4.1 Definitions
Definition 4.1.1 (Topological Space). Let $X$ be a non-empty set. A topology over $X$ is a family $\topo \subset 2^{X}$ such that
$\emptyset \in \topo$ and $X \in \topo$.
For any $U, V \in \topo$, $U \cap V \in \topo$.
For any $\seqi{U}\subset \topo$, $\bigcup_{i \in I}U_{i} \in \topo$.
The elements of $\topo$ are known as open sets, and the pair $(X, \topo)$ is known as a topological space.
Definition 4.1.2 (Closed Set). Let $(X, \topo)$ be a topological space, then $A \subset X$ is closed if $A^{c} \in \topo$.
Definition 4.1.3 (Separation Axioms). Let $(X, \topo)$ be a topological space, then $X$ may satisfy the following separation axioms:
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 $y \in U$.
For any $x, y \in X$ with $x \ne y$, there exists $U \in \topo$ with $x \in U$ and $y \not\in U$.
For any $x, y \in X$ with $x \ne y$, there exists $U, V \in \topo$ with $x \in U$, $y \in V$, and $U \cap V = \emptyset$.
$X$ 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 $U \cap V = \emptyset$.
$X$ 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 $U \cap V = \emptyset$.
Lemma 4.1.4. Let $X$ be a topological space, then $X$ satisfies (T1) if and only if $\bracs{x}$ is closed for all $x \in X$.
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.$\square$
Definition 4.1.5 (Base). Let $(X, \topo)$ be a topological space, then a base for $\topo$ is a family $\cb \subset \topo$ such that:
For every $x \in X$, there exists $U \in \cb$ such that $x \in U$.
For every $x \in X$ and $U \subset X$ open with $x \in U$, there exists $V \in \cb$ such that $x \in V \subset U$.
In which case,
Conversely, if $\cb \subset 2^{X}$ is a family such that:
For every $x \in X$, there exists $U \in \cb$ such that $x \in U$.
For every $x \in X$ and $U, V \in \cb$ such that $x \in U \cap V$, there exists $W \in \cb$ such that $x \in W \subset U \cap V$.
then $\topo(\cb)$ is a topology on $X$, and $\cb$ is a base for $\topo(\cb)$.
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 $\seqi{U}, \seqj{V}\subset \cb$,
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).$\square$
Definition 4.1.6 (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
where
is a base for $\topo(\ce)$. The topology $\topo(\ce)$ is known as the topology generated by $\ce$.
Proof. Since $\cb(\ce) \supset \ce$, $\cb(\ce)$ satisfies (TB1). In addition, $\cb(\ce)$ is closed under finite intersections, so it satisfies (TB2). By Definition 4.1.5, $\cb(\ce)$ is a base for $\topo(\ce)$.$\square$
Definition 4.1.7 (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:
For each $i \in I$, $f_{i} \in C(X; Y_{i})$.
If $\mathcal{S}$ is a topology on $X$ satisfying $(1)$, then $\mathcal{S}\supset \topo$.
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 $\topo$.
The topology $\topo$ is known a the initial/weak topology generated by the maps $\seqi{f}$.
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
For each $i \in I$, $\topo \supset \bracs{f_i^{-1}(U)|U \in \topo_i}$, so $f_{i} \in C(\topo; Y_{i})$.
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 $\mathcal{S}\supset \topo$.
By Definition 4.1.6, $\cb$ is a base for $\topo$.
$\square$