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Exercises in Set Theory: Classical Independence Results

I worked on Chapter 13 and 14 of Thomas Jech’s book ‘‘Set Theory’’.

Doing the exercises from these chapters gave me the opportunity to come back to the ‘‘classical’’ results about the independence of the Axiom of Choice and (Generalized) Continuum Hypothesis by Kurt Gödel and Paul Cohen. It’s funny to note that it’s easier to prove that AC holds in LL (essentially, the definition by ordinal induction provides the well-ordering of the class of contructible sets) than to prove that GCH holds in LL (you rely on AC in LL and on the technical condensation lemma). Actually, I believe Gödel found his proof for AC one or two years after the one for GCH. On the other hand, it is easy to make GCH fails (just add 2\aleph_{2} Cohen reals by Forcing) but more difficult to make AC fails (e.g. AC is preserved by Forcing). This can be interpreted as AC being more ‘‘natural’’ than GCH.

After reading the chapters again and now I analyzed in details the claims, I’m now convinced about the correctness of the proof. There are only two points I didn’t verify precisely about the Forcing method (namely that all axioms of predicate calculus and rules of inference are compatible with the Forcing method ; that the Forcing/Generic Model theorems can be transported from the Boolean Algebra case to the general case) but these do not seem too difficult. Here are some notes about claims that were not obvious to me at the first reading. As usual, I hope they might be useful to the readers of that blog:

  1. 1.

    In the first page of chapter 13, it is claimed that for any set MM, Mdef(M)M\in\mathrm{def}(M) and Mdef(M)𝒫(M)M\subseteq\mathrm{def}(M)\subseteq\operatorname{\mathcal{P}}(M). The first statement is always true because M={xM:(M,)x=x}def(M)M=\{x\in M:(M,\in)\models x=x\}\in\mathrm{def}(M) and (x=x)(M,){(x=x)}^{(M,\in)} is x=xx=x by definition. However, the second statement can only be true if MM is transitive (since that implies M𝒫(M)M\subseteq\operatorname{\mathcal{P}}(M)). Indeed, if MM is transitive then for all aMa\in M we have aMa\subseteq M and since xax\in a is Δ0\Delta_{0} we get a={xM:(M,)xa}def(M)a=\{x\in M:(M,\in)\models x\in a\}\in\mathrm{def}(M). If moreover we consider xXdef(M)x\in X\in\mathrm{def}(M) then xXMx\in X\subseteq M so xMdef(M)x\in M\subseteq\mathrm{def}(M) and def(M)\mathrm{def}(M) is also transitive. Hence the transitivity of the LαL_{\alpha} can still be shown by ordinal induction.

  2. 2.

    The proof of lemma 13.7 can not be done exactly by induction on the complexity of GG, as suggested. For example to prove (ii) for G=G2=×G=G_{2}=\cdot\times\cdot, we would consider uF()×H()φ(u)aF(),bH(),φ((a,b))\exists u\in F(...)\times H(...)\varphi(u)\Leftrightarrow\exists a\in F(...),% \exists b\in H(...),\varphi((a,b)) and would like to say that φ((a,b))\varphi((a,b)) is Δ0\Delta_{0}. Nevertheless, we can not deduce that from the induction hypothesis. Hence the right thing to do is to prove the lemma for G1={,}G_{1}=\{\cdot,\cdot\} first and deduce the lemma for G=(,)G=(\cdot,\cdot) (and G=(,,)G^{\prime}=(\cdot,\cdot,\cdot)). Then we can proceed by induction.

  3. 3.

    In the proof of theorem 13.18, it is mentioned that the assumption

    1. (i)

      x<αyx<_{\alpha}y implies x<βyx<_{\beta}y

    2. (ii)

      xLαx\in L_{\alpha} and yLβLαy\in L_{\beta}\setminus L_{\alpha} implies x<βyx<_{\beta}y

    implies that if xyLαx\in y\in L_{\alpha} then x<αyx<_{\alpha}y. To show that, we consider βα\beta\leq\alpha the least ordinal such that yLβy\in L_{\beta}. In particular, β\beta is not limit (L0=L_{0}=\emptyset and if yLβy\in L_{\beta} for some limit β>0\beta>0 then there is γ<β\gamma<\beta such that yLγy\in L_{\gamma}) and we can write it β=γ+1\beta=\gamma+1. We have yLβ=Lγ+1y\in L_{\beta}=L_{\gamma+1} so there is a formula φ\varphi and elements a1,,anLγa_{1},...,a_{n}\in L_{\gamma} such that xy={zLγ:(Lγ,)φ(z,a1,,an)}x\in y=\{z\in L_{\gamma}:(L_{\gamma},\in)\models\varphi(z,a_{1},...,a_{n})\}. Hence xLγx\in L_{\gamma}. Moreover by minimality of β\beta, yLβLγy\in L_{\beta}\setminus L_{\gamma} so by (ii) we have x<βyx<_{\beta}y and by (i) x<αyx<_{\alpha}y.

  4. 4.

    In lemma 14.18, we have expressions that seem ill-defined for example au(t)a_{u}(t) where tdom(au)t\notin\operatorname{dom}(a_{u}). This happens in other places, like lemma 14.17 or definition 14.27. The trick is to understand that the functions are extended by 0. Indeed, for any x,yVBx,y\in V^{B} if xyx\subseteq y and tdom(y)dom(x),y(t)=0\forall t\in\operatorname{dom}(y)\setminus\operatorname{dom}(x),y(t)=0 then

    yx=tdom(y)(-y(t)+tx)=tdom(x)(-x(t)+tx)=xx=1\begin{gathered}\displaystyle\|y\subseteq x\|\\ \displaystyle=\prod_{t\in\operatorname{dom}(y)}\left(-y(t)+{\|t\in x\|}\right)% \\ \displaystyle=\prod_{t\in\operatorname{dom}(x)}\left(-x(t)+{\|t\in x\|}\right)% \\ \displaystyle=\|x\subseteq x\|=1\end{gathered}

    and similarly we get x=y=1\|x=y\|=1. Then we can use the inequality page 207 (φ(x)=x=yφ(x)φ(y)=x=yφ(y)φ(x)\|\varphi(x)\|=\|x=y\|\cdot\|\varphi(x)\|\leq\|\varphi(y)\|=\|x=y\|\cdot\|% \varphi(y)\|\leq\|\varphi(x)\|) to replace xx by its extension yy.

  5. 5.

    In lemma 14.23, the inequality

    x is an ordinalxαˇ+x=αˇ+αˇx\|x\text{ is an ordinal}\|\leq\|x\in\check{\alpha}\|+\|x=\check{\alpha}\|+\|% \check{\alpha}\in x\|

    seems obvious but I don’t believe that it can be proved so easily at that point. For example the proof from chapter 2 requires at least the Separation axiom and the Δ0\Delta_{0} formulation from chapter 10 is based on the Axiom of Regularity. To solve that issue, it seems to me that the lemma should be moved after the proof that axioms of ZFC are valid in VBV^{B}. This is not an issue since lemma 14.23 is only used much later in lemma 14.31.

  6. 6.

    Many details could be added to the proof of theorem 14.24, but let’s just mention Powerset. For any uVBu\in V^{B}, some udom(Y)u^{\prime}\in\operatorname{dom}(Y) is defined and satisfies uXu=u\|u\subseteq X\|\leq\|u=u^{\prime}\| (this follows from the definitions, using the Boolean inequality -a+b-a+ba-a+b\leq-a+b\cdot a to conclude). Since moreover tdom(Y),Y(t)=1\forall t\in\operatorname{dom}(Y),Y(t)=1 we get

    uXuY\displaystyle\|u\subseteq X\|\implies\|u\in Y\| -u=u+tdom(Y)(u=tY(t))\displaystyle\geq-\|u=u^{\prime}\|+\sum_{t\in\operatorname{dom}(Y)}\left(\|u=t% \|\cdot Y(t)\right)
    =tdom(Y)-(utu=u)\displaystyle=\sum_{t\in\operatorname{dom}(Y)}-\left(\|u\neq t\|\cdot\|u=u^{% \prime}\|\right)
    tdom(Y)u=t=1\displaystyle\geq\sum_{t\in\operatorname{dom}(Y)}\|u^{\prime}=t\|=1
  7. 7.

    In theorem 14.34, we prove that any κ\kappa regular in VV remains regular in V[G]V[G] (the hard case is really κ\kappa uncountable and this assumption is implicitely used later to say that α<λAα\bigcup_{\alpha<\lambda}A_{\alpha} is bounded). It may not be obvious why this is enough. First recall that for any ordinal α\alpha, cfV[G](α)cfV(α)\operatorname{cf}^{V[G]}(\alpha)\leq\operatorname{cf}^{V}(\alpha), |α|V[G]|α|V{|\alpha|}^{V[G]}\leq{|\alpha|}^{V}, and any (regular) cardinal in V[G]V[G] is a (regular) cardinal in VV. Next we have,

    αOrd,cfV[G](α)cfV(α)\displaystyle\exists\alpha\in\mathrm{Ord},\operatorname{cf}^{V[G]}(\alpha)\leq% \operatorname{cf}^{V}(\alpha) αOrd,cfV[G](cfV(α))cfV[G](α)<cfV(α)\displaystyle\implies\exists\alpha\in\mathrm{Ord},\operatorname{cf}^{V[G]}(% \operatorname{cf}^{V}(\alpha))\leq\operatorname{cf}^{V[G]}(\alpha)<% \operatorname{cf}^{V}(\alpha)
    β regular cardinal in V, not regular cardinal in V[G]\displaystyle\implies\exists\beta\textrm{ regular cardinal in }V,\textrm{ not % regular cardinal in }V[G]
    βOrd,cfV[G](β)<β=cfV(β)\displaystyle\implies\exists\beta\in\mathrm{Ord},\operatorname{cf}^{V[G]}(% \beta)<\beta=\operatorname{cf}^{V}(\beta)

    that is αOrd,cfV[G](α)=cfV(α)\forall\alpha\in\mathrm{Ord},\operatorname{cf}^{V[G]}(\alpha)=\operatorname{cf% }^{V}(\alpha) is equivalent to ‘‘VV and V[G]V[G] have the same regular cardinals’’. Similarly, we can prove that αOrd,|α|V[G]=|α|V\forall\alpha\in\mathrm{Ord},{|\alpha|}^{V[G]}={|\alpha|}^{V} is equivalent to ‘‘VV and V[G]V[G] have the same cardinals’’.

    The proof of theorem 14.34 shows that ‘‘VV and V[G]V[G] have the same regular cardinals’’ and so to complete the proof, it is enough to show that α,cfV[G](α)=cfV(α)\forall\alpha,\operatorname{cf}^{V[G]}(\alpha)=\operatorname{cf}^{V}(\alpha) implies α,|α|V[G]=|α|V\forall\alpha,{|\alpha|}^{V[G]}={|\alpha|}^{V}. So suppose α,cfV[G](α)=cfV(α)\forall\alpha,\operatorname{cf}^{V[G]}(\alpha)=\operatorname{cf}^{V}(\alpha) and assume that there is α\alpha such that |α|V[G]<|α|V{|\alpha|}^{V[G]}<{|\alpha|}^{V}. Consider the least such α\alpha. If β=|α|V\beta={|\alpha|}^{V} then βα\beta\leq\alpha so |β|V[G]|α|V[G]<|α|V=β{|\beta|}^{V[G]}\leq{|\alpha|}^{V[G]}<{|\alpha|}^{V}=\beta. By minimality of α\alpha, β=α\beta=\alpha and so α\alpha is a cardinal in VV. α\alpha is actually regular in VV. Otherwise, suppose cfV(α)<α\operatorname{cf}^{V}(\alpha)<\alpha and let α=β<cfV(α)Xβ\alpha=\bigcup_{\beta<\operatorname{cf}^{V}(\alpha)}X_{\beta} such that |Xβ|V<|α|V{|X_{\beta}|}^{V}<{|\alpha|}^{V}. By minimality of α\alpha, we have |cfV(α)|V[G]=|cfV(α)|V{|\operatorname{cf}^{V}(\alpha)|}^{V[G]}={|\operatorname{cf}^{V}(\alpha)|}^{V} and |Xβ|V[G]=|Xβ|V{|X_{\beta}|}^{V[G]}={|X_{\beta}|}^{V}. Then |α|V[G]=|cfV(α)|V[G]supβ<cfV(α)|Xβ|V[G]=|cfV(α)|Vsupβ<cfV(α)|Xβ|V=|α|V{|\alpha|}^{V[G]}={|\operatorname{cf}^{V}(\alpha)|}^{V[G]}\sup_{\beta<% \operatorname{cf}^{V}(\alpha)}{|X_{\beta}|}^{V[G]}={|\operatorname{cf}^{V}(% \alpha)|}^{V}\sup_{\beta<\operatorname{cf}^{V}(\alpha)}{|X_{\beta}|}^{V}={|% \alpha|}^{V}, a contradiction. Finally, we get cfV(α)=α=|α|V>|α|V[G]cfV[G](α)\operatorname{cf}^{V}(\alpha)=\alpha={|\alpha|}^{V}>{|\alpha|}^{V[G]}\geq% \operatorname{cf}^{V[G]}(\alpha). This is again a contradiction and so α,|α|V[G]=|α|V\forall\alpha,{|\alpha|}^{V[G]}={|\alpha|}^{V}.