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\title{Open and Closed Sets in $\mathbb{R}$}
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\section{Basic Definitions and Theorems}
A central idea in topology is the notion of an open set.  We will define an open set and some related terms as they appear in the context of $\mathbb{R}$ with the usual \textit{metric}, or measure of distance 
\[
d(x,y)=|x-y|
\vspace{0.2 cm}
\]  
Although it suffices as an introduction to the basic ideas in topology, we will not discuss the algebraic properties required for a more thorough introduction to the subject.  The subset we will discuss is sometimes called \textit{point-set} topology.     
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\begin{definition}[topology] A \textbf{topology} on a set $X$ is a collection $\mathcal{T}$ of subsets of $X$ that has  the following properties:
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\begin{tabular}{ll}
(1) & $\emptyset$ and $X$ are in $\mathcal{T}$\\
(2) & The union of any subcollection of elements of $\mathcal{T}$ is in $\mathcal{T}$\\
(3) & The intersection of any \textit{finite} subcollection of the elements of $\mathcal{T}$ is in $\mathcal{T}$\\
\end{tabular}
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A set $X$ for which a topology $\mathcal{T}$ has been specified is called a \textbf{topological space}.
\end{definition} 
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We will take the set $X$ to be $\mathbb{R}$.  Now we proceed to define an $\epsilon$-neighborhood of a point, which we will use to define an open set.  We will show that the open sets that satisfy the definition form a topology on $\mathbb{R}$.  

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\begin{definition}[$\epsilon$ neighborhood] Given $a\in\mathbb{R}$, an $\epsilon$-\textbf{neighborhood} $V_{\epsilon}(a)$ is the set
\[
V_\epsilon(a)=\{x\in \mathbb{R} : |x-a|<\epsilon\}
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\]
\end{definition}
Note that $|x-a|$ is strictly less than $\epsilon$, so an $\epsilon$ neighborhood is just an open interval centered at $a$ with radius $\epsilon$.  
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\begin{definition}[open set] The set $E\subseteq\mathbb{R}$ is \textbf{open} if every point $a\in E$ has an $\epsilon$ neighborhood that is contained in $E$:
\[
V_\epsilon(a)\subseteq E\quad\mbox{for some }\quad\epsilon>0
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\]
\end{definition}
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As noted earlier, the collection $\mathcal{T}$ of sets that satisfy this definition of an open set is a topology on $\mathbb{R}$.
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Note that $E=\mathbb{R}$ is open, and $E=\emptyset$ is open vacuously.
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Also, any interval of the form $(a,b)$ with $a,b\in\mathbb{R}$ and $a<b$ is open.  
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\begin{theorem} Every interval of the form $(a,b)$ with $a,b\in\mathbb{R}$ and $a<b$ is open.
\end{theorem}
\begin{proof}
The interval $(a,b)$ is the same as the set $\{x\in\mathbb{R}:a<x<b\}$.  Choose any element $c\in(a,b)$.  Then $a<c<b$. Choose $\epsilon$ to be the smaller of $c-a$ and $b-c$, that is, take $\epsilon$ to be the distance to the closer of $a$ and $b$.  Then the interval $(c-\epsilon,c+\epsilon)$ is contained in $(a,b)$.
\end{proof} 
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The union of any collection of open sets (finite, countable, or uncountable) is also open, as is the intersection of a \textit{finite} number of open sets.
\begin{theorem} The union of an arbitrary collection of open sets is open. 
\end{theorem}
\begin{proof}
Let $I$ be an index set and suppose
\[
S=\bigcup_{\alpha\in I}E_\alpha
\]
Given an arbitrary element $s\in S$, we have to show that there is an $\epsilon$ neighborhood $V_{\epsilon}(s)$ that is contained in $S$.  Let $\epsilon>0$ be given and let $s\in S$ be an arbitrary element of $S$.  Because $s$ is in the union of the $E_\alpha$, it is in at least one of them, call it $E_s$.  By hypothesis, $E_s$ is open, so by definition there is an $\epsilon$ neighborhood $V_{\epsilon}(s)$ that is entirely contained in $E_s$.  But if $V_{\epsilon}(s)$ is contained in $E_s$, by the definition of set union it is also contained in the union of the $E_{\alpha}$, which is $S$.  Therefore, $S$ is open.
\end{proof}
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Now we consider the intersection of a collection of open sets.  As it turns out, we can only guarantee that the intersection will be open if the collection is \textbf{finite}.
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\begin{theorem} The intersection of a finite collection of open sets is open. 
\end{theorem}
\begin{proof} Let $\{E_1,E_2,\ldots,E_n\}$ be a finite collection of open sets.  Let $a$ be an arbitrary element of 
the intersection:
\[
a\in\bigcap_{i=1}^n E_i
\]
Then by definition $a\in E_i$ for $i=1,2,\ldots,n$.  Since each $E_i$ is open, for each $E_i$ there exists an $\epsilon$-neighborhood $V_{\epsilon_i}(a)$ of $a$ that is contained in $E_i$.   To show that the intersection is open, we have to find a single $\epsilon$-neighborhood that is contained in \textbf{all} of the $E_i$.  Now the $\epsilon$-neighborhoods are open intervals centered at $a$, so if we take $\epsilon$ to be the smallest of the $\epsilon_i$:
\[
\epsilon = \min\{\epsilon_1,\epsilon_2,\ldots,\epsilon_n\}
\]
then 
\[
V_{\epsilon}(a)\subseteq V_{\epsilon_i}\quad\mbox{for all } i
\]
so
\[
V_{\epsilon}(a)\subseteq\bigcap_{i-1}^n E_i
\]
Since $a$ was an arbitrary choice, we can find such an $\epsilon$-neighborhood for any $a$ in the intersection, 
so by definition the intersection is open.
\end{proof}
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Together with the assertion that $\mathbb{R}$ and $\emptyset$ are open, the previous two theorems establish that the collection of open sets $\mathcal{T}$ we defined is a topology on $\mathbb{R}$. 
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There are a great many collections of subsets of $\mathbb{R}$ that qualify as topologies on $\mathbb{R}$, such as the collection consisting only of $\mathbb{R}$ and the empty set.  Because of the way this one arises naturally from the metric $d(x,y)=|x-y|$, it is called the \textit{metric topology} on $\mathbb{R}$ with the usual metric.  
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\begin{definition}[limit point] A point $x$ is a \textbf{limit point} of a set $E$ if every $\epsilon$-neighborhood of $x$ intersects $E$ in some point other than $x$, that is,
\[
V_{\epsilon}\cap E\setminus x\;\neq\;\emptyset\quad \forall \epsilon>0
\]
\end{definition}
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\begin{example} $0$ is a limit point of $E=(0,1)$ because every $\epsilon$-neighborhood $V_\epsilon(0)$ of zero contains the point $\epsilon/2\in E$, and $\epsilon/2\neq0$.
\end{example}
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\begin{definition}[isoated point] A point $x$ is an \textbf{isolated point} of a set $E$ if it is not a limit point of $E$.
\end{definition}
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\begin{example} Every element $z$ of $\mathbb{Z}$ is an isolated point, because if we choose $\epsilon<1/2$, there are no elements of $\mathbb{Z}$ in $V_{\epsilon}(z)$ other than $z$ itself.
\end{example}
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\begin{definition}[closed set] A set $E$ is said to be \textbf{closed} if it contains all of its limit points.
\end{definition}
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\begin{definition}[closure of a set] The \textbf{closure} of a set $E$ is the union of $E$ and the set containing all of its limit points.
\end{definition}
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\begin{example} $C=[0,1]$ is the closure of $E=(0,1)$.
\end{example}
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\begin{definition}[compliment of a set] The \textbf{compliment} of a set $E$ (relative to $\mathbb{R}$) is the set of all real numbers that do not belong to $E$, that is,
\[
E^c=\{x\in\mathbb{R}: x\notin E\} = \mathbb{R}\setminus E
\]
\end{definition}
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\begin{remark} Note that the properties of being open or closed are not mutually exclusive.  A set can be open, closed, neither open nor closed, or both open and closed.  For example, $\mathbb{R}$ is both open and closed.  The half-open interval $E=(0,1]$ is neither open nor closed, because $0$ is a limit point of $E$ that does not belong to $E$, and there are no $\epsilon$-neighborhoods of $1$ that are entirely contained in $E$.
\end{remark}
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\begin{theorem} A point $x$ is a limit point of a set $E$ if and only if there exists a sequence $a_n\in E$ such that 
$a_n\neq x$ for all $n\in\mathbb{N}$ and $a_n\rightarrow x$ as $n\rightarrow\infty$.  
\end{theorem}
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\begin{theorem} A set $E$ is open if and only if its compliment $E^c$ is closed.  A set $F$ is closed if and only if its compliment $F^c$ is open.
\end{theorem}
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\begin{theorem} The intersection of an arbitrary collection of closed sets is closed.  The union of a \textit{finite} collection of closed sets is closed.
\end{theorem}
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\begin{definition}[$F_{\sigma}$] A set $E$ is called an $F_{\sigma}$ set if it can be represented as a countable union of closed sets.
\end{definition}
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\begin{definition}[$G_{\delta}$] A set $E$ is called a $G_{\delta}$ set if it can be represented as a countable intersection of open sets.
\end{definition}




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