A question about the hierarchy of sets. My textbook "Introduction to Set Theory" builds a hierarchy of sets in the following manner. It takes $V_0$ to be a set, and defines $V_1$ to be $V_0\bigcup \mathcal{P}(V_0)$, where $\mathcal{P}(V_0)$ is the power set of $V_0$. In general, $V_{n+1}=V_n\bigcup \mathcal{P}(V_n)$. Thus we obtain succesively $V_0,V_1,V_2,\dots$
We know that $\emptyset\in V_1$. Hence, $\{\emptyset\}\in V_2$, and so on. He then goes on to say that we still don't get the infinite set $\{\emptyset,\{\emptyset\},\{\{\emptyset\}\},\dots\}$. For that we have to take the infinite union $V_\omega=V_0\cup V_1\cup\dots$, and then let $V_{\omega+1}=V_\omega\cup \mathcal{P}(V_\omega)$.
Why should we have to define $V_\omega$ in this way? I don't understand why we don't get $\{\emptyset,\{\emptyset\},\{\{\emptyset\}\},\dots\}$ even without it. We are anyway assuming that $V_n$ is getting defined for all $n\in\Bbb{N}$.
Thanks in advance!
 A: Let's just take the case where $V_0 = \varnothing$.  When we do this it follows easily that $V_{n+1} = \mathcal{P} ( V_n )$ for all $n < \omega$.
So let's look at the first few levels of this hierarchy:


*

*$V_0 = \varnothing$;

*$V_1 = \mathcal{P} ( V_0 ) = \mathcal{P} ( \varnothing ) = \{ \varnothing \}$;

*$V_2 = \mathcal{P} ( V_1 ) = \mathcal{P} ( \{ \varnothing \} ) = \{ \varnothing , \{ \varnothing \} \}$;

*$V_3 = \mathcal{P} ( V_2 ) = \mathcal{P} ( \{ \varnothing , \{ \varnothing \} \} ) = \{ \varnothing , \{ \varnothing \} , \{ \{ \varnothing \} \} , \{ \varnothing , \{ \varnothing \} \} \}$.


A couple of simple facts to prove about the first $\omega$ levels of this hierarchy are


*

*$| V_{n+1} | = \overbrace{2^{2^{2^{\cdot^{\cdot^{\cdot^{2}}}}}}} ^{n\text{ }2\text{s}} $, and therefore this is the size of the largest set in $V_{n+2}$.


Therefore the infinite set $A = \{ \varnothing , \{ \varnothing \} , \{ \{ \varnothing \} \} \} , \{ \{ \{ \varnothing \} \} \} , \ldots \}$ cannot appear at in any $V_n$. But each element of $A$ appears in some $V_n$; i.e., $A \subseteq \bigcup_{n < \omega} V_n$, and so $A \in \mathcal{P} ( \bigcup_{n < \omega} V_n )$.  
The point is that at limit stages in the construction we have to do something to allow us to continue forming sets via the power set operation.  There is no real difference between using either of the following templates:


*

*$V_\lambda = \bigcup_{\alpha < \lambda} V_\alpha$;

*$V_\lambda = \mathcal{P} ( \bigcup_{\alpha < \lambda} V_\alpha )$.


(Sets are formed a bit "quicker" using the latter template, but the former template will "catch up".)  Using the former template, we know that new sets are only created at successor stages, which is perhaps a little bit cleaner, and has become the standard.
A: Here's a fun exercise. Start with $V_0=\varnothing$. Now prove that for every $n$, $V_n$ has only finitely many elements. Conclude, if so, that every set in $V_n$ is finite as well.
Therefore every set in $V_\omega$ is finite, but $V_\omega$ itself is infinite. In order to produce an infinite set as an element of the hierarchy, we have to go one more step, and consider $V_{\omega+1}$.
