$\newcommand{\scrO}{\mathcal{O}}$ $\newcommand{\scrC}{\mathcal{C}}$ Proposition: Let $\Omega$ be a metric space and let $\scrO$ and $\scrC$ be the collection of all open and closed sets in $\Omega$. Show that $\sigma(\scrO) = \sigma(\scrC)$; i.e the the Borel $\sigma$-field in $\Omega$ is equal to the minimal $\sigma$-field generated by the closed sets.
I feel like the question should require more work, but the attempt I have is really short. What is the problem with the following proof attempt?
Attempt: Because $\sigma(\scrO)$ is a $\sigma$-field it is closed under complementation, and since the complement of the open sets are the closed sets it follows that $\scrC \subseteq \sigma(\scrO)$. But then by definition of the minimal $\sigma$-field it follows that $\sigma(\scrC) \subseteq \sigma(\scrO)$. On the other hand $\sigma(\scrC)$ is a $\sigma$-field and so closed under complements, and by the same logic it follows that $\scrO \subseteq \sigma(\scrC)$ and so $\sigma(\scrO) \subseteq \sigma(\scrC)$.
I know that the minimal $\sigma$-fields always exist, so there isn't an issue there. But where would the error be? Thanks in advance for the clarification.