Math Problem Set

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\theoremstyle{definition}
\newtheorem{definition}{Definition}
\newtheorem{exercise}{Exercise}
\theoremstyle{plain}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
\newtheorem{proposition}{Proposition}
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\newcommand{\R}{\mathbb{R}}
\newcommand{\N}{\mathbb{N}}
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\newcommand{\eps}{\varepsilon}
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\begin{document}

\begin{center}
{\LARGE\bfseries Real Analysis -- Problem Set 6}\\[0.3em]
{\large Due Thursday}
\end{center}

\begin{definition}
A sequence $(a_n) \subset \R$ converges to $L \in \R$ iff for every $\eps > 0$
there exists $N \in \N$ such that $\abs{a_n - L} < \eps$ for all $n \ge N$.
\end{definition}

\begin{exercise}
Prove that $a_n = 1/n$ converges to $0$.
\end{exercise}
\begin{proof}
Given $\eps > 0$, choose $N > 1/\eps$. Then for $n \ge N$, $\abs{1/n - 0} = 1/n \le 1/N < \eps$.
\end{proof}

\begin{exercise}
Show that every convergent sequence is bounded.
\end{exercise}
\begin{proof}
Let $a_n \to L$. Choose $N$ with $\abs{a_n - L} < 1$ for $n \ge N$. Then
$\abs{a_n} \le \abs{L} + 1$ for $n \ge N$. Let $M = \max\{ \abs{a_1},\dots,\abs{a_{N-1}}, \abs{L}+1\}$.
Then $\abs{a_n} \le M$ for all $n$.
\end{proof}

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

\begin{exercise}
Prove the theorem above using the lion-hunting (bisection) method.
\end{exercise}

\begin{exercise}
Let $f: [0,1] \to \R$ be continuous with $f(0) = f(1)$. Prove there exist
$x, y \in [0,1]$ with $\abs{x-y} = 1/2$ and $f(x) = f(y)$.
\end{exercise}

\begin{remark}
Extend: does the conclusion hold for $\abs{x-y} = 1/3$? What about for $\abs{x-y} = 1/\pi$?
\end{remark}

\begin{exercise}
Evaluate or show divergence:
\begin{enumerate}[label=(\alph*)]
  \item $\displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}$
  \item $\displaystyle \sum_{n=1}^\infty \frac{n!}{n^n}$
  \item $\displaystyle \int_0^\infty e^{-x^2}\,dx$
\end{enumerate}
\end{exercise}

\end{document}