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\begin{center}\begin{large}\textbf{Math 101 - Problem Set 4\\
Due Tuesday, October 3}\end{large}\end{center}



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\textit{This problem set is slightly shorter so that you have extra time to study. The first exam will be on Thursday, October 5 in class.}

\begin{enumerate} 

\item Complete the proof of the theorem from class that if an ordered set $A$ has the least upper bound property, then it has the greatest lower bound property by proving the following claim: 

Let $A_0$ be a nonempty subset of $A$ that is bounded below. Let $B$ be the set of lower bounds of $A_0$, and let $b_0 = \sup (B)$. Then $b_0 = \inf(A_0)$.

\item Let $A$ be an ordered set. Suppose $a, b \in A$ and $a <b$. Consider the open interval $$(a,b) = \{x \mid a <x<b\}.$$ If this set is empty, we call $a$ the \textit{\textbf{immediate predecessor}} of $b$ and $b$ the \textit{\textbf{immediate successor}} of $a$. For example, in $\mathbb{Z}$, 1 is the immediate successor of 0. In $\mathbb{R}$, no element has an immediate successor or predecessor.

Show that any element in an ordered set has at most one immediate successor.

\item For each of the following sets, given the dictionary order, give a counterexample to show that they don't satisfy the least upper bound property. Make sure to prove that your example works.

\begin{enumerate}
\item $[-1,1] \times ([-1,1] - \{0\})$
\item$[0,1] \times (0,1]$
\end{enumerate}

\item Let $\mathbb{Z}_+$ denote the set of positive integers. Consider the following strict total order relations on $\mathbb{Z}_+ \times \mathbb{Z}_+$:
\begin{itemize}
\item The dictionary order.
\item $(a,b) < (x,y)$ if either $a-b < x-y$ or $a-b = x-y$ and $b<y$.
\item $(a,b) < (x,y)$  if either $a+b < x+y$ or $a+b = x+y$ and $b<y$.
\end{itemize}
In each of these relations, which elements have immediate predecessors? Does the set have a smallest element?

\item Let $f: (-1,1) \to \mathbb{R}$ be defined $$f(x) = \frac{x}{1-x^2}.$$
Show that $(-1,1)$ and $\mathbb{R}$ have the same order type by proving the following.

\begin{enumerate}
\item $f$ is a bijection.
\item $f$ is \textit{\textbf{order-preserving}}. That is,  if $a,b \in (-1,1)$ and $a <b$, then $f(a) < f(b)$.
\end{enumerate}

\item Show that the subset of $\mathbb{R}$,  $(0,1) \cup \{2\}$, has the same order type as $(1,2]$.

\item Show that there is no function $f: \mathbb{R} \to \mathbb{Z}$ such that if $a,b \in \mathbb{R}$ and $a <b$, then $f(a) < f(b)$.

\end{enumerate}
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