diff --git a/assignments/main_text.pdf b/assignments/main_text.pdf
index fee0e48..83d3b60 100644
Binary files a/assignments/main_text.pdf and b/assignments/main_text.pdf differ
diff --git a/assignments/week4/main.pdf b/assignments/week4/main.pdf
index ce63718..0deb718 100644
Binary files a/assignments/week4/main.pdf and b/assignments/week4/main.pdf differ
diff --git a/assignments/week4/main.tex b/assignments/week4/main.tex
index b63bd92..86d7f5d 100644
--- a/assignments/week4/main.tex
+++ b/assignments/week4/main.tex
@@ -16,11 +16,43 @@
     \end{enumerate}
 \end{tcolorbox}
 
-\begin{enumerate}[label=\emph{(\alph*)}
+\begin{enumerate}[label=\emph{(\alph*)}]
     \item The complement of $[a,b]$ is equal to the union of $(-\infty, a)$ and $(b, \infty)$. So, we have to prove
         that both these sets are open. But we have already done so in the assignment of the previous week, so I'll just
         leave it at that.
-    \item 
+    \item To prove that $\Z \subset \R$ is closed, we have to prove that the complement is open. Since the complement
+        consists of the union of a countably infinite number of open intervals $(a, b)$ such that $a < b$, we know
+        from a combination of earlier exercises that this is the case. This is because any interval $(a, b)$ is open
+        if $a,b \in \R$ such that $a < b$ and for any two open interval $A$ and $B$ that are open, then $A \cup B$
+        is open as well.
+    \item I claim that the set of rationals isn't closed in $\R$. This is because there doesn't exist any interval
+        $(a,b)$ where $a,b \in \Q$ such that $a < b$, since for any $a$ and $b$ you can always find a $c$ such that
+        $a < c < b$. This makes it impossible to find an $\epsilon > 0$ such that for any $x \in (a,b)$,
+        $(x - \epsilon, x + \epsilon)$ is also in $(a,b)$ but in such a way that it only contains irrational numbers.
+        This argument makes use of the fact that $\Q$ is dense in $\R$.
+\end{enumerate}
+
+\exercise*
+\begin{tcolorbox}
+    \begin{enumerate}[label=\emph{(\alph*)}]
+        \item Let $\Lambda$ be a set (not necessarily a subset of $\R$), and for each $\lambda \in \Lambda$, let
+            $F_\lambda \in \R$. Prove that if $F_\lambda$ is closed for all $\lambda \in \Lambda$ then the set
+            \begin{equation*}
+                \bigcap_{\lambda \in \Lambda} F_\lambda = \{x \in \R : x \in F_\lambda \text{ for all }
+                \lambda \in \Lambda\}
+            \end{equation*}
+            is closed.
+        \item Let $n \in \N$, and let $F_1,...,F_n \subset \R$. Prove that if $F_1,...,F_n$ are closed then the set
+            $\bigcup_{m=1}^n F_m$ is closed.
+    \end{enumerate}
+\end{tcolorbox}
+
+This exercise is very similar to an exercise of the previous assignment, in which we looked at unions and intersections
+of \textit{open} intervals, whereas in this exercise, it's all about closed intervals. As the definition of closed
+intervals is intricately linked to the definition of open intervals, the following arguments will look very similar and
+shouldn't be surprising.
+\begin{enumerate}[label=\emph{(\alph*)}]
+    \item
     \item
 \end{enumerate}