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\documentclass[12pt]{article}
\usepackage{lmodern}
\usepackage{microtype}
\usepackage{amsmath,amssymb,amsthm}
\usepackage{geometry}
\geometry{margin=2.5cm}
\usepackage{enumitem}
% Custom commands
\newcommand{\R}{\mathbb{R}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\eps}{\varepsilon}
% Theorem environments
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}[theorem]{Lemma}
\theoremstyle{definition}
\newtheorem{definition}{Definition}
\title{Problem Set 3}
\author{Your Name}
\date{Due: \today}
\begin{document}
\maketitle
\section*{Problem 1}
Prove that every bounded sequence in $\R^n$ has a convergent subsequence.
\textbf{Proof.}
We proceed by induction on $n$.
\textbf{Base case} ($n=1$): By the Bolzano-Weierstrass theorem, every bounded sequence
in $\R$ has a convergent subsequence.
\textbf{Inductive step}: Assume the result holds for $\R^{n-1}$. Let $(\mathbf{x}_k)$
be a bounded sequence in $\R^n$. The projection onto the first coordinate is bounded in $\R$,
so it has a convergent subsequence by the base case. $\square$
\section*{Problem 2}
Let $f: \R \to \R$ be differentiable. Show that if $|f'(x)| \leq M$ for all $x$, then:
\[
|f(x) - f(y)| \leq M|x - y| \quad \text{for all } x, y \in \R
\]
\textbf{Proof.}
By the Mean Value Theorem, there exists $c$ between $x$ and $y$ such that
$f(x) - f(y) = f'(c)(x-y)$.
Taking absolute values gives $|f(x) - f(y)| \leq M|x-y|$. $\square$
\end{document}
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