diff --git a/assignments/week3/main.pdf b/assignments/week3/main.pdf
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diff --git a/assignments/week3/main.tex b/assignments/week3/main.tex
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--- a/assignments/week3/main.tex
+++ b/assignments/week3/main.tex
@@ -120,9 +120,47 @@ Now for the implication to the left $(\leftarrow)$.
 
 Assume now that the right side is true, i.e. let's assume that $\forall \epsilon > 0$, there exists $a \in A$ such that
 $a_0 - \epsilon < a$. Again, let us first investigate the case where $a_0 \in A$. Well certainly still, if $a_0$ is 
-an upper bound for $A$ and it is also part of the set itself, it must be the supremum. Then, let's assume that
-$a_0 \notin A$. Now, for all positive $\epsilon$, we know there exists an $a \in A$ such that $a \neq a_0$ and
-$a_0 - \epsilon < a$. So, $a_0 < a + \epsilon$. $a_0$ is an upper bound and we can find $a$'s such that $a+\epsilon$
-is always bigger than $a_0$, so $a_0$ must be the supremum. If $a_0$ wasn't the supremum, then there must be some $b$
-such that $a < b < a_0 < a + \epsilon$. WIP
+an upper bound for $A$ and it is also part of the set itself, it must be the supremum\footnote{Proven in
+earlier exercise}. 
+
+Then, let's assume that $a_0 \notin A$. Now, for all positive $\epsilon$, we know there exists an $a \in A$ 
+such that $a \neq a_0$ and $a_0 - \epsilon < a$. Let us assume then that this implies that $a_0 \neq \sup A$ and
+try to come to a contradiction. So, then there must be some $b = \sup A$, which has as consequence that
+$a < b < a_0$, since $b$ is still an uppoer bound of $A$ (and $a_0 \notin A$). Then, since $b > a$, we can pick
+$a = b - \epsilon < b$. So, from our initial assumption we get $b - \epsilon < a_0 - \epsilon < b - \epsilon \implies
+b < a_0 < b$, which is a false statement. So, $a_0 = \sup A$.
+
+Since the implication holds both ways, the equivalence is proven. \qed
+
+\exercise*
+\begin{tcolorbox}
+    \begin{enumerate}[label=\emph{(\alph*)}]
+        \item Let $a,b \in \R$ with $a < b$. Prove that the sets $(-\infty, a), (a,b)$ and $(b, \infty)$ are open.
+        \item Let $A$ be a set (not necessarily a subset of $\R$), and for each $\lambda \in A$, let $U_\lambda \subset
+            \R$. Prove that if $U_\lambda$ is open for all $\lambda \in A$ then the set
+            \begin{equation*}
+                \bigcup_{\lambda \in A} U_\lambda = 
+                \{x \in \R : \exists \lambda \in A \text{ such that } x \in U_\lambda\}
+            \end{equation*}
+            is open.
+        \item Let $n \in \N$, and let $U_1,...,U_n \subset \R$. Prove that if $U_1,...,U_n$ are open then the set
+            \begin{equation*}
+                \bigcap_{m=1}^n U_m = \{x \in \R : x \in U_m \text{ for all } m = 1,...,n\}
+            \end{equation*}
+            is open.
+        \item Is the set of rationals $\Q \subset \R$ open? Provide a proof to substantiate your claim.
+    \end{enumerate}
+\end{tcolorbox}
+
+\begin{enumerate}[label=\emph{(\alph*)},wide]
+    \item Since $\R$ is open, it is clear that $(-\infty, a)$ and $(b, \infty)$ are open to the left and right
+        respectively as well. Also, their respective right and left side are present in $(a,b)$ as well, so we will
+        only prove it for this case. The other cases follow logically.
+
+
+    \item
+    \item
+    \item
+\end{enumerate}
+
 \end{document}