Does this set have cardinality $\beth_\omega$? And if so, how does it require the Axiom of Replacement to construct?

Question

I was reading this Wikipedia article: https://en.wikipedia.org/wiki/Von_Neumann_universe, and it mentions that the Axiom of Replacement is required to go outside of $V_{\omega+\omega}$, one of the levels of the von Neumann Hierarchy. If I'm correct, this means that the Axiom of Replacement would be required to construct a set of cardinality $\beth_\omega$, since such sets would only exist in higher levels of the hierarchy.

But now let the set $N_0 = \mathbb N$, and let $N_{i+1} = P(N_i)$, where $P(N_i)$ is the power set of $N_i$. Now let the set $A$ be the union of all $N_i$ for each $i \in \mathbb N$. Now it seems to me that $A$ has cardinality $\beth_\omega$. Certainly, $N_i$ has cardinality $\beth_i$, and $A$ cannot have cardinality $\beth_m$ for any natural number $m$, because $A$ contains $N_{i+1}$, which has a strictly larger cardinality. So the cardinality of $A$ must be higher than $\beth_m$ for all $m\in \mathbb N$. Further, sets of cardinality $\beth_{\omega + 1}$ or higher can be easily constructed by taking $P(A)$ and so on.

If $A$ does not have cardinality $\beth_\omega$, then how so? And if it does, where in this construction is the Axiom of Replacement invoked?

$N_0$ (or something analogous) exists by the Axiom of Infinity. Then all other $N_i$ exist by the Axiom of Power Set. Admittedly, there is then some subtlety in applying the Axiom of Union, since sets must be in a larger set together before a union can be taken. One might try to construct $A$, or at least a similar set, by repeatedly using the Axiom of Pairing and the Axiom of Union, but this doesn't seem to work, admittedly.

But I'm still unsure how exactly adding the Axiom of Replacement solves this problem. If we want to invoke the Axioms of ZFC explicitly in the construction of $A$, where and how does the Axiom of Replacement come into play?

Answer 1

The axiom of replacement is invoked when you take the union of all $N_i.$ Formally, that is $\bigcup\{N_i:i\in \omega\},$ so given the axiom of union, the problem is to construct $\{N_i:i\in \omega\}$.

It is true that $N_0$ exists by axiom of infinity, and that then $N_1:=P(N_0)$ exists by power set, and then $N_2 := P(N_1)$ exists by power set, and so on. It's this "and so on" where the subtlety lies. When we do this carefully, it is a definition by recursion, and when we work it out, we have proved a statement "for all $i\in \omega,$ there is a unique set $N_i$ such that $N_0=\omega$ and for all $i$, $N_{i+1} =P(N_i)$"$^*$. How do we show that the range of this "class function" $i\mapsto N_i$ is a set? Well... that exactly what replacement says.

More generally, replacement is typically used at limit stages of definitions by transfinite recursion, to collect the results thus far into a set, so that the recursion can be continued (or as in this case, if we just want to stop and collect the results into a set to take the union).

$^*$More precisely, we prove by induction $\forall i\in\omega\;\exists ! y \;\phi(i,y)$ where $\phi(i,y)$ says that there is a function $f$ with domain $i+1$ such that $f(i)=y$ and $f(0)=\omega$ and for all $j < i,$ $f(j+1) = P(f(j)).$

Answer 2

Yes, this set (if it exists) has cardinality $\beth_\omega$.

To take the union of the $N_i$, you have to show that the set $\{N_i\mid i\in\omega\}$ exists. You have defined a function with domain $\omega$, $i\mapsto N_i$ (without assuming replacement, this is just a function defined by a formula, i.e. a "class function", not a set of ordered pairs). The axiom of replacement is exactly what lets you turn this function definition into the set $\{N_i\mid i\in\omega\}$.

Answer 3

Let the set $A$ be the union of all $N_i$ for each $i\in\mathbb{N}$.

The existence of such a union is precisely what we need Replacement for! Without replacement we have no way of going from the formula assigning to each $i\in\omega$ the corresponding $N_i$ to the single set gotten by combining all the $N_i$s.