assignments/main_text.tex

@@ -1,6 +1,5 @@
 \documentclass{template/homework}
 
-\setcounter{secnumdepth}{0} % Makes sections unnumbered
 \usepackage{enumitem} % Gives access to better enumeration items
 \usepackage{tcolorbox} % Gives boxes
 \usepackage{subfiles} % Makes subfiles easier (and you want those)
@@ -16,4 +15,6 @@
 \subfile{week1/main.tex}
 \clearpage
 \subfile{week2/main.tex}
+\clearpage
+\subfile{week3/main.tex}
 \end{document}

assignments/week1/main.tex

@@ -1,7 +1,8 @@
 \documentclass[../main_text.tex]{subfiles}
 \begin{document}
+\setcounter{exercise}{0}
 
-\section{Week 1}
+\part{Assignment 1}
 
 \exercise*[0.3.6]
 % For some reason I can't put a fitted tcbox here and I really don't like it
@@ -132,7 +133,7 @@ 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*[6]
+\exercise*
 \begin{tcolorbox}
     \begin{enumerate}[label=\emph{\alph*)}, wide]
         \item Compute $f(4/15)$. Find $q$ such that $f(q) = 108$.

assignments/week2/main.tex

@@ -1,7 +1,8 @@
 \documentclass[../main_text.tex]{subfiles}
 \begin{document}
+\setcounter{exercise}{0}
 
-\section{Week 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*}

assignments/week3/main.tex

@@ -0,0 +1,234 @@
+\documentclass[../main_text.tex]{subfiles}
+\begin{document}
+\setcounter{exercise}{0}
+
+\part{Assignment 3}
+
+\exercise*
+\begin{tcolorbox}
+    Suppose $x,y \in \R$ and $x < y$. Prove that there exists $i \in \R\backslash\Q$ such that $x < i < y$.
+\end{tcolorbox}
+
+If either $x$ or $y$ (or both) are not rational numbers, we can simply take the average like so: $\frac{x+y}{2}$,
+in a similar way we did for the rationals. Since $x$ or $y$ isn't rational, the resulting fraction will also not be
+a rational and this proves the statement.
+
+Now if $x,y \in \Q$, we cannot use this average trick, because the resulting fraction will be a rational itself and
+so it doesn't satisfy the restriction that it must be in $\R\backslash\Q$. So we have to take a different approach.
+
+Let $x,y \in \Q$ with $x < y$ and $m := \frac{x+y}{2}$, so $x < m < y$. Then, let $X = \{a \in \R : x < a < m\}$ and
+let $Y = \{b \in \R : m < b < y\}$. Since $x < m$ and $m < y$, these are nonempty and they are bounded, because
+of the restrictions $x < a < m$ and $m < b < y$. So, there exists $k \in X$, that is not rational such that
+$x < k < m$ and there exists $h \in Y$ that is not rational such that $m < h < y$. Pick either $k$ or $h$ as $i$,
+since $x < k < m < h < y$. \qed
+
+\exercise*
+\begin{tcolorbox}
+    Let $E \subset (0,1)$ be the set of all real numbers with decimal representation using only the digits 1 and 2:
+    \begin{equation*}
+        E := \{x \in (0,1) : \forall j \in \N, \exists d_j \in \{1,2\} \text{ such that } x = 0.d_1d_2...\}
+    \end{equation*}
+    Prove that $|E| = |\mathcal{P}(\N)|$.
+\end{tcolorbox}
+
+As a hint to this exercise: \textit{Consider the function $f : E \rightarrow \mathcal{P}(\N)$ such that if $x \in E,
+x = 0.d_1d_2...$,
+\begin{equation*}
+    f(x) = \{j \in \N : d_j = 2\}.
+\end{equation*}}
+
+In order to prove that 2 sets are of equal cardinality, we need to prove that there is a bijective function between
+the 2 sets. In this case, the aforementioned hint function does the trick. Non-formally speaking, it is exactly what
+we are looking for: it is a (weird) representation of the power set of natural numbers, in that for every decimal,
+represented by a natural number, it is decided if that decimal is a 2 or a 1. This is similar to the actual power set
+of the natural numbers, in which for every natural number it is decided whether the number is in a subset or not.
+
+Now for a formal proof. To show that $f$ is bijective, we need to show that it is surjective and injective.
+\begin{description}
+    \item[Injectivity of $f$] In order to show that $f$ is injective, we have to show that for every $x \in E$, there
+        is a unique $y \in \mathcal{P}(\N)$ for the function, by showing that $f(a) = f(b) \implies a = b$.
+
+        So, let's assume that for some $a,b \in E$, $f(a) = f(b)$. So, there two sets of natural numbers
+        $\{a_1,a_2,...,a_n\} = \{b_1,b_2,...,b_m\}$. Equality in sets means that every element that is present in the
+        one set, is present in the other, and vice versa. No element that is present in either set, is missing in the
+        other. So, in this case, both sets will represent the same sequence of digits that are 2. Because the only
+        other option for digits is 1, that means the complete digital representation of $a$ and $b$ are known,
+        unique and the same. This concludes the proof for injectivity.
+    \item[Surjectivity of $f$] To prove surjectivity, we need to prove that for any arbitrary $y \in \mathcal{P}(\N)$,
+        there exists a corresponding $x \in E$ such that $f(x) = y$.
+
+        So, take an arbitrary $y = \{y_1, y_2, ... , y_n\}$, where each $y_i \in \N$ and thus $y \in \mathcal{P}(\N)$.
+        Then, corresponding $x \in E$ can be constructed easily as follows. Take a decimal number $0.d_1d_2...$ and 
+        turn every decimal $d_i$ for which $i \in y$ into a 2, and every other decimal into a 1. Since every decimal
+        can only be a 1 or 2, this handles every decimal correctly. Also, $f(x)$ will be in $\mathcal{P}(\N)$.
+\end{description}
+
+Since, $f$ is 1-to-1 and onto, $f$ is bijective. Then, because there exists a bijective function from
+$E$ to $\mathcal{P}(\N)$, $|E| = |\mathcal{P}(\N)|$. \qed
+
+\exercise*
+\begin{tcolorbox}
+    \begin{enumerate}[label=\emph{(\alph*)}]
+        \item Let $A$ and $B$ be two disjoint, countably infinite sets. Prove that $A \cup B$ is countably infinite.
+        \item Prove that the set of irrational numbers, $\R\backslash\Q$, is uncountable. You may use the facts
+            discussed in the lectures that $\R\backslash\Q$ is infinite and $\R$ is uncountable without proof.
+    \end{enumerate}
+\end{tcolorbox}
+\begin{enumerate}[label=\emph{(\alph*)}]
+    \item So let $A$ and $B$ be two disjoint, countably infinite sets. Since these sets are countably infinite,
+        a bijective function to $\N$ exists for both functions separately. It is then straightforward to map both these
+        function together to $\Z$ instead, in the following way. Let $f$ be the bijective function such that
+        $f : A \rightarrow \N$ and let $g$ be the bijective function such that $g : B \rightarrow \N$. Then, we can
+        define a new function $h : A \cup B \rightarrow \Z$ as
+        \begin{align*}
+            h(x) &= f(x) \text{ if } x \in A \\
+                 &= -g(x) \text{ if } x\in B.
+        \end{align*}
+        Since $A \cap B = \emptyset$, this function is unambiguously defined. Since $\Z$ is countably infinite,
+        $A \cup B$ is countably infinite as well. \qed
+
+    \item Because of part (a), we know that if we have two disjoint, countably infinite sets and join them, the result
+        is still countably infinite. The opposite must then also be true: if we have a countably infinite set and
+        we divide it into two disjoint subsets, both of which are infinite, then they still must be countable.
+
+        So then, for $\R\backslash\Q$, we know that $\R$ is uncountably infinite. So when we split it into rational
+        and irrational subsets, from which we know that $\Q$ is countably infinite, $\R\backslash\Q$ must be at least
+        and at most uncountably infinite. \qed
+\end{enumerate}
+
+\exercise*
+\begin{tcolorbox}
+    Let $A$ be a subset of $\R$ which is bounded above, and let $a_0$ be an upper bound for $A$. Prove that $a_0 =
+    \sup A$ if and only if for every $\epsilon > 0$, there exists $a \in A$ such that $a_0 - \epsilon < a$.
+\end{tcolorbox}
+
+Let $A \subset \R$, with $A$ bounded above by $a_0$.
+So, we have to prove the implication both ways. First, let's prove that the implication to the right $(\rightarrow)$.
+
+Assume that $a_0 = \sup A$, so for all $a \in A$, $a \leq a_0$. Also, let $\epsilon > 0$. If $a_0 \in A$,
+then we pick $a_0$ as $a$ and get $a_0 - \epsilon < a_0$, which holds $\forall \epsilon > 0$. If $a_0 \notin A$,
+then we choose $a$ as the average of $a_0$ and $a_0 - \epsilon$, which is definitely smaller than $a_0$. We are
+allowed to pick this as $a$, because we assume without loss of generality that $a \geq \inf A$. Then we get
+\begin{align*}
+    a_0 - \epsilon &< \frac{a_0 + a_0 - \epsilon}{2} \\
+                   &< a_0 - \frac{\epsilon}{2} \implies \\
+         -\epsilon &< -\frac{\epsilon}{2}.
+\end{align*}
+Since $\epsilon > 0$, this always holds.
+
+Now for the implication to the left $(\leftarrow)$.
+
+Assume now that the right side is true, i.e. let's assume that $\forall \epsilon > 0$, there exists $a \in A$ such that
+$a_0 - \epsilon < a$. Again, let us first investigate the case where $a_0 \in A$. Well certainly still, if $a_0$ is 
+an upper bound for $A$ and it is also part of the set itself, it must be the supremum\footnote{Proven in
+earlier exercise.}. 
+
+Then, let's assume that $a_0 \notin A$. Now, for all positive $\epsilon$, we know there exists an $a \in A$ 
+such that $a \neq a_0$ and $a_0 - \epsilon < a$. Let us assume then that this implies that $a_0 \neq \sup A$ and
+try to come to a contradiction. So, then there must be some $b = \sup A$, which has as consequence that
+$a < b < a_0$, since $b$ is still an uppoer bound of $A$ (and $a_0 \notin A$). Then, since $b > a$, we can pick
+$a = b - \epsilon < b$. So, from our initial assumption we get $b - \epsilon < a_0 - \epsilon < b - \epsilon \implies
+b < a_0 < b$, which is a false statement. So, $a_0 = \sup A$.
+
+Since the implication holds both ways, the equivalence is proven. \qed
+
+\exercise*
+\begin{tcolorbox}
+    \begin{enumerate}[label=\emph{(\alph*)}]
+        \item Let $a,b \in \R$ with $a < b$. Prove that the sets $(-\infty, a), (a,b)$ and $(b, \infty)$ are open.
+        \item Let $\Lambda$ be a set (not necessarily a subset of $\R$), and for each $\lambda \in \Lambda$,
+            let $U_\lambda \subset \R$. Prove that if $U_\lambda$ is open for all $\lambda \in \Lambda$ then the set
+            \begin{equation*}
+                \bigcup_{\lambda \in \Lambda} U_\lambda = 
+                \{x \in \R : \exists \lambda \in \Lambda \text{ such that } x \in U_\lambda\}
+            \end{equation*}
+            is open.
+        \item Let $n \in \N$, and let $U_1,...,U_n \subset \R$. Prove that if $U_1,...,U_n$ are open then the set
+            \begin{equation*}
+                \bigcap_{m=1}^n U_m = \{x \in \R : x \in U_m \text{ for all } m = 1,...,n\}
+            \end{equation*}
+            is open.
+        \item Is the set of rationals $\Q \subset \R$ open? Provide a proof to substantiate your claim.
+    \end{enumerate}
+\end{tcolorbox}
+
+\begin{enumerate}[label=\emph{(\alph*)}]
+    \item Since $\R$ is open, it is clear that $(-\infty, a)$ and $(b, \infty)$ are open to the left and right
+        respectively as well. Also, their respective right and left side are present in $(a,b)$ as well, so we will
+        only prove it for this case. The other cases follow logically.
+
+        Let $a,b \in \R$ such that $a < b$. We want to show that for all $x \in (a,b)$ there exists $\epsilon > 0$
+        such that $(x - \epsilon, x + \epsilon) \subset (a,b)$. Since for all $y \in (a,b)$ such that 
+        $x - \epsilon < y < x + \epsilon$ this statement will recursively hold, we only need to prove that there 
+        exists $\epsilon > 0$ such that $a < x - \epsilon$ and $x + \epsilon < b$. Then we can pick a fitting
+        $\epsilon$ in the following way, depending if $x$ is closer to $a$ or to $b$, formalized as follows.
+
+        If $x - a = b - x \implies 2x = b + a \implies x = \frac{b+a}{2}$, then $x$ is precisely between $a$ and $b$
+        and we can pick $\epsilon$ to be $\frac{b - a}{4}$. Then $x + \epsilon < b$ and $x - \epsilon > a$.
+
+        When $x - a < b - x$, then $x$ will be closer to $a$ then to be and $\epsilon$ is bounded more by $x$'s
+        proximity to $a$ than to $b$, i.e. $\epsilon < x - a$. So we can pick $\epsilon = \frac{x-a}{2} < x - a$.
+        Then $x - \epsilon = x - \frac{x - a}{2} = \frac{x + a}{2}$. Since $x > a$, $\frac{x + a}{2} > a$. For the
+        other side, $x + \epsilon = x + \frac{x - a}{2} < x + \frac{b - x}{2} = \frac{x + b}{2} < b$ since $x < b$.
+        So for both sides, we have shown that there exists an $\epsilon$ such that both
+        $x - \epsilon, x + \epsilon \in (a,b)$. All elements inbetween $x - \epsilon$ and $x + \epsilon$ will also
+        definitely be in $(a,b)$.
+        
+        The argument when $b - x < x - a$ is very similar and will be left to the reader. Then, for all $x \in (a,b)$,
+        the statement is proven. \qed
+    \item Non-formally speaking, in this exercise we want to prove that any union of open sets in $\R$ is open itself.
+        In order to make this formal, we will assume that $U_\lambda$ is open for all $\lambda \in \Lambda$ and
+        follow the definition as presented.
+
+        So, let us assume that $U_\lambda$ is open for all $\lambda \in \Lambda$. This means that for every $x_\lambda
+        \in U_\lambda$, there exists an $\epsilon > 0$ such that $(x_\lambda - \epsilon, x_\lambda + \epsilon) \subset
+        U_\lambda$. To prove that $\bigcup_{\lambda \in \Lambda} U_\lambda$ is open, we need to show that the same
+        property holds for all $y$ in this set. But since the union between some sets is defined as the set that holds
+        all the elements that any of these sets hold, this is trivial: for any $y \in \bigcup_{\lambda \in \Lambda}$
+        for which we want to know what $\epsilon$ we need to show that the union is open around that $y$, we just pick
+        the corresponding $\epsilon$ for the subset $U_\lambda$ which was open. Since all elements in that $U_\lambda$
+        are also in the union, this must certainly be the case. \qed
+    \item Similarly to the previous exercise, non-formally speaking we want to prove that any intersection of open sets
+        in $\R$ is open itself. This is not as trivial as in the previeous exercise however: since every set that is
+        added as an intersection poses another restriction, we don't have the immediate guarantee that every $\epsilon$
+        from the subsets will also be a well-defined element for the intersection set.
+
+        Now formally. Let $n \in \N$ and let $U_1,...,U_n \subset \R$. We assume that all $U_1,...,U_n$ are open.
+        Then we will prove that $\bigcap_{m = 1}^n U_m$ is open by induction over $n$.
+
+        For the base case, let $n = 1$. Then the intersection set is equivalent to $U_1$. Since $U_1$ is open, then
+        so is the intersection set.
+
+        For the inductive step, we assume that the intersection set is open for a certain $n = h$, i.e. there exists
+        an $\epsilon > 0$ such that $(x - \epsilon, x + \epsilon)$ is open for every $x \in \bigcap_{m = 1}^h U_m$.
+        Now, we will add one additional open set to this intersection, $U_{h+1}$, such that $n$ become $h+1$. Note
+        that $h + 1 \in \N$. Let the new intersection set be denoted as $\bigcap'$, and the old one as $\bigcap$.
+        Then in order to find an $\epsilon > 0$ for every $x \in \bigcap'$ such that $(x - \epsilon, x + \epsilon)$,
+        we take the smallest of $\epsilon$'s for that $x$ compared between $\bigcap$ and $U_{h+1}$. Since
+        $x \in \bigcap'$, we know that $x \in \bigcap$ and $x \in U_{h+1}$. Then the smallest accomponying $\epsilon$
+        always gives a well-defined open set inside of $\bigcap'$ because $|(x + \epsilon_1) - (x - \epsilon_1)| < 
+        |(x + \epsilon_2) - (x - \epsilon_2)|$ if $\epsilon_1 < \epsilon_2$, and thus $\bigcap'$ is open itself. \qed
+    \item No, $\Q$ is not open in $\R$. This is because we can't find an $\epsilon > 0$ such that for every $q \in \Q$,
+        $(q - \epsilon, q + \epsilon) \subset \Q$. We know that $\Q$ is dense in $\R$, but as we have proven in
+        exercise 1, the converse is also true. For every real numbers, we can find a real number inbetween that is not
+        a rational number. So, we cannot pick an $\epsilon > 0$ such that there is an interval around $x$ that itself
+        is completely contained in $\Q$. For every $\epsilon$ we pick, we can always find a real number $r$ such that
+        $x < r < x + \epsilon$ and $x - \epsilon < r < x$. So, $\Q$ is not open. \qed
+\end{enumerate}
+
+\exercise*
+\begin{tcolorbox}
+    Prove that
+    \begin{equation*}
+        \lim_{n \rightarrow \infty} \frac{1}{20n^2 + 20n + 2020} = 0.
+    \end{equation*}
+\end{tcolorbox}
+In order to prove that this limit holds, we need to show that a function $\{x_n\}$ converges to $x$, 
+i.e. if for all $\epsilon > 0$, $\exists M \in \N$ such that $\forall n \geq M$ the following inequality holds:
+$|x_n = x| < \epsilon$.
+
+Let $\epsilon > 0$. We choose $M \in \N$ such that $\frac{1}{M} < \epsilon$ (Archimedean Property). Then for all
+$n \geq M$, $|\frac{1}{20n^2+20n+2020} - 0| = \frac{1}{20n^2 + 20n + 2020} \leq \frac{1}{n^2 + n} \leq
+\frac{1}{n} \leq \frac{1}{M} < \epsilon$. \qed
+
+\end{document}