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		\title{Math 311 Workshop}
		\author{Yuchen Wei}
		\date{\today}	
		\maketitle

Recall: The (axiomatic)definition of real numbers $ \mathbb{R} $:

\vspace*{5pt}

Let $ \mathbb R $ be a field extension of $ \mathbb Q $, such that:
\begin{enumerate}

\item[(1)] The set $ \mathbb R $ is a field: (\textit{roughly} meaning:) For any element $ r $ in $ \mathbb R $, additive inverse exists($ -r $), multiplicative inverse exists if it is nonzero!($ 1/r, r\ne 0 $) addition and multiplication of real numebrs are commutative($ x+y=y+x, xy=yx $), associative($ x+(y+z)=(x+y)+z, (xy)z=x(yz) $), and the distibutive property($ x(y+z)=xy+xz $) holds.

\item[(2)] The field $ \mathbb R $ is ordered, meaning there exists a total order $ \ge $ such that for all real numbers $ x,y  $ and $ z $:

\subitem (a) if $ x\ge y $, then $ x+z\ge y+z $;

\subitem(b) if $ x\ge 0 $, and $ y\ge 0 $, then $ xy\ge 0 $.

\item[(3)] \textbf{Axiom of completeness. Every nonempty set of real numbers that is bounded above has a least upper bound.}

\end{enumerate}

\vspace*{10pt}
Remark: A \textbf{partial order} on a set $ S $ is a subset $ R $ of the set $ S\times S $, such that:

\begin{enumerate}

\item[1.] For any element $ x\in S, (x,x)\in R $ (\textbf{Reflexive}, in other words $ x\ge x $);

\item[2.] If for $ x,y\in S $, $ (x,y)\in R $ and $ (y,x)\in R $, then we have $ x=y $ (\textbf{Antisymmetric}, if $ x\ge y $ and $ y\ge x $, then $ x=y $ );

\item[3.] If $ (x,y)\in R $ and $ (y,z)\in R $, then $ (x,z)\in R $ (\textbf{Transitive}, if $ x\ge y $ and $ y\ge z $, then $ x\ge z $. ).

\end{enumerate}

Well, a \textbf{total order} is a partial order where every two elements in the set $ S $ are comparable, namely, it satisfies the additional condition: 
\begin{enumerate}
	\item[4*.] Either $ (x,y)\in R $ or $ (y,x)\in R $ (\textbf{Strongly connected}, either $ x\ge y $ or $ y\ge x $).
\end{enumerate}

\newpage

\textbf{Example 1.3.7.} Let $ A\subseteq\mathbb{R} $ be nonempty and bounded above, and let $ c\in\mathbb{R} $. Define the set $ c+A $ by
\[ c+A=\{c+a:a\in A\}. \]
Then $ \sup(c+A)=c+\sup A $.

\textbf{Solution:} To show that $ \sup(c+A)=c+\sup(A) $, we need to show that (1) $ (c+\sup A) $ is an upper bound; and (2) for any upper bound $ u $ of $ c+A $, we have $ u\ge c+\sup A $.

(1): For any element $ c+a\in c+A $, by the definition of $ \sup A $, we have $ \sup A\ge a $. By axiom (2)(a), for any $ a\in A $, we have  $ c+\sup A\ge c+a $(add $ c $ on both sides of the inequality $ \sup A\ge a $).

(2): Since $ u $ is an upper bound of $ c+A $, then for any $ a\in A $, $ u\ge c+a $, which implies $ (u+(-c)=)=u-c\ge a(=c+a+(-c)) .$ So, $ u-c $ is an upper bound of $ A $. But by the definition of $ \sup A $, $ u-c\ge\sup A $. Hence $ u\ge c+\sup A $. Because for any $ a\in A $ we have $ \sup A\ge a $ which implies $ u\ge c+\sup A\ge c+a $. We are done.

\qed

\noindent Remark: This is vacuously true if $ A=\emptyset $, since the infimum of the empty does not exist.

\newpage

\begin{center}
	\textbf{Workshop 1}
\end{center}

\vspace*{20pt}

\noindent Problems 1 and 2 are due tomorrow(Tuesday 9/14) at 10 pm. You must submit your work through Canvas.

\vspace*{20pt}

\noindent\textbf{Problem 1.}
Let $ A $ be a bounded set of reals and $ z $ a real number such that the following holds:
\[ (\dagger)\ \textbf{For every $ \varepsilon>0 $, there is an $ a\in A $ such that $ a<z+\varepsilon. $} \]

\noindent \underline{Hint}: (f) and $ (\dagger) $ are actually equivalent! But if you want to use this, you must prove it first! Otherwise, you have to give case-by-case reasons.



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