Let $(X, \tau)$ be a topology space, and assume the topology $\tau$ is generated by a basis $\beta$. Set $$ \mathcal{F} = \sigma(\tau) \equiv \sigma(\{O: O \in \tau\}) $$ and $$ \mathcal{G} = \sigma(\beta) \equiv \sigma(\{B: B \in \beta\}) $$
I wonder if $\mathcal{F} = \mathcal{G}$. Since $\beta \subset \tau$, we always have $\mathcal{G} \subseteq \mathcal{F}$. However, I only am be able to show $\mathcal{F} \subseteq \mathcal{G}$ if $\beta$ is a countable set. Indeed, let $U \in \tau$, then there exists $(B_i)_{i \in I} \subseteq \beta$ such that $$ U = \bigcup_{i \in I} B_i \in \sigma(\beta) = \mathcal{G} $$ Since $\beta$ is countable, the collection $(B_i)_{i \in I}$ must be also countable. Thus, $\tau \subseteq \mathcal{G}$ and therefore $\mathcal{F} \subseteq \mathcal{G}$. My question is to show $\mathcal{F} \subseteq \mathcal{G}$ if $\beta$ is uncountable, or if there is a counterexample to this case. Any hints are appreciated. Thank you