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Exercises in Set Theory: Applications of Forcing

New solutions to exercises from Thomas Jech’s book “Set Theory”:

The exercises from this chapter was a good opportunity to play a bit more with the forcing method. Exercise 15.15 seemed a straightforward generalization of Easton’s forcing but turned out to be a bit technical. I realized that the forcing notion used in that exercise provides a result in ZFC (a bit like Exercises 15.31 and 15.32 allow to prove some theorems on Boolean Algebras by Forcing).

Remember that 0=0,1=20,2=21,ω=supn,,α+1=2α,\beth_{0}=\aleph_{0},\beth_{1}=2^{\beth_{0}},\beth_{2}=2^{\beth_{1}}...,\beth_% {\omega}=\sup\beth_{n},...,\beth_{\alpha+1}=2^{\beth_{\alpha}},... is the normal sequence built by application of the continuum function at successor step. One may wonder: is α\beth_{\alpha} regular?

First consider the case where α\alpha is limit. The case α=0\alpha=0 is clear (0=0\beth_{0}=\aleph_{0} is regular) so assume α>0\alpha>0. If α\alpha is an inacessible cardinal, it is easy to prove by induction that for all β<α\beta<\alpha we have β<α\beth_{\beta}<\alpha: at step β=0\beta=0 we use that α\alpha is uncountable, at successor step that it is strong limit and at limit step that it is regular. Hence α=α\beth_{\alpha}=\alpha and so is regular. If α\alpha is not a cardinal then cf(α)=cf(α)|α|<αα\operatorname{cf}(\beth_{\alpha})=\operatorname{cf}(\alpha)\leq|\alpha|<\alpha% \leq\beth_{\alpha} so α\beth_{\alpha} is singular. If α\alpha is a cardinal but not strong limit then there is β<α\beta<\alpha such that 2βα2^{\beta}\geq\alpha. Since β<αα\beta<\alpha\leq\beth_{\alpha} there is γ<α\gamma<\alpha such that γ>β\beth_{\gamma}>\beta. Then αγ+1=2γ>2βα\beth_{\alpha}\geq\beth_{\gamma+1}=2^{\beth_{\gamma}}>2^{\beta}\geq\alpha. So cf(α)=cf(α)α<α\operatorname{cf}(\beth_{\alpha})=\operatorname{cf}(\alpha)\leq\alpha<\beth_{\alpha} and α\beth_{\alpha} is singular. Finally, if α\alpha is a singular cardinal, then again cf(α)=cf(α)<αα\operatorname{cf}(\beth_{\alpha})=\operatorname{cf}(\alpha)<\alpha\leq\beth_{\alpha} and α\beth_{\alpha} is singular.

What about the successor case i.e. α+1\beth_{\alpha+1}? By Corollary 5.3 from Thomas Jech’s book any α\alpha, we can show that α+1\aleph_{\alpha+1} is a regular cardinal. The Generalized Continuum Hypothesis says that α,α=α\forall\alpha,\aleph_{\alpha}=\beth_{\alpha}. Since it holds in LL we can not prove in ZFC that for some α\alpha, α+1\beth_{\alpha+1} is singular.

The generic extension V[G]VV[G]\supseteq V constructed in exercise 15.15 satisfies GCH and so it’s another way to show that α+1\beth_{\alpha+1} can not be proved to be singular for some α\alpha. However, it provides a better result: by construction, V[G]α+1V=(αV)+V[G]\models\beth_{\alpha+1}^{V}={(\beth_{\alpha}^{V})}^{+} and so V[G][α+1V is a regular cardinal]V[G]\models[\beth_{\alpha+1}^{V}\text{ is a regular cardinal}]. Since “regular cardinal” is a Π1\Pi_{1} notion we deduce that α+1\beth_{\alpha+1} is a regular cardinal in VV.

Now the question is: is there any “elementary” proof of the fact that α+1\beth_{\alpha+1} is regular i.e. without using the forcing method?

–update: of course, I forgot to mention that by König’s theorem, 2α=α+1cf(α+1)=cf(2α)(α)+2^{\beth_{\alpha}}=\beth_{\alpha+1}\geq\operatorname{cf}(\beth_{\alpha+1})=% \operatorname{cf}(2^{\beth_{\alpha}})\geq{(\beth_{\alpha})}^{+} so the singularity of α+1\beth_{\alpha+1} would imply the failure of the continuum hypothesis for the cardinal α\beth_{\alpha} and this is not provable in ZFC.