Computing the rank funciton of the Well founded universe of sets Definitions: $(1)$ We define the $V_\alpha$ function by transfinite recursion as:
$V_0=\varnothing$; $V_{\alpha+1}=P(V_\alpha)$; Lim$(\lambda)\rightarrow V_\lambda=\bigcup_{\alpha<\lambda}V_\alpha$
$(2)$ The rank function $\rho(x)$ is defined as the least $\tau$ such that $x\in V_{\tau+1}$
Questions: If $\rho(x)=\rho(y)=\alpha$ compute:
(i) $\rho(\{x,y\})$, (ii) $\rho(\langle x,y\rangle)$, (iii) $\rho((\alpha+\omega)\times y)$, (iv) $\rho(^{(\alpha+\omega)}y)$
Answers*(attempt): 
(i) $\rho(\{x\})=\rho(x)+1=\alpha+1$ and $\rho(\{y\})=\rho(y)+1=\alpha+1$, therefore $\rho(\{x,y\})=\alpha+1$
(ii) $\rho(\langle x,y\rangle)=\rho(\{\{x\},\{x,y\}\})$. 
$\rho(\{\{x\}\})=\rho(x)+2=\alpha+2$ and $\rho(\{\{x,y\}\}=\rho(x,y)+2=\alpha+2$
Therefore $\rho(\{\{x\},\{x,y\}\})=\alpha+2$
(iii) Havn't been able to understand this question or the one after, dont really know how to deal with the limit case, I have found that $\rho(\omega)=\omega$, $\rho(\omega\times\omega)=\omega$ which I understand since they both look like limit cases, but then $\rho((\omega+1)\times(\omega+1))=\omega+3$ which I do not understand, but I can acknowledge that it is not like the limit case. Where does the $+3$ come from?
(iv) 
Also if you could explain why $\rho(^xy)=\alpha+3$ it would help me a lot, or just a reference to material explaining it.
*Please verify if they are correct or if there are any mistakes
 A: In (i) it would be better to tie the argument more directly to the definition of rank: $x,y\in V_{\alpha+1}$, so $\{x,y\}\subseteq V_{\alpha+1}$, and therefore $\{x,y\}\in V_{\alpha+2}$. However, this shows only that $\rho\big(\{x,y\}\big)\le\alpha+1$; to complete the argument, you need to show that $\{x,y\}\notin V_{\alpha+1}$. To see this, suppose that $\{x,y\}\in V_{\alpha+1}$. Then $\{x,y\}\in\wp(V_\alpha)$, so $\{x,y\}\subseteq V_\alpha$, and $x,y\in V_\alpha$, contradicting the hypothesis that $\rho(x)=\rho(y)=\alpha$.
In (ii) you don’t want to look at the sets $\big\{\{x\}\big\}$ and $\big\{\{x,y\}\big\}$: they never enter the picture. In fact, since $\{x\}=\{x,x\}$ you can use (i) to say that $\{x\}$ and $\{x,y\}$ both have rank $\alpha+1$, and then use (i) again to say that $\big\{\{x\},\{x,y\}\big\}$ has rank $\alpha+2$.
For (iii) you want the fact that if $\beta$ is an ordinal, then $\rho(\beta)=\beta$; you can prove this by (transfinite) induction on $\beta$. Thus, $\rho(\alpha+\omega)=\alpha+\omega$. Elements of $(\alpha+\omega)\times y$ have the form $\langle\beta,z\rangle$, where $\beta\in\alpha+\omega$ and $z\in y$. For every such ordered pair we have $\rho(\beta)=\beta<\alpha+\omega$ and $\rho(z)<\alpha$. In (ii) you assumed that the components of the ordered pair had the same rank, but if you look more closely at the arguments for (i) and (ii), you’ll see that with a little work you can show in general that $\rho\big(\langle u,v\rangle\big)=\max\{\rho(u),\rho(v)\}+2$. Thus, for each $\langle\beta,z\rangle\in(\alpha+\omega)\times y$ we have
$$\rho\big(\langle\beta,z\rangle\big)=\max\{\beta,\alpha\}+2<\alpha+\omega\;.$$
It follows that $(\alpha+\omega)\times y\subseteq V_{\alpha+\omega}$, $(\alpha+\omega)\times y\in V_{\alpha+\omega+1}$, and hence $\rho\big((\alpha+\omega)\times y\big)\le\alpha+\omega$. I leave it to you to show that in fact $\rho\big((\alpha+\omega)\times y\big)=\alpha+\omega$.
To answer your question about $(\omega+1)\times(\omega+1)$, note that $\rho(\omega+1)=\omega+1$. Members of the Cartesian product are ordered pairs $\langle\alpha,\beta\rangle$, where $\alpha,\beta\le\omega$, and as above we have
$$\rho\big(\langle\alpha,\beta\rangle\big)=\max\{\alpha,\beta\}+2\le\omega+2$$
for each $\langle\alpha,\beta\rangle\in(\omega+1)\times(\omega+1)$. Thus, $(\omega+1)\times(\omega+1)\subseteq V_{\omega+3}$, meaning that $(\omega+1)\times(\omega+1)\in V_{\omega+4}$, and hence that $\rho\big((\omega+1)\times(\omega+1)\big)\le\omega+3$. Here again I’ll leave you to verify that $(\omega+1)\times(\omega+1)\notin V_{\omega+3}$, so that the rank really is $\omega+3$ and not anything smaller. (What is $\rho\big(\langle\omega,\omega\rangle\big)$?)
For (iv) I’ll just address the question about $\rho({^xy})$ when $\rho(x)=\rho(y)=\alpha$. if $f\in{^xy}$, then $f\subseteq x\times y$, so the elements of $f$ are ordered pairs $\langle u,v\rangle$, where $u\in x$ and $v\in y$, so that $\rho(u),\rho(v)<\alpha$. In (ii) we essentially saw that $\rho\big(\langle u,v\rangle\big)=\max\{\rho(u),\rho(v)\}+2$. 
If $\alpha=\beta+1$, $x\in V_{\alpha+1}\setminus V_{\alpha}=V_{\beta+2}\setminus V_{\beta+1}$, so $x\nsubseteq V_\beta$, and there is a $u\in x\setminus V_\beta$. Since $x\subseteq V_{\beta+1}$, this means that $\rho(u)=\beta$. Any $f\in{^xy}$ must contain a pair $\langle u,v\rangle$ for some $v\in y$, and $\rho\big(\langle u,v\rangle\big)=\beta+2=\alpha+1$. It’s not hard to check that this is the maximum rank of any member of $f$, and it follows that $\rho(f)=\alpha+2$ and hence that $\rho({^xy})=\alpha+3$.
If $\alpha$ is a limit ordinal, however, $\rho\big(\langle u,v\rangle\big)<\alpha$ whenever $\langle u,v\rangle\in x\times y$, so $\rho(f)\le\alpha$ for each $f\in{^xy}$. On the other hand, $x$ has elements of arbitrarily large rank less than $\alpha$, so each $f\in{^xy}$ has elements with arbitrarily large ranks less than $\alpha$, and therefore $\rho(f)=\alpha$. Thus, ${^xy}\subseteq V_{\alpha+1}$, and $\rho({^xy})\le\alpha+1$. Moreover, ${^xy}\nsubseteq V_\alpha$, so $\rho({^xy})=\alpha+1$ in this case.
