}
  {\LARGE\bfseries MATH 425 --- Real Analysis}\\[6pt]
  {\Large Lecture Notes}\\[4pt]
  {\large Prof.\ Eleanor Whitfield}\\[2pt]
  {\large Fall 2025}\\[8pt]
  \textcolor{gray}{\rule{0.5\textwidth}{0.5pt}}
\end{center}

\tableofcontents
\newpage

% ============================================================
\section{Sequences and Convergence}

\begin{definition}[Convergence of a Sequence]
A sequence $(a_n)_{n=1}^{\infty}$ of real numbers \textbf{converges} to a limit $L \in \R$ if for every $\eps > 0$, there exists $N \in \N$ such that for all $n \ge N$,
\[
\abs{a_n - L} < \eps.
\]
We write $\lim_{n\to\infty} a_n = L$ or $a_n \to L$.
\end{definition}

\begin{example}
We show that $\lim_{n\to\infty} \frac{1}{n} = 0$. Let $\eps > 0$ be given. By the Archimedean property, choose $N \in \N$ with $N > 1/\eps$. Then for all $n \ge N$,
\[
\abs*{\frac{1}{n} - 0} = \frac{1}{n} \le \frac{1}{N} < \eps. \qedhere
\]
\end{example}

\begin{theorem}[Uniqueness of Limits]
\label{thm:unique-limit}
If $(a_n)$ converges, then its limit is unique.
\end{theorem}

\begin{proof}
Suppose $a_n \to L_1$ and $a_n \to L_2$. Let $\eps > 0$. Then there exist $N_1, N_2 \in \N$ such that $\abs{a_n - L_1} < \eps/2$ for all $n \ge N_1$ and $\abs{a_n - L_2} < \eps/2$ for all $n \ge N_2$. Setting $N = \max(N_1, N_2)$, for any $n \ge N$,
\[
\abs{L_1 - L_2} \le \abs{L_1 - a_n} + \abs{a_n - L_2} < \frac{\eps}{2} + \frac{\eps}{2} = \eps.
\]
Since $\eps > 0$ was arbitrary, $\abs{L_1 - L_2} = 0$, so $L_1 = L_2$.
\end{proof}

\begin{theorem}[Algebraic Limit Theorem]
\label{thm:alg-limit}
Let $(a_n) \to a$ and $(b_n) \to b$. Then:
\begin{enumerate}[label=(\roman*)]
  \item $(a_n + b_n) \to a + b$,
  \item $(c \cdot a_n) \to c \cdot a$ for any $c \in \R$,
  \item $(a_n \cdot b_n) \to a \cdot b$,
  \item $(a_n / b_n) \to a / b$, provided $b \neq 0$.
\end{enumerate}
\end{theorem}

\begin{proof}
We prove (iii); the others are similar or simpler.

We use the identity $a_n b_n - ab = a_n(b_n - b) + b(a_n - a)$. Since $(a_n)$ converges, it is bounded: there exists $M > 0$ such that $\abs{a_n} \le M$ for all $n$.

Let $\eps > 0$. Choose $N_1$ such that $\abs{b_n - b} < \frac{\eps}{2M}$ for $n \ge N_1$, and $N_2$ such that $\abs{a_n - a} < \frac{\eps}{2(\abs{b}+1)}$ for $n \ge N_2$. Then for $n \ge \max(N_1, N_2)$,
\[
\abs{a_n b_n - ab} \le \abs{a_n}\abs{b_n - b} + \abs{b}\abs{a_n - a} < M \cdot \frac{\eps}{2M} + \abs{b} \cdot \frac{\eps}{2(\abs{b}+1)} < \eps.
\]
\end{proof}

\begin{definition}[Cauchy Sequence]
A sequence $(a_n)$ is \textbf{Cauchy} if for every $\eps > 0$, there exists $N \in \N$ such that for all $m, n \ge N$,
\[
\abs{a_m - a_n} < \eps.
\]
\end{definition}

\begin{theorem}[Cauchy Criterion]
\label{thm:cauchy}
A sequence of real numbers converges if and only if it is Cauchy.
\end{theorem}

\begin{proof}
$(\Rightarrow)$ Suppose $a_n \to L$. Given $\eps > 0$, choose $N$ such that $\abs{a_n - L} < \eps/2$ for all $n \ge N$. Then for $m, n \ge N$,
\[
\abs{a_m - a_n} \le \abs{a_m - L} + \abs{L - a_n} < \eps.
\]

$(\Leftarrow)$ Let $(a_n)$ be Cauchy. Then $(a_n)$ is bounded (taking $\eps = 1$, all but finitely many terms lie in an interval of length 2). By the Bolzano--Weierstrass theorem, $(a_n)$ has a convergent subsequence $a_{n_k} \to L$.

We claim $a_n \to L$. Given $\eps > 0$, choose $N_1$ such that $\abs{a_m - a_n} < \eps/2$ for $m, n \ge N_1$, and $N_2$ such that $\abs{a_{n_k} - L} < \eps/2$ for $k \ge N_2$. Pick $k$ with $k \ge N_2$ and $n_k \ge N_1$. Then for $n \ge N_1$,
\[
\abs{a_n - L} \le \abs{a_n - a_{n_k}} + \abs{a_{n_k} - L} < \frac{\eps}{2} + \frac{\eps}{2} = \eps. \qedhere
\]
\end{proof}

% ============================================================
\section{Continuity}

\begin{definition}[Continuity at a Point]
Let $f : A \to \R$ where $A \subseteq \R$. We say $f$ is \textbf{continuous at} $c \in A$ if for every $\eps > 0$, there exists $\delta > 0$ such that whenever $x \in A$ and $\abs{x - c} < \delta$,
\[
\abs{f(x) - f(c)} < \eps.
\]
\end{definition}

\begin{theorem}[Sequential Characterization of Continuity]
\label{thm:seq-cont}
A function $f : A \to \R$ is continuous at $c \in A$ if and only if for every sequence $(x_n)$ in $A$ with $x_n \to c$, we have $f(x_n) \to f(c)$.
\end{theorem}

\begin{proof}
$(\Rightarrow)$ Suppose $f$ is continuous at $c$ and $x_n \to c$. Let $\eps > 0$. There exists $\delta > 0$ such that $\abs{x - c} < \delta$ implies $\abs{f(x) - f(c)} < \eps$. Since $x_n \to c$, there exists $N$ with $\abs{x_n - c} < \delta$ for all $n \ge N$. Then $\abs{f(x_n) - f(c)} < \eps$ for $n \ge N$.

$(\Leftarrow)$ Suppose $f$ is \emph{not} continuous at $c$. Then there exists $\eps_0 > 0$ such that for every $\delta > 0$, there is some $x \in A$ with $\abs{x - c} < \delta$ but $\abs{f(x) - f(c)} \ge \eps_0$. Taking $\delta = 1/n$, we obtain a sequence $(x_n)$ with $x_n \to c$ but $\abs{f(x_n) - f(c)} \ge \eps_0$ for all $n$, contradicting the hypothesis.
\end{proof}

\begin{corollary}
The function $f(x) = \sin(1/x)$ has no continuous extension to $x = 0$.
\end{corollary}

\begin{proof}
Consider $x_n = \frac{1}{n\pi}$ and $y_n = \frac{1}{2n\pi + \pi/2}$. Both sequences converge to 0, but $f(x_n) = \sin(n\pi) = 0$ while $f(y_n) = \sin(2n\pi + \pi/2) = 1$. By Theorem~\ref{thm:seq-cont}, no value of $f(0)$ can make $f$ continuous at 0.
\end{proof}

% ============================================================
\section{Compactness and the Extreme Value Theorem}

\begin{definition}[Open Cover]
Let $K \subseteq \R$. An \textbf{open cover} of $K$ is a collection $\{O_\alpha\}_{\alpha \in I}$ of open sets such that $K \subseteq \bigcup_{\alpha \in I} O_\alpha$.
\end{definition}

\begin{definition}[Compact Set]
A set $K \subseteq \R$ is \textbf{compact} if every open cover of $K$ has a finite subcover.
\end{definition}

\begin{theorem}[Heine--Borel]
\label{thm:heine-borel}
A subset $K \subseteq \R$ is compact if and only if it is closed and bounded.
\end{theorem}

\begin{proof}[Proof sketch]
$(\Rightarrow)$ We show compact implies bounded and closed. If $K$ is not bounded, the cover $\{(-n, n) : n \in \N\}$ has no finite subcover. If $K$ is not closed, there exists a limit point $x \notin K$; the cover $\{y : \abs{y - x} > 1/n\}$ has no finite subcover.

$(\Leftarrow)$ This is the substantive direction. Suppose $K \subseteq [a, b]$ is closed and let $\{O_\alpha\}$ be an open cover. We use the bisection method on $[a, b]$. At each step, at least one half cannot be finitely covered (otherwise both can, giving a finite cover of the whole interval). This produces a nested sequence of closed intervals whose lengths tend to 0, converging to a point $x^* \in K$. But $x^*$ lies in some $O_{\alpha_0}$, which covers a neighborhood of $x^*$---contradicting that arbitrarily small intervals around $x^*$ cannot be finitely covered.
\end{proof}

\begin{theorem}[Extreme Value Theorem]
If $f : K \to \R$ is continuous and $K$ is compact, then $f$ attains its maximum and minimum on $K$.
\end{theorem}

\begin{proof}
Since $f$ is continuous and $K$ is compact, $f(K)$ is compact (the continuous image of a compact set is compact), hence closed and bounded by Theorem~\ref{thm:heine-borel}. Boundedness gives $M = \sup f(K) < \infty$. Closedness implies $M \in f(K)$, so there exists $x_0 \in K$ with $f(x_0) = M$. The argument for the minimum is analogous.
\end{proof}

\begin{lemma}[Lebesgue Number Lemma]
\label{lem:lebesgue}
If $K$ is a compact set and $\{O_\alpha\}$ is an open cover of $K$, then there exists $\delta > 0$ (the \textbf{Lebesgue number}) such that every subset of $K$ with diameter less than $\delta$ is contained in some $O_\alpha$.
\end{lemma}

\begin{proof}
Suppose not. Then for each $n \in \N$, there exists $C_n \subseteq K$ with $\mathrm{diam}(C_n) < 1/n$ that is not contained in any $O_\alpha$. Pick $x_n \in C_n$. By compactness (sequential compactness), $(x_n)$ has a convergent subsequence $x_{n_k} \to x^* \in K$. Since $x^* \in O_{\alpha_0}$ for some $\alpha_0$, there exists $r > 0$ with $B(x^*, r) \subseteq O_{\alpha_0}$. For large $k$, $\abs{x_{n_k} - x^*} < r/2$ and $\mathrm{diam}(C_{n_k}) < r/2$, so $C_{n_k} \subseteq B(x^*, r) \subseteq O_{\alpha_0}$, a contradiction.
\end{proof}

% ============================================================
\section{Uniform Continuity}

\begin{definition}[Uniform Continuity]
A function $f : A \to \R$ is \textbf{uniformly continuous} on $A$ if for every $\eps > 0$, there exists $\delta > 0$ such that for \emph{all} $x, y \in A$ with $\abs{x - y} < \delta$,
\[
\abs{f(x) - f(y)} < \eps.
\]
\end{definition}

\begin{remark}
The crucial difference from pointwise continuity is that $\delta$ depends only on $\eps$, not on the particular point.
\end{remark}

\begin{theorem}
If $f : K \to \R$ is continuous and $K$ is compact, then $f$ is uniformly continuous on $K$.
\end{theorem}

\begin{proof}
Let $\eps > 0$. For each $x \in K$, by continuity there exists $\delta_x > 0$ with $\abs{f(y) - f(x)} < \eps/2$ whenever $\abs{y - x} < \delta_x$. The collection $\{B(x, \delta_x/2) : x \in K\}$ is an open cover of $K$.

By compactness, there is a finite subcover: $K \subseteq \bigcup_{i=1}^{n} B(x_i, \delta_{x_i}/2)$. Set $\delta = \frac{1}{2}\min\{\delta_{x_1}, \ldots, \delta_{x_n}\} > 0$.

Now let $a, b \in K$ with $\abs{a - b} < \delta$. Then $a \in B(x_i, \delta_{x_i}/2)$ for some $i$, so $\abs{a - x_i} < \delta_{x_i}/2$. Also,
\[
\abs{b - x_i} \le \abs{b - a} + \abs{a - x_i} < \delta + \frac{\delta_{x_i}}{2} \le \frac{\delta_{x_i}}{2} + \frac{\delta_{x_i}}{2} = \delta_{x_i}.
\]
Therefore $\abs{f(a) - f(b)} \le \abs{f(a) - f(x_i)} + \abs{f(x_i) - f(b)} < \eps/2 + \eps/2 = \eps$.
\end{proof}

\begin{example}
The function $f(x) = x^2$ is uniformly continuous on $[0, 1]$ (by the theorem above, since $[0,1]$ is compact) but \emph{not} uniformly continuous on $\R$. To see the latter, note that $f(n + \tfrac{1}{n}) - f(n) = 2 + \tfrac{1}{n^2} > 2$ while $\abs{(n + \tfrac{1}{n}) - n} = \tfrac{1}{n} \to 0$.
\end{example}

\vfill
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  {\small\textit{End of Lecture Notes --- Chapters 1--4}}
\end{center}

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