Week 4 exercise 2 done by referring to earlier exercises

Jenkinsfile

@@ -0,0 +1,20 @@
+pipeline {
+    agent {label 'linux'}
+    stages {
+        stage('Build') {
+            steps {
+                echo 'Starting build step...'
+                sh './scripts/build.sh'
+            }
+        }
+
+        // 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 assignments/main_text.pdf /var/www/zwietering.eu/pdfs/real_analysis.pdf'
+            }
+        }
+    }
+}

README.md

@@ -4,4 +4,6 @@ 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 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/main_text.tex

@@ -17,4 +17,6 @@
 \subfile{week2/main.tex}
 \clearpage
 \subfile{week3/main.tex}
+\clearpage
+\subfile{week4/main.tex}
 \end{document}

assignments/week4/main.tex

@@ -0,0 +1,75 @@
+\documentclass[../main_text.tex]{subfiles}
+\begin{document}
+\setcounter{exercise}{0}
+
+\part{Assignment 4}
+
+\exercise*
+\begin{tcolorbox}
+    We say a set $F \subset \R$ is \textit{closed} if its complement $F^c := \R \backslash F$ is open. Since
+    $\emptyset$ and $\R$ are open, it follows that $\emptyset$ and $\R$ are closed as well.
+
+    \begin{enumerate}[label=\emph{(\alph*)}]
+        \item Let $a,b \in \R$ with $a < b$. Prove that $[a,b]$ is closed.
+        \item Is the set $\Z \subset \R$ closed? Provide a proof to substantiate your claim.
+        \item Is the set of rationals $\Q \subset \R$ closed? Provide a proof to substantiate your claim.
+    \end{enumerate}
+\end{tcolorbox}
+
+\begin{enumerate}[label=\emph{(\alph*)}]
+    \item The complement of $[a,b]$ is equal to the union of $(-\infty, a)$ and $(b, \infty)$. So, we have to prove
+        that both these sets are open. But we have already done so in the assignment of the previous week, so I'll just
+        leave it at that.
+    \item To prove that $\Z \subset \R$ is closed, we have to prove that the complement is open. Since the complement
+        consists of the union of a countably infinite number of open intervals $(a, b)$ such that $a < b$, we know
+        from a combination of earlier exercises that this is the case. This is because any interval $(a, b)$ is open
+        if $a,b \in \R$ such that $a < b$ and for any two open interval $A$ and $B$ that are open, then $A \cup B$
+        is open as well.
+    \item I claim that the set of rationals isn't closed in $\R$. This is because there doesn't exist any interval
+        $(a,b)$ where $a,b \in \Q$ such that $a < b$, since for any $a$ and $b$ you can always find a $c$ such that
+        $a < c < b$. This makes it impossible to find an $\epsilon > 0$ such that for any $x \in (a,b)$,
+        $(x - \epsilon, x + \epsilon)$ is also in $(a,b)$ but in such a way that it only contains irrational numbers.
+        This argument makes use of the fact that $\Q$ is dense in $\R$.
+\end{enumerate}
+
+\exercise*
+\begin{tcolorbox}
+    \begin{enumerate}[label=\emph{(\alph*)}]
+        \item Let $\Lambda$ be a set (not necessarily a subset of $\R$), and for each $\lambda \in \Lambda$, let
+            $F_\lambda \in \R$. Prove that if $F_\lambda$ is closed for all $\lambda \in \Lambda$ then the set
+            \begin{equation*}
+                \bigcap_{\lambda \in \Lambda} F_\lambda = \{x \in \R : x \in F_\lambda \text{ for all }
+                \lambda \in \Lambda\}
+            \end{equation*}
+            is closed.
+        \item Let $n \in \N$, and let $F_1,...,F_n \subset \R$. Prove that if $F_1,...,F_n$ are closed then the set
+            $\bigcup_{m=1}^n F_m$ is closed.
+    \end{enumerate}
+\end{tcolorbox}
+
+This exercise is very similar to an exercise of the previous assignment, in which we looked at unions and intersections
+of \textit{open} intervals, whereas in this exercise, it's all about closed intervals. As the definition of closed
+intervals is intricately linked to the definition of open intervals, the following arguments will look very similar and
+shouldn't be surprising.
+\begin{enumerate}[label=\emph{(\alph*)}]
+    \item Non-formally speaking, in this exercise we want to prove that any intersection of closed sets in $\R$, is
+        closed itself. In order to make this formal, we will assume that $F_\lambda$ is closed for all $\lambda \in
+        \Lambda$ and follow the definition as presented.
+
+        So, let us assume that $F_\lambda$ is closed for all $\lambda \in \Lambda$. Following the definition of closed
+        sets, this means that the complement of $F_\lambda$ is open. Let's call the complement $\R \backslash F_\lambda =
+        U_\lambda$. Similarly, proving $\bigcap_{\lambda \in \Lambda} F_\lambda$ is closed, means we need to prove its
+        complement is open. From De Morgan's laws, it follows that $\R \backslash \bigcap_{\lambda \in \Lambda}
+        F_\lambda = \bigcup_{\lambda \in \Lambda} U_\lambda$.
+
+        We already proved this statement in exercise 5.b of week 3.
+    \item Similarly to the previous exercise, non-formally speaking we want to prove that any union of closed sets in
+        $\R$ is closed itself. Again, we will assign the complement of $(F_n)^c = U_n$, since we can't directly prove
+        a set is closed; we can only use the definition of closedness by proving something on open sets.
+
+        In the same vein, in order to prove that $\bigcup_{m=1}^n F_m$ is closed, we can only try and prove that its
+        complement is open. Using De Morgan's laws, $(\bigcup_{m=1}^n F_m)^c = \bigcap_{m=1}^n U_m$. This is the same
+        exercise as exercise 5.c from week 3, which also already proves the theorem.
+\end{enumerate}
+
+\end{document}

scripts/build.sh

@@ -0,0 +1,33 @@
+#!/bin/bash
+
+# Build all directories
+subdirectory_file_name=main
+
+cd assignments
+
+for D in *; do
+    if [ "${D}" != "template" ] && [ -d "${D}" ]; then
+        echo "Building ${D}..."
+        cd "${D}"
+        pdflatex -interaction=batchmode -halt-on-error "${subdirectory_file_name}.tex"
+        pdflatex -interaction=batchmode -halt-on-error "${subdirectory_file_name}.tex"
+        cd ..
+    fi
+done
+
+# Build main PDF
+main_file_name=main_text
+
+echo "Building main PDF..."
+pdflatex -interaction=batchmode -halt-on-error "${main_file_name}.tex"
+pdflatex -interaction=batchmode -halt-on-error "${main_file_name}.tex"
+
+# Clean up
+rm -rf *.aux *.log *.out *.toc
+for D in *; do
+    if [ "${D}" != "template" ] && [ -d "${D}" ]; then
+        cd "${D}"
+        rm -rf *.aux *.log *.out *.toc
+        cd ..
+    fi
+done