Borel hierarchy

Remarks

The class $\Pi^0_1$ is the class of closed sets, $\Sigma^0_2$ is the class of $F_\sigma$ sets, and $\Pi^0_2$ is the class of $G_\delta$ sets. In $X = \R$, the set $\Q$ is in $\Sigma^0_2$, as the countable union of the closed sets $\set{q}$. It is not in $\Pi^0_2$, since by the Baire category theorem a dense $G_\delta$ subset of $\R$ is uncountable. Every class satisfies $\Sigma^0_\alpha \cup \Pi^0_\alpha \subseteq \Delta^0_{\alpha + 1}$, and the union of all the classes $\Sigma^0_\alpha$ is the Borel sigma-algebra of $X$. A non-example: an analytic set that is not Borel lies in no class of the hierarchy. If $X$ is uncountable, then $\Sigma^0_\alpha \neq \Pi^0_\alpha$ for every $\alpha$, so the hierarchy does not collapse. The same definition works in any metrizable space; metrizability is used to show that closed sets are $G_\delta$ sets.