Closure of $\mathbf{\Sigma_{n}^{0}} $ under finite cartesian products An exercise in Moschovakis' descriptive set theory, asks to show if we have pointsets $P\subset\mathcal{X}$ and $Q\subset\mathcal{Y}$ both of which are $\mathbf{\Sigma_{n}^{0}}$, then to show that $P\times Q\subset\mathcal{X}\times\mathcal{Y}$ is also $\mathbf{\Sigma_{n}^{0}}$. 
I assume the proof is by induction. The case where $P$ and $Q$ are $\mathbf{\Sigma_{1}^{0}}$ (ie open) is obvious to me, but given $\mathbf{\Sigma_{n}^{0}}$ to be closed under taking such cartesian products, how do I go about showing that $\mathbf{\Sigma_{n+1}^{0}}$ is also closed under taking such cartesian products?
To clarify, the $\mathbf{\Sigma_{n}^{0}}$ are the Borel Pointclasses of finite order, where $\mathbf{\Sigma_{1}^{0}}=$ all open pointsets and $\mathbf{\Sigma_{n+1}^{0}=\exists^{\omega}\neg\Sigma_{n}^{0}}$.
Any help is appreciated, thanks. 
 A: The case where $P\subset\mathcal{X}$ and $Q\subset\mathcal{Y}$ are in $\mathbf{\Sigma^{0}_{1}}$ is obvious since if $P$ and $Q$ are open in $\mathcal{X}$ and $\mathcal{Y}$ resp., then $P\times Q$ is open in $\mathcal{X}\times\mathcal{Y}$. 
Now suppose the claim is true for $\mathbf{\Sigma^{0}_{n}}$, and let $P\subset\mathcal{X}$ and $Q\subset\mathcal{Y}$ in $\mathbf{\Sigma^{0}_{n+1}}$. 
Then $P(x)\iff\exists n\neg A(x,n)$ and $Q(y)\iff\exists m\neg B(y,m)$ where $A\subset\mathcal{X}\times\omega$ and $B\subset\mathcal{Y}\times\omega$ are both in $\mathbf{\Sigma^{0}_{n}}$.
$(\dagger)$ So $(x,y)\in P\times Q\iff\exists n\neg A(x,n)\wedge\exists m\neg B(y,m)\iff\exists t[\neg A(x,(t)_0)\wedge\neg B(y,(t)_1)]\iff\exists t[(x,(t)_0,y,(t)_1)\in\neg A\times\neg B]$
$(\dagger\dagger)$ Now notice that $(x,n,y,m)\notin \neg A\times \neg B\iff (x,n)\notin \neg A\vee (y,m)\notin \neg B\iff (x,n)\in A\vee(y,m)\in B\iff (x,n,y,m)\in A\times\mathcal{Y}\times\omega\vee(x,n,y,m)\in\mathcal{X}\times\omega\times B$
Since $\mathcal{X}\times\omega$ and $\mathcal{Y}\times\omega$ are open in $\mathcal{X}\times\omega$ and $\mathcal{Y}\times\omega$ resp., then they are both in $\mathbf{\Sigma^{0}_{n}}$. Furthermore, since $A$ and $B$ are also in $\mathbf{\Sigma^{0}_{n}}$, then both $A\times\mathcal{Y}\times\omega$ and $\mathcal{X}\times\omega\times B$ are in $\mathbf{\Sigma^{0}_{n}}$ by our induction hypothesis. 
Therefore $A\times\mathcal{Y}\times\omega\vee\mathcal{X}\times\omega\times B$ is in $\mathbf{\Sigma^{0}_{n}}$ since $\mathbf{\Sigma^{0}_{n}}$ is closed under $\vee$. 
Hence $\neg A\times \neg B$ is in $\mathbf{\Pi^{0}_{n}}$ by $(\dagger\dagger)$; and so $P\times Q$ is $\mathbf{\Sigma^{0}_{n+1}}$ by $(\dagger)$ and closure of $\mathbf{\Pi^{0}_{n}}$ under continuous substitution.
