Created the Real Analysis Theorems and Definitions packet

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\chapter{A Glimpse Into Topology}
\subimport{./}{open-and-closed-sets-in-R.tex}
\subimport{./}{compact-sets.tex}
\subimport{./}{continuous-functions.tex}
\subimport{./}{metric-spaces.tex}
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\section{Compact Sets}
\begin{definition}
Let $A$ be a subset of $\R$. An \textbf{open cover} of $A$ is a collection $\mathcal{G}=\{G_\alpha\}$ of open sets in $\R$ whose union contains $A$; that is,
\[A \subseteq \bigcup_\alpha G_\alpha\]
If $\mathcal{G}'$ is a subcollection of sets from $\mathcal{G}$ such that the union of the sets in $\mathcal{G}'$ also contains $A$, then $\mathcal{G}'$ is called a \textbf{subcover} of $\mathcal{G}$. If $\mathcal{G}'$ consists of finitely many sets, then we call $\mathcal{G}'$ a \textbf{finite subcover} of $\mathcal{G}$.
\end{definition}
\begin{definition}
A subset $K$ of $\R$ is said to be \textbf{compact} if \textit{every} open cover of $K$ has a finite subcover.
\end{definition}
\begin{theorem}
If $K$ is a compact subset of $\R$, then $K$ is closed and bounded.
\end{theorem}
\begin{theorem}[\textbf{Heine-Borel Theorem}]
A subset $K$ of $\R$ is compact if and only if it is closed and bounded.
\end{theorem}
\begin{theorem}
A subset $K$ of $\R$ is compact if and only if every sequence in $K$ has a subsequence that converges to a point in $K$.
\end{theorem}
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\section{Continuous Functions}
\begin{lemma}
A function $f:A \to \R$ is continuous at the point $c$ in $A$ if and only if for every neighborhood $U$ of $f(c)$, there exists a neighborhood $V$ of $c$ such that if $x \in V\cap A$, then $f(x) \in U$.
\end{lemma}
\begin{theorem}[\textbf{Global Continuity Theorem}]
Let $A \subseteq \R$ and let $f:A \to \R$ be a function with domain $A$. Then the following are equivalent:
\begin{enumerate}
\item $f$ is continuous at every point of $A$.
\item For every open set $G$ in $\R$, there exists an open set $H$ in $\R$ such that $H \cap A=f^{-1}(G)$.
\end{enumerate}
\end{theorem}
\begin{corollary}
A function $f:\R\to\R$ is continuous if and only if $f^{-1}(G)$ is open in $\R$ whenever $G$ is open.
\end{corollary}
\begin{theorem}[\textbf{Preservation of Compactness}]
If $K$ is a compact subset of $\R$ and if $f:K \to \R$ is continuous on $K$, then $f(K)$ is compact.
\end{theorem}
\begin{theorem}
If $K$ is a compact subset of $\R$ and $f:K \to \R$ is injective and continuous, then $f^{-1}$ is continuous on $f(K)$.
\end{theorem}
\begin{theorem}
Let $f:A \subseteq \R \to \R$. Then the following are equivalent:
\begin{enumerate}
\item $f$ is continuous on $A$
\item \textit{Theorem 5.1.3:} $\forall$ converging sequences $x_n \to a \in A$, then $f(x_n) \to f(a)$
\item \textit{Theorem 11.3.2:} For each open set $U \subseteq \R$, $f^{-1}(U) \subseteq A$ is open relative to $A$. (i.e. inverse images of open sets are relatively open to $A$. $f^{-1}(U)=V \cap A$ for some open set $V \subseteq \R$)
\item For each closed set $F \subseteq \R$, then $f^{-1}(F) \subseteq A$ is closed relative to $A$. (i.e. inverse images of closed sets are relatively closed)
\end{enumerate}
\end{theorem}
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\section{Metric Spaces}
\begin{definition}
A \textbf{metric} on a set $S$ is a function $d:S \times S \to \R$ that satisfies the following properties:
\begin{enumerate}
\item $d(x,y)\geq 0\ \forall\ x,y\in S$ \textit{(positivity)};
\item $d(x,y)=0$ if and only if $x=y$ \textit{(definiteness)};
\item $d(x,y)=d(y,x)\ \forall\ x,y \in S$ \textit{(symmetry)};
\item $d(x,y)\leq d(x,z)+d(z,y)\ \forall\ x,y,z\in S$ \textit{(triangle inequality)}
\end{enumerate}
A \textbf{metric space} $(S,d)$ is a set $S$ together with a metric $d$ on $S$.
\end{definition}
\begin{definition}
Let $(S,d)$ be a metric space. Then for $\varepsilon>0$, the $\varepsilon$\textbf{-neighborhood} of a point $x_0$ in $S$ is the set
\[V_\varepsilon(x_0):=\{x \in S :d(x_0,x)<\varepsilon\}\]
A \textbf{neighborhood} of $x_0$ is any set $U$ that contains an $\varepsilon$-neighborhood of $x_0$ for some $\varepsilon>0$.
\end{definition}
\begin{definition}
Let $(x_n)$ be a sequence in the metric space $(S,d)$. The sequence $(x_n)$ is said to \textbf{converge} to $x$ in $S$ if for any $\varepsilon>0$, there exists $K \in \N$ such that $x_n \in V_\varepsilon(x)$ for all $n \geq K$.
\end{definition}
\begin{definition}
Let $(S,d)$ be a metric space. A sequence $(x_n)$ in $S$ is said to be a \textbf{Cauchy sequence} if for each $\varepsilon>0$, there exists $H \in \N$ such that $d(x_n,x_m)<\varepsilon$ for all $n,m \geq H$.
\end{definition}
\begin{definition}
A metric space $(S,d)$ is said to be \textbf{complete} if each Cauchy sequence in $S$ converges to a point of $S$.
\end{definition}
\begin{definition}
Let $(S,d)$ be a metric space. A subset $G$ of $S$ is said to be an \textbf{open} set in $S$ if for every point $x \in S$ there is a neighborhood $U$ of $x$ such that $U \subseteq G$. A subset $F$ of $S$ is said to be a \textbf{closed} set in $S$ if the complement $S \setminus F$ is an open set in $S$.
\end{definition}
\begin{definition}
Let $(S_1,d_1)$ and $(S_2,d_2)$ be metric spaces, and let $f:S_1 \to S_2$ be a function from $S_1$ to $S_2$. The function $f$ is said to be \textbf{continuous} at the point $c$ in $S_1$ if for every $\varepsilon$-neighborhood $V_\varepsilon(f(c))$ of $f(c)$ there exists a $\delta$-neighborhood $V_\delta(c)$ of $c$ such that if $x \in V_\delta(c)$, then $f(x)\in V_\varepsilon(f(c))$.
\end{definition}
\begin{theorem}[\textbf{Global Continuity Theorem}]
If $(S_1,d_1)$ and $(S_2,d_2)$ are metric spaces, then a function $f:S_1 \to S_2$ is continuous on $S_1$ if and only if $f^{-1}(G)$ is open in $S_1$ whenever $G$ is open in $S_2$.
\end{theorem}
\begin{theorem}[\textbf{Preservation of Compactness}]
If $(S,d)$ is a compact metric space and if the function $f:S \to \R$ is continuous, then $f(S)$ is compact in $\R$.
\end{theorem}
\begin{definition}
A \textbf{semimetric} on a set $S$ is a function $d: S\times S \to \R$ that satisfies all of the conditions of a metric, except that condition (2) is replaces by the weaker condition
\[d(x,y)=0\ \ \text{ if }\ \ x=y\]
A \textbf{semimetric space} $(S,d)$ is a set $S$ together with a semimetric $d$ on $S$.
\end{definition}
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\section{Open and Closed Sets in $\R$}
\begin{definition}
A \textbf{neighborhood} of a point $x \in \R$ is any set $V$ that contains an $\varepsilon$-neighborhood $V_\varepsilon(x):=(x-\varepsilon,x+\varepsilon)$ of $x$ for some $\varepsilon>0$.
\end{definition}
\begin{definition}
\begin{enumerate}
\item[]
\item A subset $G$ of $\R$ is \textbf{open} in $\R$ if for each $x \in G$ there exists a neighborhood $V$ of $x$ such that $V \subseteq G$.
\item A subset $F$ of $\R$ is \textbf{closed in $\R$} if the complement $\mathcal{C}(F):=\R\setminus F$ is open in $\R$.
\end{enumerate}
\end{definition}
\begin{theorem}[\textbf{Open Set Properties}]
\begin{enumerate}
\item[]
\item The union of an arbitrary collection of open subsets in $\R$ is open.
\item The intersection of any finite collection of open sets in $\R$ is open.
\end{enumerate}
\end{theorem}
\begin{theorem}[\textbf{Closed Set Properties}]
\begin{enumerate}
\item[]
\item The intersection of an arbitrary collection of closed sets in $\R$ is closed.
\item The union of any finite collection of closed sets in $\R$ is closed.
\end{enumerate}
\end{theorem}
\begin{theorem}[\textbf{Characterization of Closed Sets}]
Let $F \subset \R$; then the following assertions are equivalent:
\begin{enumerate}
\item $F$ is a closed subset of $\R$.
\item If $X=(x_n)$ is any convergent sequence of elements in $F$, then $\lim X$ belongs to $F$.
\end{enumerate}
\end{theorem}
\begin{theorem}
A subset of $\R$ is closed if and only if it contains all of its cluster points.
\end{theorem}
\begin{theorem}
A subset of $\R$ is open if and only if it is the union of countably many disjoint open intervals in $\R$.
\end{theorem}
\begin{definition}
The \textbf{Cantor Set} $\mathbb{F}$ is the intersection of the sets $F_n, n\in\N$, obtained by successive removal of open middle thirds, starting with $[0,1]$.
\end{definition}