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\textbf{Math 3100, Project 02, Due Monday April 28}
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Definition.  The sequence of rational numbers $S = \{r_i\}_{i=1}^\infty$ converges to $0$ means that if $\epsilon >0$ then there exists an integer $N_\epsilon$ so that if $n>N_\epsilon$ then
\begin{eqnarray*}
|r_n| & < & \epsilon.
\end{eqnarray*}

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Let $\hat R$ be the collection of all Cauchy sequences of rational numbers.  Define the relation $\sim$ on $\hat R$ by $\{x_i\}_{i=1}^\infty \sim \{y_i\}_{i=1}^\infty$ if and only if $\{x_i - y_i\}_{i=1}^\infty$ converges to $0$.

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Denote the collection of equivalence classes in $\hat R$ as $\mathbb{R}$.

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Definition.  We define the operation $\leq$ on $\mathbb{R}$ as follows:
\begin{eqnarray*}
\Big[\{x_i\}_{i=1}^\infty\Big] & \leq &  \Big[\{y_i\}_{i=1}^\infty\Big]
\end{eqnarray*}
if and only if for each $\epsilon >0$ there exists an integer $N_\epsilon$ such that if $n > N_\epsilon$ then
\begin{eqnarray*}
x_n - y_n & < &  \epsilon .
\end{eqnarray*}

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Prove the following theorems.

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Theorem 1.  The relation $\leq$ is well defined.

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Definition: if $M$ is a subset of the $\mathbb{R}$ then  $B$ is an upper bound of $M$ means that $x \le B$ for all $x \in M$; $B$ is said to a least upper bound of $M$ if $B$ is an upper bound of $M$ and no element less than $B$ is an upper bound of $M$.  [Note that $x \in \mathbb{R}$ means that $x=\
\Big[ \{x_i\}_{i=1}^\infty \Big]$. Also note that we may assume the $B \in \theta(\mathbb{Q})$.]

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Theorem 1.9.  [An easier version of theorem 2.]  If $M$ is a subset of $\theta(\mathbb{Q})$ and there is an upper bound of $M$ then there is a least upper bound of $M$.

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Theorem 2.  If $M$ is a subset of $\mathbb{R}$ and there is an upper bound of $M$ then there is a least upper bound of $M$.

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Homework exercises for Friday April 25.  These are supposed to be helpful for working through the project.  Use the definition of $\leq$ given above to show the following.

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A. If $\Big[\{x_i\}_{i=1}^\infty\Big]  \leq  \Big[\{y_i\}_{i=1}^\infty\Big]$ then there exists sequences $\{a_i\}_{i=1}^\infty \in \Big[\{x_i\}_{i=1}^\infty\Big]$ and $\{b_i\}_{i=1}^\infty \in \Big[\{y_i\}_{i=1}^\infty\Big]$  so that $a_i \leq b_i$ for all $i \in \mathbb{N}$.

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B. If $\{a_i\}_{i=1}^\infty \in \Big[\{x_i\}_{i=1}^\infty\Big]$ then 
\begin{eqnarray*}
\Big[\{a_i\}_{i=1}^\infty\Big] & \leq &  \Big[\{1 + a_i\}_{i=1}^\infty\Big].
\end{eqnarray*}

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C.
\begin{eqnarray*}
\Big[\Big \{\frac{100}{n} \Big \}_{n=1}^\infty\Big] & \leq &  \Big[ \Big\{1 - \frac 1n \Big \}_{n=1}^\infty\Big].
\end{eqnarray*}


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