\documentclass[../main_text.tex]{subfiles}
\begin{document}

\section{Week 2}

\exercise*[1.1.1]
\begin{tcolorbox}
    Let $F$ be an ordered field and $x,y,z \in F$. If $x < 0$ and $y < z$, then $xy > xz$.
\end{tcolorbox}

So let's assume the premise. $F$ is an ordered field and $x,y,z \in F$,
and we choose $x,y$ and $z$ such that $x < 0$ and $y < z$.

From $x < 0$ it follows that $(-x) > 0$. From $y < z$ it follows that $0 < z - y$.
From both of these, we can conclude that $0 < (-x)(z-y)$.
Working out the right side with the distributive law, gives $0 < (-x*z)-(-x*y)$.
Using $-1*-1 = 1$, gives $0 < (-xz)-(-xy)$, thus $0 < xy - xz$. The right part can be split again: $xz < xy$.
Then, the < can be flipped, which gives $xy > xz$. \qed

\exercise[1.1.2]
\begin{tcolorbox}
    Let $S$ be an ordered set. Let $A \subset S$ be a non-empty finite subset. Then A is bounded. Furthermore,
    $\text{inf} A$ exists and in in $A$ and $\text{sup} A$ exists and is in $A$.
\end{tcolorbox}

In order to prove that $A$ is bounded, we have to prove that it has an upper and a lower bound. Let us prove
that $A$ is bounded above first. 

In particular, we have to prove that $\exists a \in A$ such that $x \leq b$ for all $x \in E$.
Since $A$ is non-empty and finite, we can use induction on the cardinality of $A$, since that will always
be some natural number $n$. So, we have to prove two cases: the base case, where $|A|=1$, and the inductive step,
where we will assume that when $A$ has an upper bound when it has cardinality $m$,
then it also has an upper bound when its carindality is equal to $m+1$.

\end{document}
