\section{Definitions}
\label{section:top-definitions}

\begin{definition}[Topological Space]
\label{definition:topspace}
    Let $X$ be a non-empty set. A \textbf{topology} over $X$ is a family $\topo \subset 2^X$ such that
    \begin{enumerate}
        \item[(O1)] $\emptyset \in \topo$ and $X \in \topo$.
        \item[(O2)] For any $U, V \in \topo$, $U \cap V \in \topo$.
        \item[(O3)] For any $\seqi{U} \subset \topo$, $\bigcup_{i \in I}U_i \in \topo$.
    \end{enumerate}
    The elements of $\topo$ are known as \textbf{open sets}, and the pair $(X, \topo)$ is known as a \textbf{topological space}.
\end{definition}

\begin{definition}[Closed Set]
\label{definition:closedset}
    Let $(X, \topo)$ be a topological space, then $A \subset X$ is \textbf{closed} if $A^c \in \topo$.
\end{definition}

\begin{definition}[Separation Axioms]
\label{definition:separation}
    Let $(X, \topo)$ be a topological space, then $X$ may satisfy the following \textbf{separation axioms}:
    \begin{enumerate}
        \item[(T0)] For any $x, y \in X$ with $x \ne y$, there exists $U \in \topo$ with $x \in U$ and $y \not\in U$, or there exists $U \in \topo$ with $x \not\in U$ and $y \in U$.
        \item[(T1)] For any $x, y \in X$ with $x \ne y$, there exists $U \in \topo$ with $x \in U$ and $y \not\in U$.
        \item[(T2)] For any $x, y \in X$ with $x \ne y$, there exists $U, V \in \topo$ with $x \in U$, $y \in V$, and $U \cap V = \emptyset$.
        \item[(T3)] $X$ is (T1), and for any $x \in X$ and $A \subset X$ closed with $x \not\in A$, there exists $U, V \in \topo$ such that $x \in U$, $A \subset V$, and $U \cap V = \emptyset$.
        \item[(T4)] $X$ is (T1), and for any $A, B \subset X$ closed with $A \cap B = \emptyset$, there exists $U, V \in \topo$ such that $A \subset U$, $B \subset V$, and $U \cap V = \emptyset$.
    \end{enumerate}
\end{definition}

\begin{lemma}
\label{lemma:t1}
    Let $X$ be a topological space, then $X$ satisfies (T1) if and only if $\bracs{x}$ is closed for all $x \in X$.
\end{lemma}
\begin{proof}
    $(\Rightarrow)$: Let $x \in X$, then for each $y \in X \setminus {x}$, there exists $U_y \subset X$ open such that $x \not\in U_y$. Thus $U^c = \bigcup_{y \in X \setminus \bracs{x}}U_y$ is open.

   $(\Leftarrow)$: Let $x, y \in X$ with $x \ne y$, then $y \in \bracs{x}^c$, $x \not\in \bracs{x}^c$, and $\bracs{x}^c$ is open.
\end{proof}


\begin{definition}[Base]
\label{definition:base}
    Let $(X, \topo)$ be a topological space, then a \textbf{base} for $\topo$ is a family $\cb \subset \topo$ such that:
    \begin{enumerate}
        \item For every $x \in X$, there exists $U \in \cb$ such that $x \in U$.
        \item For every $x \in X$ and $U \subset X$ open with $x \in U$, there exists $V \in \cb$ such that $x \in V \subset U$.
    \end{enumerate}
    In which case,
    \[
    \topo = \topo(\cb) = \bracs{\bigcup_{i \in I}U_i \bigg | \seqi{U} \subset \cb, I \text{ index set}}
    \]
    Conversely, if $\cb \subset 2^X$ is a family such that:
    \begin{enumerate}
        \item[(TB1)] For every $x \in X$, there exists $U \in \cb$ such that $x \in U$.
        \item[(TB2)] For every $x \in X$ and $U, V \in \cb$ such that $x \in U \cap V$, there exists $W \in \cb$ such that $x \in W \subset U \cap V$.
    \end{enumerate}
    then $\topo(\cb)$ is a topology on $X$, and $\cb$ is a base for $\topo(\cb)$.
\end{definition}
\begin{proof}
    Let $U \in \topo$. If $U = \emptyset$, then $U$ is a union over an empty index set. Otherwise, for each $x \in U$, there exists $V_x \in \cb$ such that $x \in V_x \subset U$. In which case, $U = \bigcup_{x \in U}V_x \in \topo$ and $\topo \subset \topo(\cb)$. On the other hand, since $\cb \subset \topo$, $\topo \supset \topo(\cb)$ by (O3).

    For the converse, $\emptyset \in \topo(\cb)$ as a union over an empty index set and $X = \bigcup_{U \in \cb}U \in \topo(\cb)$ by (TB1). For any $\seqi{U}, \seqj{V} \subset \cb$,
    \[
    \paren{\bigcup_{i \in I}U_i} \cap \paren{\bigcup_{j \in J}V_j} = \bigcup_{i \in I}\bigcup_{j \in J}U_i \cap V_j
    \]
    Thus to show that $\topo(\cb)$ satisfies (O2), it is sufficient to show that $U \cap V \in \topo(\cb)$ for all $U, V \in \cb$. To this end, for every $x \in U \cap V$, there exists $W_x \in \cb$ such that $x \in W_x \subset U \cap V$. Therefore $U \cap V = \bigcup_{x \in U \cap V}W_x \in \topo(\cb)$, and $\topo(\cb)$ satisfies (O2).

    By definition, $\topo(\cb)$ satisfies (O3).
\end{proof}

\begin{definition}[Generated Topology]
\label{definition:generated-topology}
    Let $X$ be a set and $\ce \subset 2^X$ such that $\bigcup_{U \in \ce}U = X$, then the smallest topology on $X$ containing $\ce$ is given by
    \[
    \topo(\ce) = \bracs{\bigcup_{i \in I}U_i \bigg | U_i \in \cb(\ce)}
    \]
    where
    \[
    \cb(\ce) = \bracs{\bigcap_{j = 1}^n U_j \bigg | \seqf{U_j} \subset \ce, n \in \nat^+}
    \]
    is a base for $\topo(\ce)$. The topology $\topo(\ce)$ is known as the topology \textbf{generated by} $\ce$.
\end{definition}
\begin{proof}
    Since $\cb(\ce) \supset \ce$, $\cb(\ce)$ satisfies (TB1). In addition, $\cb(\ce)$ is closed under finite intersections, so it satisfies (TB2). By \ref{definition:base}, $\cb(\ce)$ is a base for $\topo(\ce)$.
\end{proof}

\begin{definition}[Initial Topology]
\label{definition:initial-topology}
    Let $X$ be a set, $\bracsn{(Y_j, \topo_i)}$ be a family of topological spaces, and $\seqi{f}$ be a family of maps such that $f_i: X \to Y_i$ for each $i \in I$, then there exists a topology $\topo$ on $X$ such that:
    \begin{enumerate}
        \item For each $i \in I$, $f_i \in C(X; Y_i)$.
        \item[(U)] If $\mathcal{S}$ is a topology on $X$ satisfying $(1)$, then $\mathcal{S} \supset \topo$.
        \item The family

        \[
        \mathcal{B} = \bracs{\bigcap_{j \in J}f_j^{-1}(U_j) \bigg | J \subset I \text{ finite}, U_j \in \topo_j}
        \]

        is a base for $\topo$.
    \end{enumerate}
    The topology $\topo$ is known a the \textbf{initial/weak topology} generated by the maps $\seqi{f}$.
\end{definition}
\begin{proof}
    Let $\topo$ be the topology genereated by sets of the form $\ce = \bracs{f_i^{-1}(U_i)| i \in I, U_i \in \topo_i}$. Let $\topo$ be the topology generated by $\ce$, then
    \begin{enumerate}
        \item For each $i \in I$, $\topo \supset \bracs{f_i^{-1}(U)|U \in \topo_i}$, so $f_i \in C(\topo; Y_i)$.
        \item If $\mathcal{S}$ is a topology such that $f_i \in C(X, \mathcal{S}; Y_i)$, then $\bracs{f_i^{-1}(U)|U \in \topo_i} \subset \mathcal{S}$. Thus $\ce \subset \mathcal{S}$ and $\mathcal{S} \supset \topo$.
        \item By \ref{definition:generated-topology}, $\cb$ is a base for $\topo$.
    \end{enumerate}
\end{proof}