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assignment
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2
.gitmodules
vendored
2
.gitmodules
vendored
@@ -1,4 +1,4 @@
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[submodule "latex-homework"]
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path = src/template
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path = assignments/template
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url = https://github.com/gijs-pennings/latex-homework.git
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branch = master
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@@ -1,2 +0,0 @@
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FROM kjarosh/latex:2025.1
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RUN tlmgr install libertinus
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5
Jenkinsfile
vendored
5
Jenkinsfile
vendored
@@ -1,11 +1,10 @@
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pipeline {
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agent any
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agent {label 'linux'}
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stages {
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stage('Build') {
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agent { dockerfile true }
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steps {
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echo 'Starting build step...'
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sh 'latexmk -pdf -outdir=out main_text.tex'
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sh './scripts/build.sh'
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}
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}
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@@ -1,4 +1,4 @@
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\documentclass{src/template/homework}
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\documentclass{template/homework}
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\usepackage{enumitem} % Gives access to better enumeration items
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\usepackage{tcolorbox} % Gives boxes
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@@ -6,15 +6,17 @@
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\title{MIT OCW Real Analysis}
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\author{Philippe H. Zwietering}
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\date{\today}
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\date{}
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\begin{document}
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\maketitle
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\tableofcontents
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\clearpage
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\subfile{src/week1.tex}
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\subfile{week1/main.tex}
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\clearpage
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\subfile{src/week2.tex}
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\subfile{week2/main.tex}
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\clearpage
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\subfile{src/week3.tex}
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\subfile{week3/main.tex}
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\clearpage
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\subfile{week4/main.tex}
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\end{document}
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BIN
assignments/week4/main.pdf
Normal file
BIN
assignments/week4/main.pdf
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Binary file not shown.
75
assignments/week4/main.tex
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75
assignments/week4/main.tex
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@@ -0,0 +1,75 @@
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\documentclass[../main_text.tex]{subfiles}
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\begin{document}
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\setcounter{exercise}{0}
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\part{Assignment 4}
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\exercise*
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\begin{tcolorbox}
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We say a set $F \subset \R$ is \textit{closed} if its complement $F^c := \R \backslash F$ is open. Since
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$\emptyset$ and $\R$ are open, it follows that $\emptyset$ and $\R$ are closed as well.
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\begin{enumerate}[label=\emph{(\alph*)}]
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\item Let $a,b \in \R$ with $a < b$. Prove that $[a,b]$ is closed.
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\item Is the set $\Z \subset \R$ closed? Provide a proof to substantiate your claim.
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\item Is the set of rationals $\Q \subset \R$ closed? Provide a proof to substantiate your claim.
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\end{enumerate}
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\end{tcolorbox}
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\begin{enumerate}[label=\emph{(\alph*)}]
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\item The complement of $[a,b]$ is equal to the union of $(-\infty, a)$ and $(b, \infty)$. So, we have to prove
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that both these sets are open. But we have already done so in the assignment of the previous week, so I'll just
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leave it at that.
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\item To prove that $\Z \subset \R$ is closed, we have to prove that the complement is open. Since the complement
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consists of the union of a countably infinite number of open intervals $(a, b)$ such that $a < b$, we know
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from a combination of earlier exercises that this is the case. This is because any interval $(a, b)$ is open
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if $a,b \in \R$ such that $a < b$ and for any two open interval $A$ and $B$ that are open, then $A \cup B$
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is open as well.
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\item I claim that the set of rationals isn't closed in $\R$. This is because there doesn't exist any interval
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$(a,b)$ where $a,b \in \Q$ such that $a < b$, since for any $a$ and $b$ you can always find a $c$ such that
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$a < c < b$. This makes it impossible to find an $\epsilon > 0$ such that for any $x \in (a,b)$,
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$(x - \epsilon, x + \epsilon)$ is also in $(a,b)$ but in such a way that it only contains irrational numbers.
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This argument makes use of the fact that $\Q$ is dense in $\R$.
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\end{enumerate}
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\exercise*
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\begin{tcolorbox}
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\begin{enumerate}[label=\emph{(\alph*)}]
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\item Let $\Lambda$ be a set (not necessarily a subset of $\R$), and for each $\lambda \in \Lambda$, let
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$F_\lambda \in \R$. Prove that if $F_\lambda$ is closed for all $\lambda \in \Lambda$ then the set
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\begin{equation*}
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\bigcap_{\lambda \in \Lambda} F_\lambda = \{x \in \R : x \in F_\lambda \text{ for all }
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\lambda \in \Lambda\}
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\end{equation*}
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is closed.
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\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
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$\bigcup_{m=1}^n F_m$ is closed.
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\end{enumerate}
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\end{tcolorbox}
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This exercise is very similar to an exercise of the previous assignment, in which we looked at unions and intersections
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of \textit{open} intervals, whereas in this exercise, it's all about closed intervals. As the definition of closed
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intervals is intricately linked to the definition of open intervals, the following arguments will look very similar and
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shouldn't be surprising.
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\begin{enumerate}[label=\emph{(\alph*)}]
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\item Non-formally speaking, in this exercise we want to prove that any intersection of closed sets in $\R$, is
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closed itself. In order to make this formal, we will assume that $F_\lambda$ is closed for all $\lambda \in
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\Lambda$ and follow the definition as presented.
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So, let us assume that $F_\lambda$ is closed for all $\lambda \in \Lambda$. Following the definition of closed
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sets, this means that the complement of $F_\lambda$ is open. Let's call the complement $\R \backslash F_\lambda =
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U_\lambda$. Similarly, proving $\bigcap_{\lambda \in \Lambda} F_\lambda$ is closed, means we need to prove its
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complement is open. From De Morgan's laws, it follows that $\R \backslash \bigcap_{\lambda \in \Lambda}
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F_\lambda = \bigcup_{\lambda \in \Lambda} U_\lambda$.
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We already proved this statement in exercise 5.b of week 3.
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\item Similarly to the previous exercise, non-formally speaking we want to prove that any union of closed sets in
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$\R$ is closed itself. Again, we will assign the complement of $(F_n)^c = U_n$, since we can't directly prove
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a set is closed; we can only use the definition of closedness by proving something on open sets.
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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
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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
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exercise as exercise 5.c from week 3, which also already proves the theorem.
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\end{enumerate}
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\end{document}
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