If
- $(X, \mathscr S, \mathscr {B(S)}$, and $(Y, \mathscr T, \mathscr {B(T)}$, are two homeomorphic topological spaces with their associated Borel sigma algebras,
- $\phi: X \to Y$ is a homeomorphism and
- $ \mathscr G$ , is a generating set for $\mathscr {B(S)}$
then is $ \phi(\mathscr G) = \{\phi (G): G \in \mathscr G \} $ a generating set for $\mathscr {B(T)}$ ?
(I have slightly abused notation by keeping the same function name for $\phi(x); \phi(G), \phi(\mathscr G)$, but I think the meaning is clear.)
Why in particular I ask this is it would complete a proof that the complex Borel sigma algebra can be generated from sets of the form
C = {$a + ib: a \in A \in \mathscr A $ and $b \in B \in \mathscr B $ and $\mathscr A, \mathscr B $ generating sets for $\mathscr B( \mathbb R) $ }.
If $Z$ is a set and $\mathscr{A} \subseteq Z$ is a subset, let $\sigma(\mathscr{A})$ denote the $\sigma$-algebra generated by $\mathscr{A}$. Thus $\mathscr{B}(\mathscr{S}) = \sigma(\mathscr{S}) = \sigma(\mathscr{G})$ and $\mathscr{B}(\mathscr{T}) = \sigma(\mathscr{T})$.
Now $\phi(\mathscr{S}) = \mathscr{T}$ because $\phi$ is a homeomorphism, so we can apply this fact twice (with $f = \phi^{-1}$) to obtain the desired result:
$$\sigma(\phi(\mathscr{G})) = \phi(\sigma(\mathscr{G})) = \phi(\sigma(\mathscr{S})) = \sigma(\phi(\mathscr{S})) = \sigma(\mathscr{T}).$$