Borel $\sigma$-algebra of initial topology Let $X$ be a set and $h_i : X \to \mathbb{R}$, $i \in I$ a collection of maps. We can consider on $X$ the initial topology $\tau$ generated by all the $h_i$ and the induced Borel $\sigma$-algebra $\mathcal{B}(\tau)$. We can also consider the $\sigma$-algebra $\Sigma$ on $X$ generated by the maps $h_i$.
Under which topological conditions on the space $(X, \tau)$ can we deduce that $\mathcal{B}(\tau) = \Sigma$? I am searching for rather general conditions that can be used for applications. In my situation, I know that $(X,\tau)$ is at least separable and Hausdorff.
 A: This doesn't answer the question. At first I thought it was clear that the two algebras are the same just by definition; this is an example showing that they can be different.
Say $X$ is uncountable and let $I=X$. For $x\in X$ let $h_x=\chi_{\{x\}}$.
Then singletons are open in the topology generated by the $h_x$; hence every subset of $X$ is open, so the Borel algebra is the power set of $X$.
Otoh the $\sigma$-algebra generated by the $h_x$ is the minimal $\sigma$-algebra such that every singleton is measurable, which is the countable/co-countable algebra.
Hmm. The problem with the trivial argument I had in mind showing that the two are equal is that arbitrary unions of open sets are open. The two algebras will be equal in any situation where you can show that the countable unions of $h_i^{-1}(O)$ ($O$ open) form a topology.
A: In general if $\mathcal A$ is some collection of subsets of some set then let $\rho(\mathcal A)$ denote the topology and let $\sigma(\mathcal A)$ denote the $\sigma$-algebra generated by $\mathcal A$
Denote the topology on $\mathbb R$ by $\mathcal V$ so that $\sigma(\mathcal V)$ denotes the Borel $\sigma$ algebra on $\mathbb R$.
Then  $$\mathcal B(\tau)=\sigma(\rho(\bigcup_{i\in I}h_i^{-1}(\mathcal V)))$$
and:$$\Sigma=\sigma(\bigcup_{i\in I}h_i^{-1}(\sigma(\mathcal V))$$
It can be shown that $h_i^{-1}(\sigma(\mathcal V))=\sigma(h_i^{-1}(\mathcal V))$ for every $i$ and based on that we find a simpler form for $\Sigma$:$$\Sigma=\sigma(\bigcup_{i\in I}h_i^{-1}(\mathcal V))$$
So it is immediate that $\Sigma\subseteq\mathcal B(\tau)$.
The other inclusion will be valid if and only if:  $$\rho(\bigcup_{i\in I}h_i^{-1}(\mathcal V))\subseteq\sigma(\bigcup_{i\in I}h_i^{-1}(\mathcal V))$$
If $\mathcal W$ denotes a subbase for $\mathcal V$ then $\rho(\bigcup_{i\in I}h_i^{-1}(\mathcal V))=\rho(\bigcup_{i\in I}h_i^{-1}(\mathcal W))$.
We can choose $\mathcal W$ to be countable so if $I$ is countable then in $\bigcup_{i\in I}h_i^{-1}(\mathcal W)$ we have a countable subbase of the topology which on its turn implies a countable base of the topology. That implies that every open set can be written as a countable union of sets in  $\sigma(\bigcup_{i\in I}h_i^{-1}(\mathcal V))$.
So by countable $I$ the equality $\mathcal B(\tau)=\Sigma$ is valid.
