Suppose $X$ is a metrizable space. Define $\Sigma_1^0$ to be the set of open sets in $X$ and $\prod_1^0$ to be the set of closed sets in $X$.
In this post, I asked whether the class $\Sigma_{\alpha}^0$ closed under arbitrary union. By Pedro's answer, we know the answer is negative. But he did not provide a proof for a general $\alpha$, instead he provided a proof for $\alpha =2.$
I am wondering how would one prove that the class $\Sigma_{\alpha}^0$ is not closed under arbitrary union.
The answer in general depends on $X$. For instance, if we take $X$ to be the metric space with only one point, then each $\Sigma^0_\alpha$ class is closed under arbitrary union. More generally, in any discrete metric space every set is $\Sigma^0_1$, so each $\Sigma^0_\alpha$ class is closed under arbitrary unions.
However, as long as $X$ is sufficiently "rich" (say, $X$ is a perfect Polish space like $\mathbb{R}$), then for $\alpha>1$ the $\Sigma^0_\alpha$ sets are not closed under arbitrary union. This is because every singleton is $\Sigma^0_2$, so every subset of $X$ is a union of $\Sigma^0_2$ sets. More generally, for any $\alpha>1$, every subset of $X$ is a union of $\Sigma^0_\alpha$ sets. So for a given $\alpha$, there are exactly two possibilities:
OR
So if we want to show that the second possibility does hold, it's enough to show that the first possibility doesn't hold. And the easiest way to do this is a cardinality argument. For instance, take $X=\mathbb{R}$. Then (exercise) for each countable ordinal $\alpha$, there are only continuum-many $\Sigma^0_\alpha$ sets; but there are more than continuum-many subsets of $\mathbb{R}$.