Artifacts opslaan

This reverts commit d2b0965281.

.gitmodules

@@ -1,4 +1,4 @@
 [submodule "latex-homework"]
-	path = assignments/template
+	path = src/template
 	url = https://github.com/gijs-pennings/latex-homework.git
 	branch = master

Jenkinsfile

@@ -0,0 +1,27 @@
+pipeline {
+    agent any
+    stages {
+        stage('Build') {
+            steps {
+                echo 'Starting build step...'
+                sh 'latexmk -pdf -outdir=out main_text.tex'
+
+                archiveArtifacts artifacts: 'out/main_text.pdf'
+            }
+        }
+
+        // Testing latex isn't really a thing, but we could do basic sanity checks in the future?
+
+        stage('Deploy') {
+            steps{
+                echo 'Starting deploy step...'
+                sh 'cp out/main_text.pdf /var/www/zwietering.eu/pdfs/real_analysis.pdf'
+            }
+        }
+    }
+    post {
+        cleanup {
+            cleanWs()
+        }
+    }
+}

README.md

@@ -5,3 +5,5 @@ Exercises and maybe (just maybe, like, very probably not) lecture notes on the R
 [Course link](https://ocw.mit.edu/courses/18-100a-real-analysis-fall-2020/)
 
 The text book is included in the directory root. The `.tex` files can be compiled individually in the corresponding subdirectories or collectively from the assignments directory. I use `pdflatex` with texlive on Arch (through WSL2, [check this guide](https://gist.github.com/ld100/3376435a4bb62ca0906b0cff9de4f94b) on how to do this (why use Ubuntu if you're only going to use the CLI right?)).
+
+The complete generated PDF file can be downloaded from [this address](https://zwietering.eu/pdfs/real_analysis.pdf).

assignments/week2/main.tex

@@ -1,9 +0,0 @@
-\documentclass[../main_text.tex]{subfiles}
-\begin{document}
-
-\section{Week 2}
-
-\exercise*[1.1.1]
-Exercise 1.1.1 comes here.
-
-\end{document}

main_text.tex

@@ -1,19 +1,20 @@
-\documentclass{template/homework}
+\documentclass{src/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)
 
 \title{MIT OCW Real Analysis}
 \author{Philippe H. Zwietering}
-\date{}
+\date{\today}
 
 \begin{document}
 \maketitle
 \tableofcontents
 \clearpage
-\subfile{week1/main.tex}
+\subfile{./src/week1.tex}
 \clearpage
-\subfile{week2/main.tex}
+\subfile{./src/week2.tex}
+\clearpage
+\subfile{./src/week3.tex}
 \end{document}

src/week1.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

src/week2.tex

@@ -0,0 +1,191 @@
+\documentclass[../main_text.tex]{subfiles}
+\begin{document}
+\setcounter{exercise}{0}
+
+\part{Assignment 2}
+
+\exercise*[1.1.1]
+\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*
+\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}

src/week3.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}