Basic material on bases for topological spaces.
Definition 5.5.1. Let $X$ be a topological space. A collection of subsets $\mathcal{B}$ of $X$ is called a base for the topology on $X$ or a basis for the topology on $X$ if the following conditions hold:
Every element $B \in \mathcal{B}$ is open in $X$.
For every open $U \subset X$ and every $x \in U$, there exists an element $B \in \mathcal{B}$ such that $x \in B \subset U$.
The following lemma is sometimes used to define a topology.
Lemma 5.5.2. Let $X$ be a set and let $\mathcal{B}$ be a collection of subsets. Assume that $X = \bigcup _{B \in \mathcal{B}} B$ and that given $x \in B_1 \cap B_2$ with $B_1, B_2 \in \mathcal{B}$ there is a $B_3 \in \mathcal{B}$ with $x \in B_3 \subset B_1 \cap B_2$. Then there is a unique topology on $X$ such that $\mathcal{B}$ is a basis for this topology.
Proof. Let $\sigma (\mathcal B)$ be the set of subsets of $X$ which can be written as unions of elements of $\mathcal B$. We claim $\sigma (\mathcal{B})$ is a topology. Namely, the empty set is an element of $\sigma (\mathcal{B})$ (as an empty union) and $X$ is an element of $\sigma (\mathcal{B})$ (as the union of all elements of $\mathcal{B}$). It is clear that $\sigma (\mathcal{B})$ is preserved under unions. Finally, if $U, V \in \sigma (\mathcal{B})$ then write $U = \bigcup _{i \in I} U_ i$ and $V = \bigcup _{j \in J} V_ j$ with $U_ i, V_ j \in \mathcal{B}$ for all $i \in I$ and $j \in J$. Then
The assumption in the lemma tells us that $U_ i \cap V_ j \in \sigma (\mathcal{B})$ hence we see that $U \cap V$ is too. Thus $\sigma (\mathcal{B}$ is a topology. Properties (1) and (2) of Definition 5.5.1 are immediate for this topology. To prove the uniqueness of this topology let $\mathcal T$ be a topology on $X$ such that $\mathcal B$ is a base for it. Then of course every element of $\mathcal{B}$ is in $\mathcal{T}$ by (1) of Definition 5.5.1 and hence $\sigma (\mathcal{B} \subset \mathcal{T}$. Conversely, part (2) of Definition 5.5.1 tells us that every element of $\mathcal{T}$ is a union of elements of $\mathcal{B}$, i.e., $\mathcal{T} \subset \sigma (\mathcal{B})$. This finishes the proof. $\square$
Lemma 5.5.3. Let $X$ be a topological space. Let $\mathcal{B}$ be a basis for the topology on $X$. Let $\mathcal{U} : U = \bigcup _ i U_ i$ be an open covering of $U \subset X$. There exists an open covering $U = \bigcup V_ j$ which is a refinement of $\mathcal{U}$ such that each $V_ j$ is an element of the basis $\mathcal{B}$.
Proof. If $ x \in U = \bigcup _{i\in I} U_ i $, there is an $ i_ x \in I $ such that $ x \in U_{i_ x} $. Thus we have a $ B_{i_ x} \in \mathcal{B}$ verifying $ x \in B_{i_ x} \subset U_{i_ x}$. Set $J = \{ i_ x | x \in U\} $ and for $j = i_ x \in J$ set $V_ j = B_{i_ x}$. This gives the desired open covering of $U$ by $\{ V_ j\} _{j \in J}$. $\square$
Definition 5.5.4. Let $X$ be a topological space. A collection of subsets $\mathcal{B}$ of $X$ is called a subbase for the topology on $X$ or a subbasis for the topology on $X$ if the finite intersections of elements of $\mathcal{B}$ form a basis for the topology on $X$.
In particular every element of $\mathcal{B}$ is open.
Lemma 5.5.5. Let $X$ be a set. Given any collection $\mathcal{B}$ of subsets of $X$ there is a unique topology on $X$ such that $\mathcal{B}$ is a subbase for this topology.
Proof. By convention $\bigcap _\emptyset B = X $. Thus we can apply Lemma 5.5.2 to the set of finite intersections of elements from $\mathcal B$. $\square$
Lemma 5.5.6. Let $X$ be a topological space. Let $\mathcal{B}$ be a collection of opens of $X$. Assume $X = \bigcup _{U \in \mathcal{B}} U$ and for $U, V \in \mathcal{B}$ we have $U \cap V = \bigcup _{W \in \mathcal{B}, W \subset U \cap V} W$. Then there is a continuous map $f : X \to Y$ of topological spaces such that
for $U \in \mathcal{B}$ the image $f(U)$ is open,
for $U \in \mathcal{B}$ we have $f^{-1}(f(U)) = U$, and
the opens $f(U)$, $U \in \mathcal{B}$ form a basis for the topology on $Y$.
Proof. Define an equivalence relation $\sim $ on points of $X$ by the rule
Let $Y$ be the set of equivalence classes and $f : X \to Y$ the natural map. Part (2) holds by construction. The assumptions on $\mathcal{B}$ exactly mirror the assumptions in Lemma 5.5.2 on the set of subsets $f(U)$, $U \in \mathcal{B}$. Hence there is a unique topology on $Y$ such that (3) holds. Then (1) is clear as well. $\square$
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