Set theory - Chapter 6: The Axiom of Regularity

6.1

zx(rank(z)<rank(x))\forall z\in x(\operatorname{rank}(z)<\operatorname{rank}(x)) so zx(rank(z)+1rank(x))\forall z\in x(\operatorname{rank}(z)+1\leq\operatorname{rank}(x)) and finally supzx(rank(z)+1)rank(x){\sup_{{z\in x}}(\operatorname{rank}(z)+1)}\leq\operatorname{rank}(x). Conversely, for all zxz\in x, zVrank(z)+1Vsupzx(rank(z)+1)z\in V_{{\operatorname{rank}(z)+1}}\subseteq V_{{\sup_{{z\in x}}(\operatorname% {rank}(z)+1)}}. Hence xVsupzx(rank(z)+1)+1x\in V_{{{\sup_{{z\in x}}(\operatorname{rank}(z)+1)}+1}} i.e. rank(x)supzx(rank(z)+1)\operatorname{rank}(x)\leq{\sup_{{z\in x}}(\operatorname{rank}(z)+1)}.

6.2

By induction on n<ωn<\omega, we show that |Vn||V_{n}| is finite (using 3.1 (iii)). Hence |Vω|n<ω|Vn|=0supn<ω|Vn|02=0|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, |Vω|0|V_{\omega}|\geq\aleph_{0} since ωVω\omega\subseteq V_{\omega}. Finally |Vω|=0|V_{\omega}|=\aleph_{0}.

Now we shall show by induction on α\alpha that Vω+α=αV_{{\omega+\alpha}}=\beth_{\alpha}. We saw above that it is true for α=0\alpha=0. If it is the case for an ordinal α\alpha, then |Vω+α+1|=|𝒫(Vω+α)|=2α=α+1|V_{{\omega+\alpha+1}}|=|\operatorname{\mathcal{P}}(V_{{\omega+\alpha}})|=2^{{% \beth_{\alpha}}}=\beth_{{\alpha+1}}. Finally, consider a limit ordinal λ\lambda such that the property holds below it. Then |Vω+λ|=|α<λVω+αα<λ|Vω+α|=λsupα<λα=λλ|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 λλ\beth_{\lambda}\geq\lambda and |Vω+λ|=λ|V_{{\omega+\lambda}}|=\beth_{\lambda}.

6.3

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

|Vκ|α<κ|Vω+α|=κsupα<κακ2=κ|V_{\kappa}|\leq\sum_{{\alpha<\kappa}}|V_{{\omega+\alpha}}|=\kappa\sup_{{% \alpha<\kappa}}\beth_{\alpha}\leq\kappa^{2}=\kappa

6.4

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

6.5

We can write =2/R\mathbb{R}=\mathbb{N}^{2}/R, =×*/R2\mathbb{Q}=\mathbb{R}\times\mathbb{R}^{*}/R_{2} and =/I\mathbb{R}=\mathbb{Q}^{\mathbb{N}}/I where R1,R2R_{1},R_{2} are relations of equivalence on 2\mathbb{N}^{2} and ×*\mathbb{R}\times\mathbb{R}^{*} and II is the ideal of {u:unn+0}\{u\in\mathbb{Q}^{\mathbb{N}}:u_{n}\underset{n\rightarrow+\infty}{% \longrightarrow}0\}. Hence we have:

  1. (1)

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

  2. (2)

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

  3. (3)

    rank()rank(𝒫(𝒫(×*)))(ω+2+2)+2=ω+6\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)

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

6.6

Let BB be the class of all xx that are hereditarily in AA. We have B={xA:xB}B=\{x\in A:x\subseteq B\}.

  1. (i))

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

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

  2. (ii))

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

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