diff --git a/assignments/main_text.pdf b/assignments/main_text.pdf index 8262e63..9048868 100644 Binary files a/assignments/main_text.pdf and b/assignments/main_text.pdf differ diff --git a/assignments/week1/main.pdf b/assignments/week1/main.pdf index fe5a43a..a8cb92f 100644 Binary files a/assignments/week1/main.pdf and b/assignments/week1/main.pdf differ diff --git a/assignments/week1/main.tex b/assignments/week1/main.tex index 43b0ce2..ef4ea6c 100644 --- a/assignments/week1/main.tex +++ b/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 $f(q) = 108$. diff --git a/assignments/week2/main.pdf b/assignments/week2/main.pdf index 2858bb7..0ddca3f 100644 Binary files a/assignments/week2/main.pdf and b/assignments/week2/main.pdf differ diff --git a/assignments/week2/main.tex b/assignments/week2/main.tex index 62a1eac..69ee920 100644 --- a/assignments/week2/main.tex +++ b/assignments/week2/main.tex @@ -4,6 +4,187 @@ \section{Week 2} \exercise*[1.1.1] -Exercise 1.1.1 comes here. +\begin{tcolorbox} + Let $F$ be an ordered field and $x,y,z \in F$. If $x < 0$ and $y < z$, then $xy > xz$. +\end{tcolorbox} +So let's assume the premise. $F$ is an ordered field and $x,y,z \in F$, +and we choose $x,y$ and $z$ such that $x < 0$ and $y < z$. + +From $x < 0$ it follows that $(-x) > 0$. From $y < z$ it follows that $0 < z - y$. +From both of these, we can conclude that $0 < (-x)(z-y)$. +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] +\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 $A$. +\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 +that $A$ is bounded above first. + +In particular, we have to prove that $\exists a \in A$ such that $x \leq b$ for all $x \in E$. +Since $A$ is non-empty and finite, we can use induction on the cardinality of $A$, since that will always +be some natural number $n$. So, we have to prove two cases: the base case, where $|A|=1$, and the inductive step, +where we will assume that when $A$ has an upper bound when it has cardinality $m$, +then it also has an upper bound when its cardinality is equal to $m+1$. + +The base case is quite simple; if $A = \{x\}$, then $x$ is the greatest element and $A$ has an upper bound. +Now for the inductive step. We assume that for some set $B \subset S$ with cardinality $m$, +$B$ is bounded above. Thus, there is some $b \in B$ such that $b$ is greater than all other +elements in $B$. Now, let's add a new element $h \in S$ to $B$, such that $h$ is distinct from all elements already in +$B$ and the cardinality of $B$ is now $m+1$. Then, since $S$ is well ordered, we can compare $h$ also to $b$. +Either $h$ is greater than this $b$, in which case $h$ is the new greatest element, or it is less than $b$, in which +case $b$ stays the greatest element of $B$. In both cases however, $B$ remains bounded above. \qed + +A similar argument can be made to prove the existence of the lower bound, the supremum of $A$ in $A$ and the infimum +of $A$ in $A$ \footnote{It might even be that I have already proven that $A$ has a supremum present in $A$. +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] +\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 $b = \sup A$. +\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. + +\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}