As mentioned in a previous blog post, I've been working on the exercises of chapter 1 of Kunen's Set Theory book. I finally uploaded my solutions this morning.
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Wednesday, July 24 2013
Exercises in Set Theory: Large Cardinals
By fredw on Wednesday, July 24 2013, 22:37
New solutions to exercises from Thomas Jech’s book “Set Theory”:
Saturday, June 1 2013
Exercises in Set Theory: Iterated Forcing and Martin’s Axiom
By fredw on Saturday, June 1 2013, 18:29
New solutions to exercises from Thomas Jech’s book “Set Theory”:
Last November, I tried to provide some details of the proof given in chapter 7, regarding the fact that the continuum hypothesis implies the existence of a Ramsey ultrafilter. Peter Krautzberger pointed out that the proof could probably work assuming only Martin’s Axiom. This was indeed proved by Booth in 1970 and the missing argument is actually given in exercise 16.16. For completeness, I copy the details on this blog post.
Remember that the proof involves contructing a sequence ${(X_{\alpha})}_{{\alpha<2^{{\aleph_{0}}}}}$ of infinite subsets of $\omega$. The induction hypothesis is that at step $\alpha<2^{{\aleph_{0}}}$, for all $\beta_{1},\beta_{2}<\alpha$ we have $\beta_{1}<\beta_{2}\implies X_{{\beta_{2}}}\setminus X_{{\beta_{1}}}$ is finite. It is then easy to show the result for the successor step, since the construction satisfies $X_{{\alpha+1}}\subseteq X_{\alpha}$. However at limit step, to ensure that $X_{\alpha}\setminus X_{\beta}$ is finite for all $\beta<\alpha$, the proof relies on the continunum hypothesis. This is the only place where it is used.
Assume instead Martin’s Axiom and consider a limit step $\alpha<2^{{\aleph_{0}}}$. Define the forcing notion $P_{\alpha}=\{(s,F):s\in{[\omega]}^{{<\omega}},F\in{[\alpha]}^{{<\omega}}\}$ and $(s^{{\prime}},F^{{\prime}})\leq(s,F)$ iff $s\subseteq s^{{\prime}}$, $F\subseteq F^{{\prime}}$ and $s^{{\prime}}\setminus s\subseteq X_{\beta}$ for all $\beta\in F$. It is clear that the relation is reflexive and antisymmetric. The transitivity is almost obvious, just note that if $(s_{3},F_{3})\leq(s_{2},F_{2})$ and $(s_{2},F_{2})\leq(s_{1},F_{1})$ then for all $\beta\in F_{1}\subseteq F_{2}$ we have $s_{3}\setminus s_{1}\subseteq s_{3}\setminus s_{2}\cup s_{2}\setminus s_{1}% \subseteq X_{\beta}$.
The forcing notion satisfies ccc or even property (K): since ${[\omega]}^{{<\omega}}$ is countable, for any uncountable subset $W$ there is $t\in{[\omega]}^{{<\omega}}$ such that $Z=\{(s,F)\in W:s=t\}$ is uncountable. Then any $(t,F_{1}),(t,F_{2})\in Z$ have a common refinement $(t,F_{1}\cup F_{2})$.
For all $n<\omega$, define $D_{n}=\{(s,F):s\geq n\}$. Let $\beta_{1}>\beta_{2}>...>\beta_{k}$ the elements of $F$. We show by induction on $1\leq m\leq k$ that $\bigcap_{{i=1}}^{m}X_{{\beta_{i}}}$ is infinite. This is true for $m=1$ by assumption. If it is true for $m1$ then
$\bigcap_{{i=1}}^{{m1}}X_{{\beta_{i}}}=\bigcap_{{i=1}}^{m}X_{{\beta_{i}}}\cup% \bigcap_{{i=1}}^{m}X_{{\beta_{i}}}\setminus X_{{\beta_{m}}}$ 
The left hand side is infinite by induction hypothesis. The second term of the right hand side is included in $X_{{\beta_{1}}}\setminus X_{{\beta_{m}}}$ and thus is finite. Hence the first term is infinite and the result is true for $m$. Finally, for $m=k$, we get that $\bigcap_{{\beta\in F}}X_{\beta}$ is infinite. Pick $x_{1},x_{2},...,x_{n}$ distinct elements from that set and define $(s^{{\prime}},F^{{\prime}})=(s\cup\{x_{1},...x_{n}\},F)$. We have $(s^{{\prime}},F^{{\prime}})\in D_{n}$, $s\subseteq s^{{\prime}}$, $F\subseteq F^{{\prime}}$ and for all $\beta\in F$, $s^{{\prime}}\setminus s=\{x_{1},...x_{n}\}\subseteq X_{\beta}$. This shows that $D_{n}$ is dense. For each $\beta<\alpha$, the set $E_{\beta}=\{(s,F):\beta\in F\}$ is also dense: for any $(s,F)$ consider $(s^{{\prime}},F^{{\prime}})=(s,F\cup\{\beta\})$.
By Martin’s Axiom there is a generic filter $G$ for the family $\{D_{n}:n<\omega\}\cup\{E_{\beta}:\beta<\alpha\}$ of size $\alpha<2^{{\aleph_{0}}}$. Let $X_{\alpha}=\{n<\omega:\exists(s,F)\in G,n\in s\}$. For all $n<\omega$, there is $(s,F)\in G\cap D_{n}$ and so $X_{\alpha}\geqs\geq n$. Hence $X_{\alpha}$ is infinite. Let $\beta<\alpha$ and $(s_{1},F_{1})\in G\cap E_{\beta}$. For any $x\in X_{\alpha}$, there is $(s_{2},F_{2})\in G$ such that $x\in s_{2}$. Hence there is $(s_{3},F_{3})\in G$ a refinement of $(s_{1},F_{2}),(s_{2},F_{2})$. We have $x\in s_{2}\subseteq s_{3}$ and $s_{3}\subseteq s_{1}\subseteq X_{\beta}$. Hence $X_{\alpha}\setminus X_{\beta}\subseteq s_{1}$ is finite and the induction hypothesis is true at step $\alpha$.
Friday, May 3 2013
Exercises in Set Theory: Applications of Forcing
By fredw on Friday, May 3 2013, 22:04
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 $\beth_{0}=\aleph_{0},\beth_{1}=2^{{\beth_{0}}},\beth_{2}=2^{{\beth_{1}}}...,% \beth_{\omega}=\sup\beth_{n},...,\beth_{{\alpha+1}}=2^{{\beth_{{\alpha}}}},...$ is the normal sequence built by application of the continuum function at successor step. One may wonder: is $\beth_{\alpha}$ regular?
First consider the case where $\alpha$ is limit. The case $\alpha=0$ is clear ($\beth_{0}=\aleph_{0}$ is regular) so assume $\alpha>0$. If $\alpha$ is an inacessible cardinal, it is easy to prove by induction that for all $\beta<\alpha$ we have $\beth_{\beta}<\alpha$: at step $\beta=0$ we use that $\alpha$ is uncountable, at successor step that it is strong limit and at limit step that it is regular. Hence $\beth_{\alpha}=\alpha$ and so is regular. If $\alpha$ is not a cardinal then $\operatorname{cf}(\beth_{\alpha})=\operatorname{cf}(\alpha)\leq\alpha<\alpha% \leq\beth_{\alpha}$ so $\beth_{\alpha}$ is singular. If $\alpha$ is a cardinal but not strong limit then there is $\beta<\alpha$ such that $2^{\beta}\geq\alpha$. Since $\beta<\alpha\leq\beth_{\alpha}$ there is $\gamma<\alpha$ such that $\beth_{\gamma}>\beta$. Then $\beth_{\alpha}\geq\beth_{{\gamma+1}}=2^{{\beth_{\gamma}}}>2^{\beta}\geq\alpha$. So $\operatorname{cf}(\beth_{\alpha})=\operatorname{cf}(\alpha)\leq\alpha<\beth_{\alpha}$ and $\beth_{\alpha}$ is singular. Finally, if $\alpha$ is a singular cardinal, then again $\operatorname{cf}(\beth_{\alpha})=\operatorname{cf}(\alpha)<\alpha\leq\beth_{\alpha}$ and $\beth_{\alpha}$ is singular.
What about the successor case i.e. $\beth_{{\alpha+1}}$? By Corollary 5.3 from Thomas Jech’s book any $\alpha$, we can show that $\aleph_{{\alpha+1}}$ is a regular cardinal. The Generalized Continuum Hypothesis says that $\forall\alpha,\aleph_{\alpha}=\beth_{\alpha}$. Since it holds in $L$ we can not prove in ZFC that for some $\alpha$, $\beth_{{\alpha+1}}$ is singular.
The generic extension $V[G]\supseteq V$ constructed in exercise 15.15 satisfies GCH and so it’s another way to show that $\beth_{{\alpha+1}}$ can not be proved to be singular for some $\alpha$. However, it provides a better result: by construction, $V[G]\models\beth_{{\alpha+1}}^{V}={(\beth_{{\alpha}}^{V})}^{+}$ and so $V[G]\models[\beth_{{\alpha+1}}^{V}\text{ is a regular cardinal}]$. Since ‘‘regular cardinal’’ is a $\Pi_{1}$ notion we deduce that $\beth_{{\alpha+1}}$ is a regular cardinal in $V$.
Now the question is: is there any ‘‘elementary’’ proof of the fact that $\beth_{{\alpha+1}}$ is regular i.e. without using the forcing method?
update: of course, I forgot to mention that by König’s theorem, $2^{{\beth_{{\alpha}}}}=\beth_{{\alpha+1}}\geq\operatorname{cf}(\beth_{{\alpha+% 1}})=\operatorname{cf}(2^{{\beth_{{\alpha}}}})\geq{(\beth_{{\alpha}})}^{+}$ so the singularity of $\beth_{{\alpha+1}}$ would imply the failure of the continuum hypothesis for the cardinal $\beth_{\alpha}$ and this is not provable in ZFC.
Wednesday, March 20 2013
Exercises in Set Theory: Classical Independence Results
By fredw on Wednesday, March 20 2013, 00:01
Here are new solutions to exercises from 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 $L$ (essentially, the definition by ordinal induction provides the wellordering of the class of contructible sets) than to prove that GCH holds in $L$ (you rely on AC in $L$ 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 $\aleph_{2}$ 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:

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

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

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

$x<_{\alpha}y$ implies $x<_{\beta}y$

$x\in L_{\alpha}$ and $y\in L_{\beta}\setminus L_{\alpha}$ implies $x<_{\beta}y$
implies that if $x\in y\in L_{\alpha}$ then $x<_{\alpha}y$. To show that, we consider $\beta\leq\alpha$ the least ordinal such that $y\in L_{\beta}$. In particular, $\beta$ is not limit ($L_{0}=\emptyset$ and if $y\in L_{\beta}$ for some limit $\beta>0$ then there is $\gamma<\beta$ such that $y\in L_{\gamma}$) and we can write it $\beta=\gamma+1$. We have $y\in L_{\beta}=L_{{\gamma+1}}$ so there is a formula $\varphi$ and elements $a_{1},...,a_{n}\in L_{\gamma}$ such that $x\in y=\{z\in L_{\gamma}:(L_{\gamma},\in)\models\varphi(z,a_{1},...,a_{n})\}$. Hence $x\in L_{\gamma}$. Moreover by minimality of $\beta$, $y\in L_{\beta}\setminus L_{\gamma}$ so by (ii) we have $x<_{\beta}y$ and by (i) $x<_{\alpha}y$.


In lemma 14.18, we have expressions that seem illdefined for example $a_{u}(t)$ where $t\notin\operatorname{dom}(a_{u})$. 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 $x,y\in V^{B}$ if $x\subseteq y$ and $\forall t\in\operatorname{dom}(y)\setminus\operatorname{dom}(x),y(t)=0$ then
$\displaystyle\y\subseteq x\$ $\displaystyle=\prod_{{t\in\operatorname{dom}(y)}}\left(y(t)+{\t\in x\}\right)$ $\displaystyle=\prod_{{t\in\operatorname{dom}(x)}}\left(x(t)+{\t\in x\}\right)$ $\displaystyle=\x\subseteq x\=1$ and similarly we get $\x=y\=1$. Then we can use the inequality page 207 ($\\varphi(x)\=\x=y\\cdot\\varphi(x)\\leq\\varphi(y)\=\x=y\\cdot\% \varphi(y)\\leq\\varphi(x)\$) to replace $x$ by its extension $y$.

In lemma 14.23, the inequality
$\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 $\Delta_{0}$ 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 $V^{B}$. This is not an issue since lemma 14.23 is only used much later in lemma 14.31.

Many details could be added to the proof of theorem 14.24, but let’s just mention Powerset. For any $u\in V^{B}$, some $u^{{\prime}}\in\operatorname{dom}(Y)$ is defined and satisfies $\u\subseteq X\\leq\u=u^{{\prime}}\$ (this follows from the definitions, using the Boolean inequality $a+b\leqa+b\cdot a$ to conclude). Since moreover $\forall t\in\operatorname{dom}(Y),Y(t)=1$ we get
$\displaystyle\u\subseteq X\\implies\u\in Y\$ $\displaystyle\geq\u=u^{{\prime}}\+\sum_{{t\in\operatorname{dom}(Y)}}\left(% \u=t\\cdot Y(t)\right)$ $\displaystyle=\sum_{{t\in\operatorname{dom}(Y)}}\left(\u\neq t\\cdot\u=u^{% {\prime}}\\right)$ $\displaystyle\geq\sum_{{t\in\operatorname{dom}(Y)}}\u^{{\prime}}=t\=1$ 
In theorem 14.34, we prove that any $\kappa$ regular in $V$ remains regular in $V[G]$ (the hard case is really $\kappa$ uncountable and this assumption is implicitely used later to say that $\bigcup_{{\alpha<\lambda}}A_{\alpha}$ is bounded). It may not be obvious why this is enough. First recall that for any ordinal $\alpha$, $\operatorname{cf}^{{V[G]}}(\alpha)\leq\operatorname{cf}^{V}(\alpha)$, ${\alpha}^{{V[G]}}\leq{\alpha}^{{V}}$, and any (regular) cardinal in $V[G]$ is a (regular) cardinal in $V$. Next we have,
$\displaystyle\exists\alpha\in\mathrm{Ord},\operatorname{cf}^{{V[G]}}(\alpha)% \leq\operatorname{cf}^{{V}}(\alpha)$ $\displaystyle\implies\exists\alpha\in\mathrm{Ord},\operatorname{cf}^{{V[G]}}(% \operatorname{cf}^{V}(\alpha))\leq\operatorname{cf}^{{V[G]}}(\alpha)<% \operatorname{cf}^{{V}}(\alpha)$ $\displaystyle\implies\exists\beta\textrm{ regular cardinal in }V,\textrm{ not % regular cardinal in }V[G]$ $\displaystyle\implies\exists\beta\in\mathrm{Ord},\operatorname{cf}^{{V[G]}}(% \beta)<\beta=\operatorname{cf}^{V}(\beta)$ that is $\forall\alpha\in\mathrm{Ord},\operatorname{cf}^{{V[G]}}(\alpha)=\operatorname{% cf}^{{V}}(\alpha)$ is equivalent to ‘‘$V$ and $V[G]$ have the same regular cardinals’’. Similarly, we can prove that $\forall\alpha\in\mathrm{Ord},{\alpha}^{{V[G]}}={\alpha}^{{V}}$ is equivalent to ‘‘$V$ and $V[G]$ have the same cardinals’’.
The proof of theorem 14.34 shows that ‘‘$V$ and $V[G]$ have the same regular cardinals’’ and so to complete the proof, it is enough to show that $\forall\alpha,\operatorname{cf}^{{V[G]}}(\alpha)=\operatorname{cf}^{{V}}(\alpha)$ implies $\forall\alpha,{\alpha}^{{V[G]}}={\alpha}^{{V}}$. So suppose $\forall\alpha,\operatorname{cf}^{{V[G]}}(\alpha)=\operatorname{cf}^{{V}}(\alpha)$ and assume that there is $\alpha$ such that ${\alpha}^{{V[G]}}<{\alpha}^{{V}}$. Consider the least such $\alpha$. If $\beta={\alpha}^{{V}}$ then $\beta\leq\alpha$ so ${\beta}^{{V[G]}}\leq{\alpha}^{{V[G]}}<{\alpha}^{{V}}=\beta$. By minimality of $\alpha$, $\beta=\alpha$ and so $\alpha$ is a cardinal in $V$. $\alpha$ is actually regular in $V$. Otherwise, suppose $\operatorname{cf}^{V}(\alpha)<\alpha$ and let $\alpha=\bigcup_{{\beta<\operatorname{cf}^{V}(\alpha)}}X_{\beta}$ such that ${X_{\beta}}^{V}<{\alpha}^{V}$. By minimality of $\alpha$, we have ${\operatorname{cf}^{V}(\alpha)}^{{V[G]}}={\operatorname{cf}^{V}(\alpha)}^{% {V}}$ and ${X_{\beta}}^{{V[G]}}={X_{\beta}}^{{V}}$. Then ${\alpha}^{{V[G]}}={\operatorname{cf}^{V}(\alpha)}^{{V[G]}}\sup_{{\beta<% \operatorname{cf}^{V}(\alpha)}}{X_{\beta}}^{{V[G]}}={\operatorname{cf}^{V}(% \alpha)}^{{V}}\sup_{{\beta<\operatorname{cf}^{V}(\alpha)}}{X_{\beta}}^{{V}}% ={\alpha}^{{V}}$, a contradiction. Finally, we get $\operatorname{cf}^{V}(\alpha)=\alpha={\alpha}^{V}>{\alpha}^{{V[G]}}\geq% \operatorname{cf}^{{V[G]}}(\alpha)$. This is again a contradiction and so $\forall\alpha,{\alpha}^{{V[G]}}={\alpha}^{{V}}$.
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