Additive Class is closed under arbitrary union?

Question

Suppose $X$ is a metrizable space. In Kechris's Descriptive Set Theory book, page $68$, he defined that, for any countable $\xi,$ the additive class as $\Sigma_{\xi}^0 = \{ \cup_{n\in \mathbb{N}}A_n: A_n \in \prod_{\zeta_n}^0, \zeta_n < \xi, n \in \mathbb{N} \}$ .

Note that $\Sigma_1^0$ is the set of open sets in $X$, while $\prod_1^0$ is the set of closed sets in $X$.

Question: Is the additive class closed under arbitrary union?

Answer 1

No. It is enough to note the following two facts:

1. $\Sigma_2^0(\mathbb{R})=F_\sigma$ contains all the singletons;
2. $\Pi_2^0(\mathbb{R})=G_\delta = \{X\setminus A : A \in \Sigma_2^0(\mathbb{R})\}$ does not comprise all subsets of $\mathbb{R}$.

This leads to contradiction if you assume $\Sigma_2^0(\mathbb{R})$ is closed under arbitrary union.

To see 2. above, notice that $\mathbb{Q}\notin \Pi_2^0(\mathbb{R})$: any countable subset of $\mathbb{R}$ (or any perfect space) is first category, but a dense $G_\delta$ should not.