Finished first exercise of assignment 3 and did some refactoring on numbering etc

assignments/week2/main.tex

@@ -1,7 +1,8 @@
 \documentclass[../main_text.tex]{subfiles}
 \begin{document}
+\setcounter{exercise}{0}
 
-\section{Assignment 2}
+\part{Assignment 2}
 
 \exercise*[1.1.1]
 \begin{tcolorbox}
@@ -17,7 +18,7 @@ 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]
+\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 $\sup A$ exists and is in $A$.
@@ -45,7 +46,7 @@ of $A$ in $A$ \footnote{It might even be that I have already proven that $A$ has
 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]
+\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 $b = \sup A$.
@@ -63,7 +64,7 @@ not an upper bound of $A$ and thus also not its supremum. If $c > b$, then $b$ i
 $c$ and therefore $c$ cannot be the supremum. The only option left is that $c=b$, and therefore $b = \sup A$.
 \qed
 
-\exercise[1.1.6]
+\exercise*[1.1.6]
 \begin{tcolorbox}
     Let $S$ be an ordered set. Let $A \subset S$ be nonempty and bounded above. Suppose $\sup A$ exists
     and $\sup A \notin A$. Show that $A$ contains a countably infinite subset.
@@ -79,7 +80,7 @@ $a < b$ and $x < a $ $\forall x \in A$. So $b$ is in fact not the least upper bo
 which is in contradiction with our assumption earlier. Ergo, $|X| \geq |\N|$. Since $X$ then is at least of the same
 cardinality as $\N$, it must also contain a countably infinite subset. \qed
 
-\exercise[1.2.7]
+\exercise*[1.2.7]
 \begin{tcolorbox}
     Prove the arithmetic-geometric mean inequality. That is, for two positive real numbers $x, y$, we have
     \begin{equation*}
@@ -103,7 +104,7 @@ Now to prove the second statement. We assume $x = y$ is a positive real number.
     \sqrt{xy}=\sqrt{x^2}=x=\frac{2x}{2}=\frac{x + y}{2}. \qed
 \end{equation*}
 
-\exercise[1.2.9]
+\exercise*[1.2.9]
 \begin{tcolorbox}
     Let $A$ and $B$ be two nonempty bounded sets of real numbers. Define the set $C := \{a + b : a \in A,
     b \in B\}$. Show that $C$ is a bounded set and that
@@ -133,7 +134,7 @@ $y$ can be given by the supremum; $x \leq \sup A$ and $y \leq \sup B$. So, $\sup
 
 A similar argument can be given for the infimum, which is left to the reader.
 
-\exercise[7]
+\exercise*
 \begin{tcolorbox}
     Let
     \begin{equation*}