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\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}

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