assignments/week1/main.tex

@@ -132,7 +132,7 @@ So each set $A_i := \{i\}  \forall  i \in \N$. Since $\N$ is countably infinite,
 Each set is definitely finite, because they all contain just one element.
 Finally, the union of the collection of sets is equal to $\N$, which is not a finite set.
 
-\exercise*
+\exercise*[6]
 \begin{tcolorbox}
     \begin{enumerate}[label=\emph{\alph*)}, wide]
         \item Compute $f(4/15)$. Find $q$ such that $f(q) = 108$.

assignments/week2/main.tex

@@ -132,4 +132,59 @@ $y$ can be given by the supremum; $x \leq \sup A$ and $y \leq \sup B$. So, $\sup
 \sup A + \sup B \leq \sup C$, completing the equality. \qed
 
 A similar argument can be given for the infimum, which is left to the reader.
+
+\exercise*[7]
+\begin{tcolorbox}
+    Let
+    \begin{equation*}
+        E = \{ x \in \R : x > 0 \; \text{and} \; x^3 < 2\}.
+    \end{equation*}
+
+    \begin{enumerate}[label=\emph{\alph*)}]
+        \item Prove that $E$ is bounded above.
+        \item Let $r = \sup E$ (which exists by part a)). Prove that $r > 0$ and $r^3=2$.
+
+            \textit{Hint:} Adapt the proof used in Example 1.2.3.
+    \end{enumerate}
+\end{tcolorbox}
+
+So, let $E$ and $r$ be defined as in the exercise statement. Then:
+\begin{enumerate}[label=\emph{\alph*)}]
+    \item $x \leq 2$ is an upper bound for $E$, as $2 * 2 * 2 = 8$. So, $E$ is bounded above.
+    \item As $1 \in E$, $r \geq 1 > 0$, so the first part of the statement holds. In order to show that $r^3=2$,
+        we want to show that $r^3 \leq 2$ and $r^3 \geq 2$ hold.
+
+        First, let's show that $r^3 \geq 2$. We will take a similar approach as in Example 1.2.3 from the textbook.
+        So, take a positive number $s$ such that $s^3 < 2$. We wish to find an $h > 0$ such that $(s + h)^3 < 2$.
+        As $2 - s^3 > 0$, we have $\frac{2 - s^3}{3s^2 + 3s + 1} > 0$. Choose an $h \in \R$ such that $0 < h <
+        \frac{2 - s^3}{3s^2+3s+1}$. Furthermore, assume $h < 1$. Estimate,
+        
+        \begin{align*}
+            (s+h)^3 - s^3 &= h(3s^2 + 3sh + h^2) \\
+                          &< h(3s^2 + 3s + 1) \quad &\text{(since } h < 1) \\
+                          &< 2 - s^3 \quad &\text{(since } h < \frac{2 - s^3}{3s^2 + 3s + 1}).
+        \end{align*}
+
+        Therefore, $(s+h)^3 < 2$. Hence $s + h \in E$, but as $h > 0$, we have $s + h > s$. So $s < r = \sup E$.
+        As $s$ was an arbitrary positive number such that $s^3 < 2$, it follows that $r^3 \geq 2$.
+
+        Now take an arbitrary positive number $s$ such that $s^3 > 2$. We wish to find an $h > 0$ such that 
+        $(s-h)^3 > 2$ and $s - h$ is still positive. As $s^3 - 2 > 0$, we have that $\frac{s^3 - 2}{3s^2+1} > 0$.
+        Let $h := \frac{s^3-2}{3s^2+1}$, and check that $s - h = s - \frac{s^3-2}{3s^2+1} = 
+        \frac{2s^3+s+2}{3s^2+1} > 0$. Assume that $h < 1$. Estimate,
+
+        \begin{align*}
+            s^3 - (s-h)^3 &= h(3s^2 - 3sh + h^2) \\
+                          &< h(3s^2+h^2) \quad &\text{(since } s>0 \; \text{ and } h > 0) \\
+                          &< h(3s^2+1) \quad &\text{(since } h < 1) \\
+                          &= s^3 - 2 &\text{(because of the definition of } h).
+        \end{align*}
+
+        By subtracting $s^3$ from both sides and multiplying by -1, we find $(s-h)^3>2$. Therefore, $s-h \notin E$.
+        Moreover, if $x \geq s-h$, then $x^3 \geq (s-h)^3 > 2$ (as $x > 0$ and $s-h > 0$) and so $x \notin E$. Thus,
+        $s - h$ is an upper bound for $E$. However, $s-h < s$, or in other words, $s > r = \sup E$. 
+        Hence, $r^3 \leq 2$.
+
+        Together, $r^3 \geq 2$ and $r^3 \leq 2$ imply $r^3 = 3$. \qed
+\end{enumerate}
 \end{document}