diff --git a/assignments/week2/main.pdf b/assignments/week2/main.pdf
index 6fe7864..7c63797 100644
Binary files a/assignments/week2/main.pdf and b/assignments/week2/main.pdf differ
diff --git a/assignments/week2/main.tex b/assignments/week2/main.tex
index c4a2bd2..3dc243c 100644
--- a/assignments/week2/main.tex
+++ b/assignments/week2/main.tex
@@ -5,8 +5,6 @@
 
 \exercise*[1.1.1]
 \begin{tcolorbox}
-    Prove:
-
     Let $F$ be an ordered field and $x,y,z \in F$. If $x < 0$ and $y < z$, then $xy > xz$.
 \end{tcolorbox}
 
@@ -19,4 +17,19 @@ 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}