Notes

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\booknotes{Sample Document}{Joshua Motoaki Lau}{Summer 2025}
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\section{Introduction to Real Analysis}

This chapter covers the foundations of real analysis, beginning with properties of the real number system.

\begin{keyidea}
    The completeness property of real numbers states that every non-empty set of real numbers that is bounded above has a least upper bound.
\end{keyidea}

\begin{example}
    Consider the set $S = \{x \in \mathbb{R} : x^2 < 2\}$. 
    This set is bounded above, and its least upper bound is $\sqrt{2}$.
    
    We can verify this by noting that:
    \begin{itemize}
        \item If $x^2 < 2$, then $x < \sqrt{2}$, so $\sqrt{2}$ is an upper bound.
        \item For any $\varepsilon > 0$, $(\sqrt{2} - \varepsilon)^2 < 2$, so $\sqrt{2} - \varepsilon \in S$.
    \end{itemize}
    
    Therefore, $\sqrt{2}$ is the least upper bound of $S$.
\end{example}

\begin{remark}
    The completeness property distinguishes the real numbers from the rational numbers. For instance, the set $\{x \in \mathbb{Q} : x^2 < 2\}$ has no least upper bound in $\mathbb{Q}$.
\end{remark}

A proper understanding of completeness\footnote{This concept was first rigorously formulated by Dedekind in the 19th century.} is essential for developing the theory of limits and continuity.

\newpage
\section{Sequences and Series}

In this chapter, we explore infinite sequences and series, their convergence properties, and tests for convergence.

\begin{keyidea}
    A sequence $\{a_n\}$ converges to a limit $L$ if for every $\varepsilon > 0$, there exists $N \in \mathbb{N}$ such that $|a_n - L| < \varepsilon$ for all $n \geq N$.
\end{keyidea}

\begin{example}
    Let's prove that the sequence $a_n = \frac{1}{n}$ converges to 0.
    
    Given $\varepsilon > 0$, we need to find $N$ such that $|\frac{1}{n} - 0| < \varepsilon$ for all $n \geq N$.
    
    Since $|\frac{1}{n} - 0| = \frac{1}{n}$, we need $\frac{1}{n} < \varepsilon$, which is equivalent to $n > \frac{1}{\varepsilon}$.
    
    Therefore, we can choose $N = \lceil\frac{1}{\varepsilon}\rceil$, and for all $n \geq N$, we have $|a_n - 0| < \varepsilon$.
\end{example}

\begin{remark}
    While many sequences converge, there are important examples of divergent sequences like $a_n = (-1)^n$ and $a_n = n$.
\end{remark}


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