For some background: I'm reading up on the transfinite hierarchy of Borel sets.
Given a Polish space $(X,d)$, put:
$$\Sigma^0_1=\{U\subset X: U \ \text{is open}\}$$
$$\Pi^0_1=\{F\subset X: F \ \text{is closed}\}$$
$$\Sigma_\alpha^0=\left\{ \bigcup_{n=1}^\infty A_n: \text{each}\ A_n \in \Pi_\beta^0 \ \text{for some $\beta<\alpha$}\right \}$$
$$\Pi_\alpha^0=\left\{ \bigcap_{n=1}^\infty A_n:\text{each}\ A_n \in \Sigma_\beta^0 \ \text{for some $\beta<\alpha$}\right \}$$
Most books go on to claim that if $\alpha <\beta$, then:
$$\Sigma_\alpha^0\subset \Sigma_\beta^0\quad \Sigma_\alpha^0\subset\Pi_\beta^0\quad \Pi_\alpha^0\subset\Pi_\beta^0\quad \Pi_\alpha^0\subset \Sigma_\beta^0$$
However, they just prove it for $\alpha=1,\beta=2$, and then just say "it follows by induction". I've tried to prove the inclusion $\Sigma_\alpha^0\subset \Sigma_\beta^0$ by myself but I don't see the need for induction. My reasoning is as follows:
If $1<\alpha<\beta$, then it is trivial, for if $A\in\Sigma_\alpha^0$, then $A=\bigcup_n A_n$, with $A_n\in \Pi_\alpha^0$ for some $\gamma<\alpha$. But then certainly $A\in \Sigma^o_\beta$.
If $\alpha=1$ and $\beta=2$, then $\Sigma_1^0\subset\Sigma_2^0$ follows from the fact that every open set is $F_\sigma$.
Finally, if $1=\alpha<2<\beta$, then use the preceeding two cases to write: $\Sigma_1^0\subset\Sigma_2^0\subset\Sigma_\beta^0$.
The inclusion $\Pi_\alpha^0\subset \Pi_\beta^0$ is similar. To see $\Pi_\alpha^0\subset \Sigma_\beta^0$, take $A\in \Pi_\alpha^0$ and put $A_n=A$ for $n\in \mathbb{Z}^+$. Then $A=\bigcup_nA_n$, and hence $A\in\Sigma_\alpha^0$. An entirely analogous argument shows that $\Sigma_\alpha^0\subset\Pi_\beta^0$.
Is this correct? Or have I missed something that makes an inductive argument necessary? I see how the first part has a certain inductive taste to it, so maybe that’s what the book meant.
Thanks in advance!