Mathematical Document

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\lhead{Introduction to Real Analysis}
\rhead{Lecture Notes}
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\title{\textbf{Foundations of Mathematical Analysis}\\[6pt]\large A Survey of Key Results in Real Analysis and Linear Algebra}
\author{Prof.\ Katherine Yang\\Department of Mathematics, University of Chicago}
\date{Spring 2026}

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\maketitle

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\newpage

\section{Metric Spaces and Topology}

\begin{definition}[Metric Space]
A \emph{metric space} is a pair $(X, d)$ where $X$ is a set and $d: X \times X \to \mathbb{R}$ is a function satisfying for all $x, y, z \in X$:
\begin{enumerate}[label=(\roman*)]
  \item $d(x, y) \geq 0$ with equality if and only if $x = y$ \quad (positive definiteness),
  \item $d(x, y) = d(y, x)$ \quad (symmetry),
  \item $d(x, z) \leq d(x, y) + d(y, z)$ \quad (triangle inequality).
\end{enumerate}
\end{definition}

\begin{example}
The Euclidean space $\mathbb{R}^n$ with the standard metric
\[
  d(\mathbf{x}, \mathbf{y}) = \sqrt{\sum_{i=1}^{n} (x_i - y_i)^2}
\]
is a complete metric space. More generally, for $1 \leq p < \infty$, the $\ell^p$ metric is defined by
\[
  d_p(\mathbf{x}, \mathbf{y}) = \left( \sum_{i=1}^{n} |x_i - y_i|^p \right)^{1/p}.
\]
\end{example}

\begin{definition}[Open and Closed Sets]
Let $(X, d)$ be a metric space. A set $U \subseteq X$ is \emph{open} if for every $x \in U$, there exists $\varepsilon > 0$ such that $B(x, \varepsilon) \subseteq U$, where $B(x, \varepsilon) = \{y \in X : d(x, y) < \varepsilon\}$ is the open ball of radius $\varepsilon$ centered at $x$. A set $F \subseteq X$ is \emph{closed} if its complement $X \setminus F$ is open.
\end{definition}

\begin{theorem}[Characterization of Closed Sets]
\label{thm:closed-char}
A set $F \subseteq X$ is closed if and only if for every convergent sequence $(x_n)_{n=1}^{\infty}$ in $F$, the limit $\lim_{n \to \infty} x_n$ belongs to $F$.
\end{theorem}

\begin{proof}
$(\Rightarrow)$ Suppose $F$ is closed and let $(x_n)$ be a sequence in $F$ with $x_n \to x$. If $x \notin F$, then $x \in X \setminus F$ which is open. Hence there exists $\varepsilon > 0$ with $B(x, \varepsilon) \subseteq X \setminus F$. But $x_n \to x$ implies there exists $N$ with $d(x_n, x) < \varepsilon$ for all $n \geq N$, contradicting $x_n \in F$. Therefore $x \in F$.

$(\Leftarrow)$ Suppose every convergent sequence in $F$ has its limit in $F$. We show $X \setminus F$ is open. Let $x \in X \setminus F$. If no $\varepsilon > 0$ satisfies $B(x, \varepsilon) \subseteq X \setminus F$, then for each $n \in \mathbb{N}$, there exists $x_n \in F \cap B(x, 1/n)$. Then $x_n \to x$, so $x \in F$ by hypothesis---a contradiction. Thus $X \setminus F$ is open, and $F$ is closed.
\end{proof}

\section{Sequences and Series}

\begin{theorem}[Bolzano--Weierstrass]
Every bounded sequence in $\mathbb{R}^n$ has a convergent subsequence.
\end{theorem}

\begin{lemma}[Squeeze Lemma]
\label{lem:squeeze}
If $(a_n)$, $(b_n)$, and $(c_n)$ are real sequences satisfying $a_n \leq b_n \leq c_n$ for all $n$ sufficiently large, and if $\lim_{n \to \infty} a_n = \lim_{n \to \infty} c_n = L$, then $\lim_{n \to \infty} b_n = L$.
\end{lemma}

\begin{proof}
Let $\varepsilon > 0$. There exists $N_1$ such that $|a_n - L| < \varepsilon$ for all $n \geq N_1$, and $N_2$ such that $|c_n - L| < \varepsilon$ for all $n \geq N_2$. Let $N_3$ be such that $a_n \leq b_n \leq c_n$ for all $n \geq N_3$. Set $N = \max(N_1, N_2, N_3)$. For $n \geq N$:
\[
  L - \varepsilon < a_n \leq b_n \leq c_n < L + \varepsilon,
\]
so $|b_n - L| < \varepsilon$.
\end{proof}

\begin{definition}[Absolute and Conditional Convergence]
A series $\sum_{n=1}^{\infty} a_n$ \emph{converges absolutely} if $\sum_{n=1}^{\infty} |a_n|$ converges. It \emph{converges conditionally} if it converges but does not converge absolutely.
\end{definition}

\begin{example}
The series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots = \ln 2$ converges conditionally, since the harmonic series $\sum 1/n$ diverges.
\end{example}

\section{Linear Algebra}

\begin{definition}[Eigenvalues and Eigenvectors]
Let $A \in \mathbb{R}^{n \times n}$. A scalar $\lambda \in \mathbb{C}$ is an \emph{eigenvalue} of $A$ if there exists a nonzero vector $\mathbf{v} \in \mathbb{C}^n$ such that $A\mathbf{v} = \lambda \mathbf{v}$. The vector $\mathbf{v}$ is called an \emph{eigenvector} corresponding to $\lambda$.
\end{definition}

\begin{theorem}[Spectral Theorem for Symmetric Matrices]
\label{thm:spectral}
If $A \in \mathbb{R}^{n \times n}$ is symmetric (i.e., $A = A^\top$), then:
\begin{enumerate}[label=(\roman*)]
  \item All eigenvalues of $A$ are real.
  \item There exists an orthogonal matrix $Q \in \mathbb{R}^{n \times n}$ such that $A = Q \Lambda Q^\top$, where $\Lambda = \diag(\lambda_1, \ldots, \lambda_n)$.
\end{enumerate}
\end{theorem}

Consider the matrix
\[
  A = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 2 \end{pmatrix}.
\]
The characteristic polynomial is
\[
  \det(A - \lambda I) = \det \begin{pmatrix} 2 - \lambda & 1 & 0 \\ 1 & 3 - \lambda & 1 \\ 0 & 1 & 2 - \lambda \end{pmatrix} = -\lambda^3 + 7\lambda^2 - 14\lambda + 8 = -(\lambda - 1)(\lambda - 2)(\lambda - 4).
\]
Hence the eigenvalues are $\lambda_1 = 1$, $\lambda_2 = 2$, $\lambda_3 = 4$.

\begin{proposition}[Trace and Determinant]
For any $A \in \mathbb{R}^{n \times n}$ with eigenvalues $\lambda_1, \ldots, \lambda_n$ (counted with multiplicity):
\begin{align}
  \tr(A) &= \sum_{i=1}^{n} \lambda_i, \\
  \det(A) &= \prod_{i=1}^{n} \lambda_i.
\end{align}
\end{proposition}

\section{Integral Calculus}

\begin{theorem}[Fundamental Theorem of Calculus]
Let $f: [a, b] \to \mathbb{R}$ be continuous. Define $F(x) = \int_a^x f(t)\, dt$ for $x \in [a,b]$. Then:
\begin{enumerate}[label=(\roman*)]
  \item $F$ is continuous on $[a,b]$ and differentiable on $(a,b)$ with $F'(x) = f(x)$.
  \item If $G$ is any antiderivative of $f$ on $[a,b]$, then $\displaystyle\int_a^b f(t)\, dt = G(b) - G(a)$.
\end{enumerate}
\end{theorem}

\begin{example}
Evaluate the integral $\displaystyle\int_0^1 \frac{x^4(1-x)^4}{1+x^2}\, dx$.

Expanding and performing polynomial long division:
\begin{align*}
  \frac{x^4(1-x)^4}{1+x^2} &= x^6 - 4x^5 + 5x^4 - 4x^2 + 4 - \frac{4}{1+x^2}.
\end{align*}
Integrating term by term:
\begin{align*}
  \int_0^1 \frac{x^4(1-x)^4}{1+x^2}\, dx &= \left[\frac{x^7}{7} - \frac{2x^6}{3} + x^5 - \frac{4x^3}{3} + 4x - 4\arctan x\right]_0^1 \\
  &= \frac{1}{7} - \frac{2}{3} + 1 - \frac{4}{3} + 4 - \pi \\
  &= \frac{22}{7} - \pi.
\end{align*}
This provides the classical proof that $\frac{22}{7} - \pi > 0$, i.e., $\pi < \frac{22}{7}$.
\end{example}

\section{Set Theory Notation}

We collect common set-theoretic notation used throughout:
\begin{align*}
  A \cup B &= \{x : x \in A \text{ or } x \in B\}, \\
  A \cap B &= \{x : x \in A \text{ and } x \in B\}, \\
  A \setminus B &= \{x : x \in A \text{ and } x \notin B\}, \\
  A \times B &= \{(a, b) : a \in A,\, b \in B\}, \\
  \bigcup_{i \in I} A_i &= \{x : x \in A_i \text{ for some } i \in I\}, \\
  \bigcap_{i \in I} A_i &= \{x : x \in A_i \text{ for all } i \in I\}.
\end{align*}

\begin{remark}
De Morgan's laws hold in any universe $X$: for a family $\{A_i\}_{i \in I}$ of subsets of $X$,
\[
  X \setminus \bigcup_{i \in I} A_i = \bigcap_{i \in I} (X \setminus A_i) \quad\text{and}\quad X \setminus \bigcap_{i \in I} A_i = \bigcup_{i \in I} (X \setminus A_i).
\]
\end{remark}

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