Finished a few more exercies of week 2

assignments/week2/main.tex

@@ -20,7 +20,7 @@ Then, the < can be flipped, which gives $xy > xz$. \qed
 \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 $\text{sup} A$ exists and is in $A$.
+    $\text{inf} A$ exists and in in $A$ and $\sup A$ exists and is in $A$.
 \end{tcolorbox}
 
 In order to prove that $A$ is bounded, we have to prove that it has an upper and a lower bound. Let us prove
@@ -48,7 +48,88 @@ it must also be bounded.}. This will be left to the reader.
 \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 $b = \text{sup} A$.
+    Show that $b = \sup A$.
 \end{tcolorbox}
 
+So, let $S$ be an ordered set, with $A \subset S$ and $b \in A$ being an upper bound for $A$. Since $b$ is an upper
+bound, $a \leq b$ for all $a \in A$. Since $b \in A$ as well, we know that there is some element in $A$ which is
+the greatest element of them all, and all other elements are smaller. 
+
+Now let's assume that $b \neq \sup A$. Then either some other element of $A$ is the supremum,
+which would imply that $b$ is not larger than this element, which is a contradiction.
+The other possibility is that there is an element $c \in S\backslash A$ that is the
+supremum. Because $S$ is ordered, $c$ must either be greater than, smaller than or equal to $b$. If $c < b$, c is 
+not an upper bound of $A$ and thus also not its supremum. If $c > b$, then $b$ is an upper bound that is smaller than
+$c$ 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]
+\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.
+\end{tcolorbox}
+
+Let $S$ be an ordered set, with $A \subset S$ nonempty and bounded above. We assume that $b = \sup A$ exists and
+$b \notin A$ $(\implies b \in S\backslash A)$. We are asked to show this then implies that $\exists X \subset A$
+such that $|X| \geq |\N|$. We will prove this with a proof by contradiction.
+
+We assume that no such set $X$ exists, i.e. $|X| < |\N|$. So, $A$ also doesn't have to countably infinite anymore.
+Since $b \notin A$ and $A$ is ordered, finite and nonempty, there is a greatest element $a \in A$ such that
+$a < b$ and $x < a $ $\forall x \in A$. So $b$ is in fact not the least upper bound of $A$,
+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]
+\begin{tcolorbox}
+    Prove the arithmetic-geometric mean inequality. That is, for two positive real numbers $x, y$, we have
+    \begin{equation*}
+        \sqrt{xy} \leq \frac{x + y}{2}.
+    \end{equation*}
+    Furthermore, equality occurs if and only if $x = y$.
+\end{tcolorbox}
+
+Let us prove the first statement first. So we let $x,y \in \R$ such that $x,y > 0$. Then we will prove the statement
+by contradiction. Hence, we assume that
+\begin{equation*}
+    \sqrt{xy} > \frac{x + y}{2}.
+\end{equation*}
+
+We can multiply both sides with 2. This results in $2\sqrt{xy} > x + y$. We can pull the left part into the right,
+so we get $0 > x - 2\sqrt{xy} + y$. We can restructure the right side to $0 > (\sqrt{x} - \sqrt{y})^2$.
+We know that $0 \leq z^2$, $\forall z \in \R$, so this is a contradiction. \qed
+
+Now to prove the second statement. We assume $x = y$ is a positive real number. Then,
+\begin{equation*}
+    \sqrt{xy}=\sqrt{x^2}=x=\frac{2x}{2}=\frac{x + y}{2}. \qed
+\end{equation*}
+
+\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
+    \begin{align*}
+        \sup C &= \sup A+\sup B \; \text{and}\\
+        \inf C &= \inf A+\inf B.
+    \end{align*}
+\end{tcolorbox}
+
+First, let us show that $C$ is a bounded set. Since $A$ and $B$ are both subsets of $\R$, which is an ordered field, 
+all elements of $C$ must also be real numbers. Let $a$ be an upper bound for $A$ and $b$ be an upper bound for $B$.
+So $x \leq a \; \forall x \in A$ and $y \leq b \; \forall y \in B$. Since $C$ is defined as the sum of any element in
+$A$ with any element in $B$, an upper bound of $C$, $c$, can be found as $c \leq a + b$. A similar argument can be
+made for the lower bound of $C$, which makes $C$ bounded. \qed
+
+To prove that $\sup C = \sup A + \sup B$, we will show that $\sup C \geq \sup A + \sup B$ and
+$\sup C \leq \sup A + \sup B$. 
+
+Let $a = \sup A$ and $b = \sup B$. So $x \leq a$ for all $x \in A$ and $y \leq b$ for all $y \in B$. Then,
+$x + y \leq a + b$. Since $z \leq x + y$ for all $z \in C$ because of the definition of $C$,
+$z \leq a + b$. In other words, $\sup C \leq \sup A + \sup B$.
+
+Now to prove the other direction. Let $c = \sup C$. So $z \leq c$ for all $c \in C$. Since all elements in $C$ are
+the sum of an element $x \in A$ and $y \in B$, $x + y \leq c$ for all $x,y$. The least upper bound for these $x$ and
+$y$ can be given by the supremum; $x \leq \sup A$ and $y \leq \sup B$. So, $\sup A + \sup B \leq c \implies
+\sup A + \sup B \leq \sup C$, completing the equality. \qed
+
+A similar argument can be given for the infimum, which is left to the reader.
 \end{document}