Another exercise done week 2

assignments/week2/main.tex

@@ -30,6 +30,25 @@ In particular, we have to prove that $\exists a \in A$ such that $x \leq b$ for
 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 $m+1$.
+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}
 
 \end{document}