If $A \subset \mathcal{N}^2$ is a $\mathbf{\Sigma}^0_\alpha$ set, then $\{x : (x,x) \in A\}$ is also $\mathbf{\Sigma}^0_\alpha$.

Question

This is the boldface Borel hierarchy on Baire space.

Jech states this with a "clearly". What am I missing that makes the statement completely obvious?

I clearly have zero intuition for this material.

Answer 1

HINT: Show by induction on $\alpha$ that the preimage of a $\Sigma^0_\alpha$ set under a continuous function is again $\Sigma^0_\alpha$. This uses continuity trivially for the base case, and for the induction step the fact that preimage commutes with $\cap $ and $\cup$; note that this is false for image!

---

Note that by contrast, $\{(x, x): (x, x)\in A\}$ is not $\Sigma^0_1$ necessarily even if $A$ is; but this does work for $\alpha>1$. The key point is that for $\beta<\alpha$, $\Pi^0_\beta\subseteq \Sigma^0_\alpha$, and each pointclass $\Sigma^0_\alpha$ is closed under finite intersections.