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\begin{document}

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\textsc{Julius Elinson} \hfill \textsc{\today}\\[.1cm]
\textsc{\Large{Math 131 Notes}}\\[-.1cm] % Title
\HRule \\[.9cm]
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\textsc{Disclaimer: }\small{Please feel free to use the pdf notes and/or the provided source code. However, the accuracy of these notes is in no way guaranteed. Furthermore, please do not use these notes instead of going to class, except in extenuating circumstances. Honor code applies.
Feel free to send any and all feedback, suggestions, or corrections to \href{mailto:jelinson@g.hmc.edu}{jelinson@hmc.edu}. }\\
\begin{rec}
 $\cl{E}=E\cup E'$.
\end{rec}
\begin{thm}
 $E\subseteq X$. $\cl{E}$ is closed.
\end{thm}
\begin{proof}
 We will show $X\bslash \cl{E}$ is open. Let $x\in X\bslash \cl{E}$. We must show $x$ is in an interior point of $X\bslash \cl{E}$. Since $x\in X\bslash \cl{E}$, we know $x$ is not a limit point of $E$. Thus there exists an open neighborhood $U$ of $x$ such that $U$ is completely contained in $X\bslash E$. Also note that $U$ contains no limit points of $E$. Indeed, let $p\in E'$. Suppose $p\in U$. Thus there exists a point $q\neq p, \;q\in U, \;q\in E$. However, we know $U\subseteq X\bslash E$. Thus there does not exist a point $p\in E'\cap U$. Thus $U\subseteq X\bslash E \cap X\bslash E'=X\bslash (E\cap E')=X\bslash \cl{E}$. Thus $x$ is an interior point of $X\bslash \cl{E}$ This holds for all $x\in X\bslash\cl{E}$. Thus $X\bslash \cl{E}$ is open. Therefore $\cl{E}$ is closed.
\end{proof}
\begin{thm}
 $E=\cl{E}\iff E$ is closed.
\end{thm}
\begin{proof}
 $(\Rightarrow)$ Done, by previous theorem. $(\Leftarrow).$ Suppose $E$ is closed. Therefore it contains all of its limit points, i.e. $E'\subseteq E$. Thus $\cl{E}=E\cup \cl{E}=E$.
\end{proof}

\begin{thm}
 Let $E\subseteq F\subseteq X$. If $F$ is closed, then $\cl{E}\subseteq F$.
\end{thm}

\begin{proof}
 Let $p\in E'$. Let $U$ be an open neighborhood of $p$. There exists $q\neq p$ such that $q\in U$ and $q\in E$. Since $E\subseteq F$, we know $q\in F$. Therefore $p$ is a limit point of $F$. Since $F$ is closed, $p\in F$. Thus $E'\subseteq F$.  Therefore $\cl{E}=E\cup E'\seq F$.
\end{proof}
\begin{rem}
 $$\cl{E}=\bigcap_{\substack{F=\cl{F},\\E\seq F\seq X}}F.$$
\end{rem}
\begin{question}
 Is $(a,b)$ open? Consider this interval in $\R$. Consider any $c\in (a,b)$. Indeed we can find a neighborhood of $c$ completely contained in $(a,b)$. Now consider the interval in $\R^2$. In this case it does not. Thus the question is ambiguous. 
\end{question}
\begin{defn}
 Let $Y\seq X$ be metric spaces. A subset $E\seq Y$ is \emph{open relative to $Y$} if and only if there is an open set $U\seq X$ such that:
 $$
 E=U\cap Y.
 $$
\end{defn}
\begin{ex}
 Let $Y=(a,b]$ and $X=\R$. Then $(c,b]$ is open in $Y$ for any $c\in (a,b)$.
\end{ex}
\topic{Compact Sets}
\begin{defn}
 An \emph{open cover} of $E\seq X$ is a collection $\{U_\alpha\}$ of open sets of $X$ ($U_\alpha$ is open in $X$) such that $E\seq \displaystyle\bigcup_\alpha U_\alpha$.
\end{defn}
\newpage
\begin{ex}
 Consider the following examples:
\begin{itemize}
 \item Let  $E=X=\R$. Consider $U_n=(n-1,n+1)$.
 \item Let $E=\Z, X=\R$. Consider $U_n=(n-\sfrac{1}{2},n+\sfrac{1}{2})$.
 \item Let $E=[0,1], X=\R$. Consider $U_n=(\sfrac{-2}{n},\sfrac{2}{n}).$
 \item Let $E=(0,1), X=\R$. Consider $U_n=(0,\sfrac{1}{n}), V_n=(\sfrac{1}{n},1)$. Consider then $\{U_n\}\cup\{V_n\}$.
 \item Let $E=\{1,3,4\}, X=\R$. Let $U_n=(n-\sfrac{1}{2},n+\sfrac{1}{2})$.
\end{itemize}

\end{ex}
\begin{defn}
 A set $K\seq X$ is \emph{compact} if and only if every open cover $\{U_\alpha\}$ contains a finite sub cover. $(K\seq U_{\alpha_1}\cup\dots\cup U_{\alpha_k})$
\end{defn}
\end{document}