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\title{MTH435 - Homework 2}

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 \maketitle
 
 Solutions to the questions in Section B should be submitted by the start of class on 09/26/18. %You are encouraged to work in groups, but the solutions to the submitted work must be written up on your own. As per University policy, if you did a significant amount of work with another person, this should be indicated on your assignment. You may find the questions in Section A helpful in the later questions. Since this is an upper level course, your answers are supposed to be clearly written. Unclear or poorly written proofs may not receive full credit.
 
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 \section{Warm-up Questions}
 
 \begin{que} We investigate the necessity of the assumptions in the Nested Intervals Property.
  \begin{itemize}
   \item[(i)] \emph{Closed}: Let $I_n = \left( 0 , \frac{1}{n} \right)$. Prove that $ \bigcap I_n = \varnothing$.
   \item[(ii)] \emph{Bounded}: Let $I_n = [n,\infty)$. Prove that $\bigcap I_n = \varnothing$.
   \item[(iii)] \emph{Nested}: Let $I_n = [n,n+1]$. Prove that $\bigcap I_n = \varnothing$.
   \item[(iv)] \emph{Completeness}: Let $I_n = \left[\sqrt{2}, \sqrt{2} + \frac{1}{n}\right] \cap \Q$. Prove that $\bigcap I_n = \varnothing$.
  \end{itemize}

 \end{que}

 \begin{que}
  Use the definition of convergence to prove the following limits.
  \begin{itemize}
   \item[(i)] $\displaystyle \lim_{n \to \infty} \frac{n}{n+1} = 1$ 
   \vskip 0.1cm
   \item[(ii)] $\displaystyle \lim_{n \to \infty} \frac{1}{n} (\cos(n^2+5) - \sin(2n^3)) = 0$
   \vskip 0.1cm
   \item[(iii)] $\displaystyle \lim_{n \to \infty} \sqrt{4 + \frac{1}{n}} = 2$
   \item[(iv)] (\emph{Harder?}) $\displaystyle \lim_{n \to \infty} \sqrt{n(n+1)} - n = \frac{1}{2}$
  \end{itemize}
 \end{que}

 
 \begin{que}
  Prove or provide a counterexample to the following.
  \begin{itemize}
   \item[(i)] If $x_n \geq 0$ for all $n \in \N$ and $\lim_{n \to \infty} x_n = x$ then $x \geq 0$.
   \item[(ii)] If $x_n > 0$ for all $n \in \N$ and $\lim_{n \to \infty} x_n = x$ then $x > 0$.
  \end{itemize}
 \end{que}
  
  \begin{que}
   Prove that if $x_n \geq 0$ for all $n \in \N$ and $x_n \to x$ then $\sqrt{x_n} \to \sqrt{x}$. You may like to use the following steps.
   \begin{enumerate}
    \item[(i)] Show that if $x_n \to 0$ then $\sqrt{x_n} \to 0$.
    \item[(ii)] Prove that if $a,b \geq 0$ then
    \[
     \sqrt{a} - \sqrt{b} = \frac{a - b}{\sqrt{a}+\sqrt{b}}.
    \]
    \item[(iii)] Use part (ii) to show that if $x > 0$ and $x_n \to x$ then $\sqrt{x_n} \to \sqrt{x}$.
   \end{enumerate}

  \end{que}

 
 
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 \section{Submitted Questions}
 
 \begin{que}
  Suppose $I_n = [a_n,b_n]$ is a nested sequence of closed, bounded intervals. Define $\xi = \sup \{ a_n \mid n \in \N \}$ and $\eta = \inf \{b_n \mid n \in N\}$. Prove that $\displaystyle \bigcap_{n \in \N} I_n = [\xi,\eta]$.
 \end{que}

 \begin{que} Prove or give a counterexample to the following. Let $(x_n)$ be a sequence.
  \begin{itemize}
   \item[(i)] If $(x_n)$ converges then $(|x_n|)$ converges.
   \item[(ii)] If $(|x_n|)$ converges then $x_n$ converges.
  \end{itemize}

 \end{que}
 
 
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 \section{Challenge Questions}
 
 \begin{que}
  Show that $x_n \to x$ if and only if for every neighbourhood $U$ of $x$ there exist only finitely many $n \in \N$ such that $x_n \notin U$. 
 \end{que}

 \begin{que}
  Let $A \subset \R$. We call $x \in \R$ a \emph{limit point} of $A$ if for every neighbourhood $U$ of $x$ there exists $a \in A$ such that $a \in U$. Denote the set of limit points of $A$ by $\overline{A}$\footnote{This set is called the \emph{closure} of $A$}.
  \begin{itemize}
   \item[(i)] Compute $\overline{(0,1)}, \overline{[0,1]}, \overline{\{1/n \mid n \in \N \}}$ and $\overline{\Q}$.
   \item[(ii)] Show that $x \in \overline{A}$ if and only if there exists a sequence $(x_n) \to x$ with $x_n \in A$ for all $n \in \N$.
   \item[(iii)] Prove that $A$ is closed if and only if $A = \overline{A}$.
  \end{itemize}
 \end{que}

 \begin{que}
  Let $a > 0$ and $n \in \N$. By following the argument that $\sqrt{2}$ exists, or otherwise, prove that there exists $b \in \R$ such that $b^n = a$.
 \end{que}

 
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