Set theory - Chapter 6: The Axiom of Regularity

$\forall z\in x(\operatorname{rank}(z)<\operatorname{rank}(x))$ so $\forall z\in x(\operatorname{rank}(z)+1\leq\operatorname{rank}(x))$ and finally ${\sup_{{z\in x}}(\operatorname{rank}(z)+1)}\leq\operatorname{rank}(x)$. Conversely, for all $z\in x$, $z\in V_{{\operatorname{rank}(z)+1}}\subseteq V_{{\sup_{{z\in x}}(\operatorname{rank}(z)+1)}}$. Hence $x\in V_{{{\sup_{{z\in x}}(\operatorname{rank}(z)+1)}+1}}$ i.e. $\operatorname{rank}(x)\leq{\sup_{{z\in x}}(\operatorname{rank}(z)+1)}$.

By induction on $n<\omega$, we show that $|V_{n}|$ is finite (using 3.1 (iii)). Hence $|V_{\omega}|\leq\sum_{{n<\omega}}|V_{n}|=\aleph_{0}\sup_{{n<\omega}}|V_{n}|\leq\aleph_{0}^{2}=\aleph_{0}$. (Note that we can apply Lemma 5.8, since only the first term is equal to zero). Moreover, $|V_{\omega}|\geq\aleph_{0}$ since $\omega\subseteq V_{\omega}$. Finally $|V_{\omega}|=\aleph_{0}$.

Now we shall show by induction on $\alpha$ that $V_{{\omega+\alpha}}=\beth_{\alpha}$. We saw above that it is true for $\alpha=0$. If it is the case for an ordinal $\alpha$, then $|V_{{\omega+\alpha+1}}|=|\operatorname{\mathcal{P}}(V_{{\omega+\alpha}})|=2^{{\beth_{\alpha}}}=\beth_{{\alpha+1}}$. Finally, consider a limit ordinal $\lambda$ such that the property holds below it. Then $|V_{{\omega+\lambda}}|=|\bigcup_{{\alpha<\lambda}}V_{{\omega+\alpha}}\leq\sum_{{\alpha<\lambda}}|V_{{\omega+\alpha}}|=\lambda\sup_{{\alpha<\lambda}}\beth_{\alpha}=\lambda\beth_{\lambda}$. But Beth is a normal function, so $\beth_{\lambda}\geq\lambda$ and $|V_{{\omega+\lambda}}|=\beth_{\lambda}$.

For all $\kappa$ we have $\kappa\subseteq V_{\kappa}$ so $\kappa\leq|V_{\kappa}|$. Suppose that $\kappa$ is inaccessible. Then by induction on $\alpha<\kappa$ we have $\beth_{\alpha}<\kappa$. For $\alpha=0$, it is true because $\kappa$ is uncountable. If the property holds for an $\alpha$ then it also holds for $\alpha+1$ since $\kappa$ is strong limit. The regularity of $\kappa$ allows to prove the property at limit step. Finally we have :

Let $x,y$ be two sets of rank at most $\alpha$. Then by 6.1, $\operatorname{rank}(\{x,y\})=\sup\{\operatorname{rank}(x)+1,\operatorname{rank}(y)+1\}\leq\alpha+1$. $\operatorname{rank}(x,y)=\operatorname{rank}(\{\{x\},\{x,y\}\})$. But we’ve just seen that the rank of $\{x\}$ and $\{x,y\}$ is at most $\alpha+1$. So $\operatorname{rank}(x,y)\leq\alpha+2$. If $t\in x\cup y$, then $\operatorname{rank}(t)<\inf{\{\operatorname{rank}(x),\operatorname{rank}(y)\}}\leq\alpha$ so $\operatorname{rank}(t)+1\leq\alpha$ and $\operatorname{rank}(x\cup y)\leq\alpha$. If $t\in\bigcup x$ then there is $z\in x$ such that $t\in z$. So $\operatorname{rank}(t)+1\leq\operatorname{rank}(z)<\operatorname{rank}(x)\leq\alpha$ that is $\operatorname{rank}(\bigcup x)<\alpha$. If $z\in\operatorname{\mathcal{P}}(x)$ then for all $t\in z$, $t\in x$ and so $\operatorname{rank}(t)+1\leq\operatorname{rank}(x)\leq\alpha$. Hence $\operatorname{rank}z\leq\alpha$ and finally $\operatorname{rank}(\operatorname{\mathcal{P}}(z))\leq\alpha+1$. $x^{y}\subseteq y\times x=\{(a,b):a\in y\wedge a\in x\}$. For $(a,b)\in x^{y}$, we have $\operatorname{rank}(a,b)\leq\alpha+1$ and finally $\operatorname{rank}(x^{y})\leq\alpha+2$.

We can write $\mathbb{R}=\mathbb{N}^{2}/R$, $\mathbb{Q}=\mathbb{R}\times\mathbb{R}^{*}/R_{2}$ and $\mathbb{R}=\mathbb{Q}^{\mathbb{N}}/I$ where $R_{1},R_{2}$ are relations of equivalence on $\mathbb{N}^{2}$ and $\mathbb{R}\times\mathbb{R}^{*}$ and $I$ is the ideal of $\{u\in\mathbb{Q}^{\mathbb{N}}:u_{n}\underset{n\rightarrow+\infty}{\longrightarrow}0\}$. Hence we have:

1. (1)

   $\operatorname{rank}(\mathbb{N})=\operatorname{rank}(\omega)=\omega$

2. (2)

   $\operatorname{rank}(\mathbb{R})\leq\operatorname{rank}(\operatorname{\mathcal{P}}(\operatorname{\mathcal{P}}(\mathbb{N}^{2})))\leq(\omega+2)+2=\omega+4$.

3. (3)

   $\operatorname{rank}(\mathbb{Q})\leq\operatorname{rank}(\operatorname{\mathcal{P}}(\operatorname{\mathcal{P}}(\mathbb{R}\times\mathbb{R}^{*})))\leq(\omega+2+2)+2=\omega+6$

4. (4)

   $\operatorname{rank}(\mathbb{R})\leq\operatorname{rank}(\operatorname{\mathcal{P}}(\operatorname{\mathcal{P}}(\mathbb{Q}^{\mathbb{N}})))\leq(\omega+6+2)+2=\omega+10$

Let $B$ be the class of all $x$ that are hereditarily in $A$. We have $B=\{x\in A:x\subseteq B\}$.

1. (i))

   The statement in the book is incorrect (for example if $A=\{\{\emptyset\}\}$ then $B=\emptyset$ but $TC(\emptyset)=\emptyset\subseteq A$). Instead, we will prove that $x\in B$ iff $TC({\{x\}})\subseteq A$.

   First, if $x\in B$ then by definition $x\in A$ and, because $B$ is transitive, $TC(x)\subseteq B\subseteq A$. Hence $TC({\{x\}})=\{x\}\cup TC(x)\subseteq A$. Conversely, we show by induction that if $TC({\{x\}})\subseteq A$ then $x\in B$. For $x=\emptyset$, we have $TC({\{\emptyset\}})={\{\emptyset\}}$. So $TC({\{\emptyset\}})\subseteq A$ means $\emptyset\in A$. In that case, because $\emptyset\subseteq B$ is always true so we have $\emptyset\in B$. Otherwise suppose that the result is true for all elements of $x$ and that $TC({\{x\}})=\{x\}\cup TC(x)\subseteq A$. Then $x\in A$ and $\forall y\in x,TC({\{y\}})\subseteq TC(x)\subseteq A$. Hence by induction hypothesis, $\forall y\in x,y\in B$ and so $x\subseteq B$. Finally, $x\in B$.

2. (ii))

   By definition, $B$ is a transitive class included in $A$. Suppose that $T$ is another such class. If $x\in T$ then $TC({\{x\}})\subseteq T\subseteq A$ and by (i) we have $x\in B$. Finally $T\subseteq B$ and $B$ is the largest transitive class included in $A$.