Show that for $\alpha \geq 3$, the class $\Sigma_{\alpha}^0$ in Borel Hierachy is not closed under arbitrary union 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. 
 A: 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:


*

*Every subset of $X$ is $\Sigma^0_\alpha$.


OR


*

*The class of $\Sigma^0_\alpha$ sets is not closed under arbitrary unions. 


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 $\mathbb{R}$. 
