Proposition 5.26.4.label Let $X$ be a set, $(Y, \topo)$ be a topological space, and $\mathcal{B}$ be a base for $Y$ with $\emptyset \in \mathcal{B}$, then:
- (1)
For any open preimage function $P: \topo \to 2^{X}$, $P|_{\mathcal{B}}$ is a basic preimage function.
- (2)
For any basic preimage function $p: \mathcal{B}\to 2^{X}$, there exists a unique open preimage function $P: \topo \to 2^{X}$ such that $p = P|_{\mathcal{B}}$.
Proof. (2): For each $U \in \topo$, let $\mathcal{B}(U) = \bracs{V \in \mathcal{B}|V \subset U}$, and
then $P(\emptyset) = \emptyset$. For any $U \in\topo$, $U = \bigcup_{V \in \mathcal{B}(U)}V$, so if $P$ is an extension of $p$ to $\topo$ as an open preimage function, then $P$ must be unique by (PF2’).
Let $\mathcal{U}\subset \topo$. Since $\bigcup_{U \in \mathcal{U}}\mathcal{B}(U) \subset \mathcal{B}\paren{\bigcup_{U \in \mathcal{U}}U}$, $\bigcup_{U \in \mathcal{U}}P(U) \subset P\paren{\bigcup_{U \in \mathcal{U}}U}$. On the other hand, for each $V \in \mathcal{B}\paren{\bigcup_{U \in \mathcal{U}}U}$, $V \subset \bigcup_{U \in \mathcal{U}}\bigcup_{W \in \mathcal{B}(U)}W$. By (PF2’),
so
Finally, let $U, V \in \topo$, then since $\mathcal{B}(U \cap V) = \mathcal{B}(U) \cap \mathcal{B}(V)$,
On the other hand, by (PF3’),
so $P(U \cap V) = P(U) \cap P(V)$. Therefore $P$ is an open preimage function.$\square$
Post a Comment