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\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 cardinality is equal to $m+1$.

The base case is quite simple; if $A = \{x\}$, then $x$ is the greatest element and $A$ has an upper bound.
Now for the inductive step. We assume that for some set $B \subset S$ with cardinality $m$,
$B$ is bounded above. Thus, there is some $b \in B$ such that $b$ is greater than all other
elements in $B$. Now, let's add a new element $h \in S$ to $B$, such that $h$ is distinct from all elements already in
$B$ and the cardinality of $B$ is now $m+1$. Then, since $S$ is well ordered, we can compare $h$ also to $b$.
Either $h$ is greater than this $b$, in which case $h$ is the new greatest element, or it is less than $b$, in which
case $b$ stays the greatest element of $B$. In both cases however, $B$ remains bounded above. \qed

A similar argument can be made to prove the existence of the lower bound, the supremum of $A$ in $A$ and the infimum
of $A$ in $A$ \footnote{It might even be that I have already proven that $A$ has a supremum present in $A$.
Then that's also good enough to show that $A$ is bounded, since in order for $A$ to have a supremum,
it must also be bounded.}. This will be left to the reader.

\exercise[1.1.5]
\begin{tcolorbox}
    Let $S$ be an ordered set. Let $A \subset S$ and suppose $b$ is an upper bound for $A$. Suppose $b \in A$.
    Show that $b = \text{sup} A$.
\end{tcolorbox}

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