Finished git submodule test

assignments/main_text.tex

@@ -17,6 +17,4 @@
 \subfile{week2/main.tex}
 \clearpage
 \subfile{week3/main.tex}
-\clearpage
-\subfile{week4/main.tex}
 \end{document}

assignments/week4/main.tex

@@ -1,75 +0,0 @@
-\documentclass[../main_text.tex]{subfiles}
-\begin{document}
-\setcounter{exercise}{0}
-
-\part{Assignment 4}
-
-\exercise*
-\begin{tcolorbox}
-    We say a set $F \subset \R$ is \textit{closed} if its complement $F^c := \R \backslash F$ is open. Since
-    $\emptyset$ and $\R$ are open, it follows that $\emptyset$ and $\R$ are closed as well.
-
-    \begin{enumerate}[label=\emph{(\alph*)}]
-        \item Let $a,b \in \R$ with $a < b$. Prove that $[a,b]$ is closed.
-        \item Is the set $\Z \subset \R$ closed? Provide a proof to substantiate your claim.
-        \item Is the set of rationals $\Q \subset \R$ closed? Provide a proof to substantiate your claim.
-    \end{enumerate}
-\end{tcolorbox}
-
-\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 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 Non-formally speaking, in this exercise we want to prove that any intersection of closed sets in $\R$, is
-        closed itself. In order to make this formal, we will assume that $F_\lambda$ is closed for all $\lambda \in
-        \Lambda$ and follow the definition as presented.
-
-        So, let us assume that $F_\lambda$ is closed for all $\lambda \in \Lambda$. Following the definition of closed
-        sets, this means that the complement of $F_\lambda$ is open. Let's call the complement $\R \backslash F_\lambda =
-        U_\lambda$. Similarly, proving $\bigcap_{\lambda \in \Lambda} F_\lambda$ is closed, means we need to prove its
-        complement is open. From De Morgan's laws, it follows that $\R \backslash \bigcap_{\lambda \in \Lambda}
-        F_\lambda = \bigcup_{\lambda \in \Lambda} U_\lambda$.
-
-        We already proved this statement in exercise 5.b of week 3.
-    \item Similarly to the previous exercise, non-formally speaking we want to prove that any union of closed sets in
-        $\R$ is closed itself. Again, we will assign the complement of $(F_n)^c = U_n$, since we can't directly prove
-        a set is closed; we can only use the definition of closedness by proving something on open sets.
-
-        In the same vein, in order to prove that $\bigcup_{m=1}^n F_m$ is closed, we can only try and prove that its
-        complement is open. Using De Morgan's laws, $(\bigcup_{m=1}^n F_m)^c = \bigcap_{m=1}^n U_m$. This is the same
-        exercise as exercise 5.c from week 3, which also already proves the theorem.
-\end{enumerate}
-
-\end{document}