580 -Cardinal arithmetic (10)

Let me begin with a couple of comments that may help clarify some of the results from last lecture.

First, I want to show a different proof of Lemma 21.2, that I think is cleaner than the argument I gave before. (The argument from last lecture, however, will be useful below, in the proof of Kunen’s theorem.)

Lemma 1 If ${\kappa}$ is measurable, ${{\mathcal U}}$ is a ${\kappa}$ -complete nonprincipal ultrafilter over ${\kappa,}$ and ${j_{\mathcal U}:V\rightarrow M}$ is the corresponding ultrapower embedding, then ${{}^\kappa M\subset M.}$

Proof: Recall that if ${\pi}$ is Mostowski’s collapsing function and ${[\cdot]}$ denotes classes in ${V^\kappa/{\mathcal U},}$ then ${M=\{\pi([f]):f\in{}^\kappa V\}.}$ To ease notation, write ${\langle f\rangle}$ for ${\pi([f]).}$

Let ${h:\kappa\rightarrow M.}$ Pick ${f:\kappa\rightarrow V}$ such that for all ${\alpha<\kappa,}$ ${h(\alpha)=\langle f(\alpha)\rangle.}$

Lemma 2 With notation as above, ${\langle f\rangle=j_{\mathcal U}(f)(\langle{\rm id}\rangle)}$ for any ${f:\kappa\rightarrow V.}$

Proof: For a set ${X}$ let ${c_X:\kappa\rightarrow V}$ denote the function constantly equal to ${X.}$ Since ${\pi}$ is an isomorphism, ${\mbox{\L o\'s}}$ ‘s lemma gives us that the required equality holds iff

$\displaystyle \{\alpha<\kappa : f(\alpha)=((c_f)(\alpha))({\rm id}(\alpha))\}\in{\mathcal U},$

but this last set is just ${\{\alpha<\kappa:f(\alpha)=f(\alpha)\}=\kappa.}$ $\Box$

From the nice representation just showed, we conclude that ${\langle f(\alpha)\rangle=j_{\mathcal U}(f(\alpha))(\langle{\rm id}\rangle)}$ for all ${\alpha<\kappa.}$ But for any such ${\alpha,}$ ${j_{\mathcal U}(f(\alpha))=j_{\mathcal U}(f)(\alpha)}$ because ${{\rm cp}(j_{\mathcal U})=\kappa}$ by Lemma 21 from last lecture. Hence, ${h=(j_{\mathcal U}(f)(\alpha)(\langle{\rm id}\rangle):\alpha<\kappa),}$ which is obviously in ${M,}$ being definable from ${j_{\mathcal U}(f),}$ ${\langle{\rm id}\rangle,}$ and ${\kappa.}$ $\Box$

The following was shown in the proof of Lemma 20, but it deserves to be isolated.

Lemma 3 If ${{\mathcal U}}$ is a normal nonprincipal ${\kappa}$ -complete ultrafilter over the measurable cardinal ${\kappa,}$ then ${{\mathcal U}=\{X\subseteq\kappa:\kappa\in i_{\mathcal U}(X)\},}$ i.e., we get back ${{\mathcal U}}$ when we compute the normal measure derived from the embedding induced by ${{\mathcal U}.}$ ${\Box}$

Finally, the construction in Lemma 10 and preceeding remarks is a particular case of a much more general result.

Definition 4 Given ${f:I\rightarrow J}$ and an ultrafilter ${{\mathcal D}}$ over ${I,}$ the projection ${f_*({\mathcal D})}$ of ${{\mathcal D}}$ over ${J}$ is the set of ${X\subseteq J}$ such that ${f^{-1}(X)\in{\mathcal D}.}$

Clearly, ${f_*({\mathcal D})}$ is an ultrafilter over ${J.}$

Notice that if ${\kappa={\rm add}({\mathcal D}),}$ ${(X_\alpha:\alpha<\kappa)}$ is a partition of ${I}$ into sets not in ${{\mathcal D},}$ and ${f:I\rightarrow\kappa}$ is given by ${f(x)=}$ the unique ${\alpha}$ such that ${x\in X_\alpha,}$ then ${f_*({\mathcal D})}$ is a ${\kappa}$ -complete nonprincipal ultrafilter over ${\kappa.}$ (Of course, ${\kappa=\omega}$ is possible.)

For a different example, let ${{\mathcal U}}$ be a ${\kappa}$ -complete nonprincipal ultrafilter over the measurable cardinal ${\kappa,}$ and let ${f:\kappa\rightarrow\kappa}$ represent the identity in the ultrapower by ${{\mathcal U},}$ ${\langle f\rangle=\kappa.}$ Then ${f_*({\mathcal U})}$ is the normal ultrafilter over ${\kappa}$ derived from the embedding induced by ${{\mathcal U}.}$

Definition 5 Given ultrafilters ${{\mathcal U}}$ and ${{\mathcal V}}$ (not necessarily over the same set), say that ${{\mathcal U}}$ is Rudin-Keisler below ${{\mathcal V},}$ in symbols, ${{\mathcal U}\le_{RK}{\mathcal V},}$ iff there are sets ${S\in{\mathcal U},}$ ${T\in{\mathcal V},}$ and a function ${f:T\rightarrow S}$ such that ${{\mathcal U}\upharpoonright S=f_*({\mathcal V}\upharpoonright T).}$

Theorem 6 Let ${{\mathcal U}}$ be an ultrafilter over a set ${X}$ and ${{\mathcal V}}$ an ultrafilter over a set ${Y.}$ Suppose that ${{\mathcal U}\le_{RK}{\mathcal V}.}$ Then there is an elementary embedding ${j:V^X/{\mathcal U}\rightarrow V^Y/{\mathcal V}}$ such that ${j\circ i_{\mathcal U}=i_{\mathcal V}.}$

Proof: Fix ${T\in{\mathcal U}}$ and ${S\in{\mathcal V}}$ for which there is a map ${f:S\rightarrow T}$ such that ${{\mathcal U}\upharpoonright T=f_*({\mathcal V}\upharpoonright S).}$ Clearly, ${V^X/{\mathcal U}\cong V^T/({\mathcal U}\upharpoonright T)}$ as witnessed by the map ${[f]_{\mathcal U}\mapsto[f\upharpoonright T]_{{\mathcal U}\upharpoonright T},}$ and similarly ${V^Y/{\mathcal V}\cong V^S/({\mathcal V}\upharpoonright S),}$ so it suffices to assume that ${S=Y}$ and ${T=X.}$

Given ${h:X\rightarrow V,}$ let ${h_*:Y\rightarrow V}$ be given by ${h_*=h\circ f.}$ Then ${j([h]_{\mathcal U})=[h_*]_{\mathcal V}}$ is well-defined, elementary, and ${j\circ i_{\mathcal U}=i_{\mathcal V}.}$

In effect, ${h=_{\mathcal U}h'}$ iff ${\{x\in X:h(x)=h'(x)\}\in{\mathcal U}}$ iff ${\{y\in Y:h\circ f(y)=h'\circ f(y)\}\in{\mathcal V}}$ iff ${h_*=_{\mathcal V}h'_*,}$ where the second equivalence holds by assumption, and it follows that ${j}$ is well-defined.

If ${c_B^A}$ denotes the function with domain ${A}$ and constantly equal to ${B,}$ then for any ${x,}$ ${j\circ i_{\mathcal U}(x)=j([c^X_x]_{\mathcal U})=[(c^X_x)_*]_{\mathcal V}=[c^Y_x]_{\mathcal V}=i_{\mathcal V}(x)}$ since ${(c^X_x)_*=c^Y_x}$ by definition of the map ${h\mapsto h_*.}$ This shows that ${j\circ i_{\mathcal U}=i_{\mathcal V}.}$

Elementarity is a straightforward modification of the proof of Lemma 10 from last lecture. $\Box$

One can show that Theorem 6 “very nearly” characterizes the Rudin-Keisler ordering, see for example Proposition 0.3.2 in Jussi Ketonen, Strong compactness and other cardinal sins, Annals of Mathematical Logic 5 (1972), 47–76.

3. Large cardinals and ${{\sf GCH}}$

With the preliminary technical machinery established last lecture, we are ready to show how the presence of large cardinals in the universe may influence its cardinal arithmetic.

Theorem 7 (Scott) Let ${\kappa}$ be measurable and let ${{\mathcal U}}$ be a normal nonprincipal, ${\kappa}$ -complete measure on ${\kappa.}$ If ${\alpha<\kappa}$ and ${\{\rho<\kappa:2^\rho\le\rho^{+\alpha}\}\in{\mathcal U},}$ then ${2^\kappa\le\kappa^{+\alpha}.}$ In particular, a measurable cardinal cannot be the first counterexample to ${{\sf GCH}.}$ Similarly, if ${\{\rho<\kappa:2^\rho\le\rho^{+\rho}\}\in{\mathcal U},}$ then ${2^\kappa\le\kappa^{+\kappa}}$ and if ${\{\rho<\kappa:2^\rho<\rho^{+\rho}\}\in{\mathcal U},}$ then ${2^\kappa<\kappa^{+\kappa}.}$

Proof: Let ${j:V\rightarrow M}$ be the ultrapower embedding derived from ${{\mathcal U}.}$ If ${\alpha<\kappa}$ and ${\{\rho<\kappa:2^\rho\le\rho^{+\alpha}\}\in{\mathcal U},}$ then ${M\models 2^{\pi([{\rm id}])}\le\pi([{\rm id}])^{+j(\alpha)},}$ by ${\mbox{\L o\'s}}$ ‘s lemma. But ${j(\alpha)=\alpha}$ since ${{\rm cp}(j)=\kappa,}$ and ${\pi([{\rm id}])=\kappa}$ by Lemma 20 from last lecture, since ${{\mathcal U}}$ is normal.

This shows that, in ${M,}$ ${2^\kappa\le\kappa^{+\alpha}.}$ By Lemma 21 from last lecture, we know that ${{\mathcal P}(\kappa)\in M,}$ so ${2^\kappa\le(2^\kappa)^M,}$ and ${(\kappa^{+\alpha})^M\le\kappa^{+\alpha}}$ since ${M\subseteq V.}$ It follows that ${2^\kappa\le\kappa^{+\alpha}}$ in ${V.}$

The other two statements are shown similarly. $\Box$

The assumption that ${{\mathcal U}}$ is normal cannot be eliminated in the previous theorem. For example, Gitik showed that it is consistent to have a measurable cardinal ${\kappa}$ and a stationary set ${S\subset\kappa}$ such that ${2^\nu=\nu^+}$ for all ${\nu\in S}$ but ${2^\kappa=\kappa^{++}.}$ On the other hand, if ${2^\nu=\nu^+}$ holds in a club below ${\kappa}$ , then it also holds at ${\kappa,}$ since any normal ultrafilter extends the club filter.

Definition 8 Denote by ${(\aleph'_\alpha:\alpha\in{\sf ORD})}$ the increasing enumeration of the fixed points of the ${\aleph}$ function.

The following is the first result we obtain about fixed points of the aleph sequence:

Theorem 9 Suppose that ${\kappa}$ is measurable and that ${\lambda=\aleph'_{\kappa+\kappa}}$ is strong limit. Then ${2^\lambda<\aleph'_{(2^\kappa)^+}.}$

Proof: Let ${{\mathcal U}}$ be a ${\kappa}$ -complete nonprincipal ultrafilter over ${\kappa,}$ and let ${j:V\rightarrow M={\rm Ult}(V,{\mathcal U})}$ be the corresponding elementary embedding. The result follows from general facts about continuity points of ${j.}$

Lemma 10 With notation as above, assume that ${\tau}$ is a limit ordinal.

If ${{\rm cf}(\tau)=\kappa,}$ then ${j(\tau)>\sup_{\alpha<\tau}j(\alpha)}$ while if ${{\rm cf}(\tau)\ne\kappa,}$ then ${j(\tau)=\sup_{\alpha<\tau}j(\alpha).}$

If ${\tau}$ is strong limit and ${{\rm cf}(\tau)\ne\kappa,}$ then ${j(\tau)=\tau,}$ while if ${{\rm cf}(\tau)=\kappa,}$ then ${2^\tau<j(\tau).}$

Proof: To prove 1, we need to determine whether there is some function ${f:\kappa\rightarrow\lambda}$ such that ${\sup_{\alpha<\tau}j(\alpha)\le\langle f\rangle < j(\tau).}$

If ${{\rm cf}(\tau)=\kappa,}$ let ${(\tau_\alpha:\alpha<\kappa)}$ be increasing and cofinal in ${\tau,}$ and take ${f(\alpha)=\tau_\alpha}$ for all ${\alpha<\kappa.}$

If ${{\rm cf}(\tau)>\kappa,}$ then any ${f:\kappa\rightarrow\tau}$ is bounded.

If ${{\rm cf}(\tau)<\kappa,}$ the ${\kappa}$ -completeness of ${{\mathcal U}}$ implies that there is some ${\beta<\tau}$ such that ${\{\alpha<\kappa:f(\alpha)<\beta\}\in{\mathcal U}.}$

To prove 2, notice first that ${\tau}$ is closed under ${j}$ : If ${\alpha<\tau,}$ then ${j(\alpha)<|{}^\kappa\alpha|^+<\tau,}$ where the first inequality follows from the fact that any ordinal below ${j(\alpha)}$ is ${\langle f\rangle}$ for some ${f:\kappa\rightarrow\alpha.}$

The result now follows from item 1: If ${{\rm cf}(\tau)\ne\kappa,}$ then ${j(\tau)=\sup_{\alpha<\tau}j(\alpha)\le\tau,}$ but ${\tau\le j(\tau)}$ always holds.

If ${{\rm cf}(\tau)=\kappa,}$ note that ${\tau\le\sup_{\alpha<\tau}j(\alpha)< j(\tau)}$ so, in particular, ${\tau<j(\tau).}$ Also, ${2^\tau=\tau^\kappa,}$ since ${\tau}$ is strong limit. Since ${{}^\kappa M\subset M,}$ then ${\tau^\kappa\le(\tau^\kappa)^M\le(\tau^{j(\kappa)})^M.}$ If ${\tau=\kappa,}$ that ${2^\kappa<j(\kappa)}$ was shown in Lemma 21 from last lecture. If ${\kappa<\tau,}$ then ${j(\kappa)<\tau,}$ and ${(\tau^{j(\kappa)})^M<j(\tau),}$ since ${j(\tau)}$ is strong limit in ${M.}$ $\Box$

Let now ${\lambda=\aleph'_{\kappa+\kappa}}$ and suppose that ${\lambda}$ is strong limit. Then, by item 2 of Lemma 10, ${2^\lambda<j(\lambda).}$ But ${j(\lambda)=(\aleph'_{j(\kappa)+j(\kappa)})^M\le\aleph'_{j(\kappa)+j(\kappa)}<\aleph'_{(2^\kappa)^+}}$ since ${j(\kappa)<(2^\kappa)^+}$ by Lemma 21 from last lecture. $\Box$

Before we continue, let me remark that the continuity results above depend in an essential way on the fact that the embedding ${j}$ being considered is an ultrapower embedding, and they do not need to hold for other elementary maps.

As mentioned at the end of last lecture, the embeddings ${j:V\rightarrow M}$ that measurability provides are rather restrictive since, for example, ${V_{{\rm cp}(j)+2}\not\subset M.}$ The results above show that measurability influences cardinal arithmetic, so it is reasonable to expect that stronger large cardinal assumptions will provide us with stronger results. This is indeed the case, as the following example illustrates. Next lecture, I will show another result along these lines, deriving a significantly stronger conclusion from the existence of a significantly stronger cardinal.

Definition 11 Let ${\alpha}$ be an ordinal. A cardinal ${\kappa}$ is ${\alpha}$ -strong iff there is an elementary embedding ${j:V\rightarrow M}$ with ${{\rm cp}(j)=\kappa}$ and ${V_{\kappa+\alpha}\subset M.}$

${\kappa}$ is strong iff it is ${\alpha}$ -strong for all ${\alpha.}$

Clearly, ${\kappa}$ is measurable iff it is ${0}$ -strong iff it is ${1}$ -strong, but already ${2}$ -strength of ${\kappa}$ is more significant than measurability as it implies that ${{\mathcal U}\in M}$ for any ultrafilter on ${\kappa,}$ so in particular the measurable cardinals below ${\kappa}$ form a stationary set.

The standard presentation of strength adds the extra requirement that ${\alpha<j(\kappa).}$ This is not necessary; in fact, if ${\kappa}$ is ${(\alpha+2)}$ -strong as defined above, then it is ${(\alpha+2)}$ -strong under the stronger requirement of this paragraph. This is a consequence of the following famous result of Kunen:

Theorem 12 (Kunen) Suppose that ${j:V\rightarrow M}$ is elementary. Then ${V\ne M.}$

Proof: There are several nice proofs of this result. Kunen’s argument uses that any infinite set ${x}$ admits an ${\omega}$ -Jónsson function, that is, a map ${f:[x]^{\aleph_0}\rightarrow x}$ such that whenever ${A\subseteq x}$ has the same size as ${x,}$ then ${f''[A]^{\aleph_0}=x.}$ Here, as usual, ${[t]^\tau=\{y\subseteq t:|y|=\tau\}.}$ This is a result of ${\mbox{Erd\H os}}$ and Hajnal.

Zapletal has found a proof that uses a modicum of pcf theory, namely, Shelah’s result that any singular cardinal ${\lambda}$ admits a scale, that is, there is an increasing sequence ${(\lambda_\alpha:\alpha<{\rm cf}(\lambda))}$ of regular cardinals cofinal in ${\lambda,}$ and a sequence ${(f_\beta:\beta<\lambda^+)}$ of functions such that ${f_\beta\in\prod_\alpha\lambda_\alpha}$ for all ${\beta<\lambda^+;}$ whenever ${\beta<\gamma<\lambda^+,}$ then ${f_\beta<_{b,{\rm cf}(\lambda)}f_\gamma;}$ and whenever ${f\in\prod_\alpha\lambda_\alpha,}$ then ${f<_{b,{\rm cf}(\lambda)}f_\beta}$ for some ${\beta.}$ Here, ${<_{b,{\rm cf}(\lambda^+)}}$ is the partial order induced by the ideal of bounded subsets of ${{\rm cf}(\lambda^+),}$ see Definition 13 in lecture II.6.

There are a few other arguments as well. I present here a proof due to Woodin. All arguments begin by noticing that if

$\displaystyle \lambda=\sup_{n<\omega}j^n(\kappa)$

then ${\lambda=j(\lambda).}$ Here, ${j^0(\kappa)=\kappa}$ and ${j^{n+1}(\kappa)=j(j^n(\kappa)).}$ To see this, apply ${j}$ to both sides of the displayed equation:

$\displaystyle j(\lambda)=\sup_{n<\omega}j(j^n(\kappa))=\lambda.$

It follows that ${j(\lambda^+)=\lambda^+,}$ because ${\lambda^+\le j(\lambda^+)=(\lambda^+)^M\le\lambda^+.}$ Therefore, ${j[\lambda^+]}$ is cofinal in ${\lambda^+.}$

Let

$\displaystyle C=\{\xi<\lambda^+:\xi=j(\xi)\mbox{ and }{\rm cf}(\xi)=\omega\}.$

The same argument that shows that ${\lambda}$ is fixed shows that ${C}$ is unbounded: For any ${\alpha<\lambda^+}$ consider the sequence ${\alpha+\kappa,j(\alpha+\kappa),j^2(\alpha+\kappa),\dots}$ Its supremum is in ${C.}$

Also, ${C}$ is ${\omega}$ -closed. (A set ${A\subseteq\lambda^+}$ is ${\omega}$ -closed iff whenever ${\alpha\cap A}$ is unbounded in ${\alpha<\lambda^+}$ and ${{\rm cf}(\alpha)=\omega,}$ then ${\alpha\in A.}$ ) This is simply because the supremum of an ${\omega}$ -sequence of fixed points of ${j}$ is necessarily fixed by ${j.}$

But any ${\omega}$ -closed unbounded subset ${S}$ of an ordinal ${\gamma}$ of uncountable cofinality contains the restriction to ${S^\gamma_\omega}$ of a club, namely, the club of limit points of elements of ${S.}$ In particular, this holds of ${C.}$ It follows that if ${T\subseteq S^{\lambda^+}_\omega}$ is stationary, then ${C\cap T\ne\emptyset.}$

Woodin’s argument now invokes Ulam’s result that every stationary subset of a successor cardinal ${\tau}$ can be split into ${\tau}$ many disjoint stationary sets, this is Theorem 13 in lecture II.5. Consequently, we can partition ${S^{\lambda^+}_\omega=\{\alpha<\lambda^+:{\rm cf}(\alpha)=\omega\}}$ into ${\kappa}$ many stationary pieces. Let ${f:\kappa\rightarrow{\mathcal P}(\lambda^+)}$ be such a partition. By elementarity, ${j(f):j(\kappa)\rightarrow{\mathcal P}^M(\lambda^+)}$ is (in ${M}$ ) a partition of ${S^{\lambda^+}_\omega}$ into ${j(\kappa)}$ many stationary pieces. The result will follow by deriving a contradiction from the assumption that ${j(f)(\kappa)}$ is stationary in ${V.}$

For suppose that it is. It follows that ${j(f)(\kappa)\cap f(\alpha)}$ is stationary for some ${\alpha<\kappa,}$ since the club filter on ${\lambda^+}$ is ${\lambda^+}$ -complete and ${\lambda^+>\kappa.}$ But then ${j(f)(\kappa)\cap f(\alpha)}$ must meet ${C.}$ Let ${\xi}$ be in this intersection. Then ${\xi=j(\xi),}$ ${\xi\in f(\alpha)}$ and ${\xi\in j(f)(\kappa).}$ But then ${\xi=j(\xi)\in j(f(\alpha))=j(f)(\alpha),}$ so ${j(f)(\alpha)\cap j(f)(\kappa)\ne\emptyset,}$ contradicting that ${j(f)}$ is a partition. $\Box$

The argument can actually be tightened to prove the following two stronger results:

Theorem 13 (Kunen) There is no elementary ${j:V\rightarrow M}$ such that ${V_{\lambda+2}\subset M,}$ where ${\lambda}$ is the first fixed point of ${j}$ past its critical point.

Proof: It is easy to code (in an absolute way) subsets of ${\lambda^+}$ by elements of ${V_{\lambda+2},}$ so Woodin’s proof of Kunen’s Theorem 12 goes through for ${j:V\rightarrow M,}$ since the sets it makes reference to are subsets of ${\lambda^+,}$ and we have a contradiction.

To see that there is such a coding, simply notice (as in the appendix in lecture I.5) that there is a definable bijection between ${\lambda}$ and ${\lambda\times\lambda,}$ that the map from a well-ordering of ${\lambda}$ to its order type gives via this bijection a (definable) surjection from ${{\mathcal P}(\lambda)}$ onto ${\lambda^+,}$ and that the inverse of this surjection is a (definable) injection of ${{\mathcal P}(\lambda^+)}$ into ${{\mathcal P}({\mathcal P}(\lambda))\subset V_{\lambda+2}.}$ $\Box$

Theorem 14 (Kunen) If ${j:V\rightarrow M}$ is elementary and ${\lambda}$ is the first fixed point of ${j}$ past ${\kappa={\rm cp}(j),}$ then ${j[\lambda]\notin M.}$

Proof: Suppose otherwise, and define an ultrafilter ${{\mathcal U}}$ over ${Z={\mathcal P}(\lambda)}$ by

$\displaystyle {\mathcal U}=\{A\subseteq {\mathcal P}(\lambda):j[\lambda]\in j(A)\}.$

Then ${{\mathcal U}}$ is a ${\sigma}$ -complete ultrafilter, and we can form the ultrapower embedding ${i=i_{\mathcal U}:V\rightarrow N={\rm Ult}(V,{\mathcal U}).}$

Lemma 15 ${{\rm cp}(i)=\kappa,}$ ${i(\lambda)=\lambda,}$ ${i[\lambda]\in N}$ and ${i[{\mathcal P}(\lambda)]\in N.}$

Proof: Begin by noticing that ${j[\lambda]=\langle {\rm id}\rangle.}$ In effect: Let ${f:{\mathcal P}(\lambda)\rightarrow V}$ be arbitrary such that ${\langle f\rangle\in\langle{\rm id}\rangle.}$ This occurs iff ${\{B\subseteq\lambda: f(B)\in B\}\in{\mathcal U},}$ which (by definition) holds iff ${j(f)(j[\lambda])=j(\gamma)}$ for some ${\gamma<\lambda.}$ But then ${\{B\subseteq \lambda:f(B)=\gamma\}\in{\mathcal U},}$ and ${\langle f\rangle=j(\gamma).}$ We easily conclude that ${j[\lambda]=\langle {\rm id}\rangle,}$ and ${\langle f\rangle=j(f)(j[\lambda])}$ for all ${f.}$

Letting ${k:N\rightarrow M}$ be given by ${k(\langle f\rangle)=j(f)(j[\lambda])}$ is now easily checked to be well-defined and elementary, and ${k\circ i_{\mathcal U}=j}$ .

Note now that if ${\alpha\le\lambda,}$ and ${f:{\mathcal P}(\lambda)\rightarrow V}$ is given by ${f(B)={\rm ot}(B\cap\alpha),}$ then ${\langle f\rangle=j(f)(j[\lambda])={\rm ot}(j[\lambda]\cap j(\alpha))=\alpha.}$

It follows that ${k\upharpoonright(\lambda+1)={\rm id}}$ by the definition of ${k}$ and the same argument. Therefore, ${i\upharpoonright(\lambda+1)=j\upharpoonright(\lambda+1).}$ In particular, ${{\rm cp}(i)=\kappa,}$ ${i(\lambda)=\lambda,}$ and ${i[\lambda]=j[\lambda]\in N.}$

Finally, let ${X\subseteq\lambda.}$ Then ${i(X)=i(\bigcup_{\alpha<\lambda}X\cap\alpha)=\bigcup_{\alpha<\lambda}i(X\cap\alpha),}$ since ${i(\lambda)=\lambda.}$ Note that ${i[\lambda]\in N}$ implies that ${{}^\lambda N\subset N,}$ by the proof of Lemma 21.2 from last lecture, so ${{\mathcal P}(\lambda)\in N}$ and ${i\upharpoonright V_\lambda\in N,}$ since ${|V_\lambda|=\lambda.}$ But then

$\displaystyle i[{\mathcal P}(\lambda)]=\{\bigcup_{\alpha<\lambda}i(X\cap\alpha):X\in{\mathcal P}(\lambda)\}\in N,$

as required. $\Box$

It follows from the lemma that ${{}^{2^\lambda}N\subset N,}$ again by the proof of Lemma 21.2 from last lecture. But then ${V_{\lambda+2}\subset N,}$ since any element of ${V_{\lambda+2}}$ has size at most ${2^\lambda,}$ and we have a contradiction by Theorem 13. $\Box$

Kunen’s result shows that the definition of strong cardinals given above coincides with the standard one.

Corollary 16 A cardinal ${\kappa}$ is strong iff for all ${\alpha}$ there is an elementary embedding ${j:V\rightarrow M}$ with ${{\rm cp}(j)=\kappa,}$ ${V_\alpha\subset M,}$ and ${\alpha<j(\kappa).}$ In fact, if ${\kappa}$ is ${(\alpha+2)}$ -strong, then there is an elementary ${j:V\rightarrow M}$ with critical point ${\kappa}$ such that ${V_{\kappa+\alpha+2}\subset M}$ and ${\kappa+\alpha+2<j(\kappa).}$

Proof: Suppose that ${\kappa}$ is ${(\alpha+2)}$ -strong as witnessed by ${j:V\rightarrow M.}$ By Theorem 13, ${\kappa+\alpha<\lambda,}$ where ${\lambda}$ is the first fixed point of ${j}$ past ${\kappa,}$ so ${\kappa+\alpha}$ (and therefore also ${\kappa+\alpha+2}$ ) is strictly below ${j^n(\kappa)}$ for some ${n<\omega.}$

Extend ${j}$ to proper classes by setting ${j(N)=\bigcup_\beta j(N\cap V_\beta)}$ for all classes ${N.}$ Define ${M_k}$ for ${k\le n}$ by ${M_0=V}$ and ${M_{l+1}=j(M_l),}$ so ${M_1=M.}$ Define ${j_k}$ for ${1\le k\le n}$ by ${j_1=j}$ and ${j_{l+1}=j(j_l)\circ j,}$ so each ${j_l:V\rightarrow M_l}$ is elementary, ${V_{\kappa+\alpha+2}\subset M_n,}$ ${{\rm cp}(j_n)=\kappa}$ and ${j_n(\kappa)>\kappa+\alpha+2.}$ $\Box$

Corollary 17 If ${\kappa}$ is ${\beta+2}$ -strong for all ${\beta<\alpha,}$ and ${{\sf GCH}}$ holds below ${\kappa,}$ then it holds for all cardinals below ${\kappa+\alpha.}$ Thus if ${\kappa}$ is strong and ${{\sf GCH}}$ holds below ${\kappa}$ then it holds everywhere.

Proof: Let ${\tau}$ be a cardinal below ${\kappa+\alpha.}$ Then ${\tau<\kappa+\rho+2}$ for some ${\rho<\alpha.}$ By Corollary 16, there is an embedding ${j:V\rightarrow M}$ such that ${{\rm cp}(j)=\kappa,}$ ${V_{\kappa+\rho+2}\subset M}$ and ${\kappa+\rho+2<j(\kappa).}$ Hence, ${{\sf GCH}}$ holds in ${M}$ below ${j(\kappa);}$ in particular it holds at ${\tau.}$ But ${{\mathcal P}(\tau)\in V_{\kappa+\rho+2},}$ so in fact ${2^\tau=\tau^+}$ in ${V.}$ $\Box$

As mentioned above, it is consistent that ${{\sf GCH}}$ fails at a measurable. This is closely related to the failure of ${{\sf SCH}:}$ Prikry forcing allows us to change the cofinality of a measurable ${\kappa}$ to ${\omega}$ while preserving all cardinals and without adding any bounded subsets of ${\kappa.}$ Hence, if ${2^\kappa>\kappa^+,}$ then after Prikry forcing we obtain a singular strong limit of cofinality ${\omega}$ with ${2^\kappa>\kappa^+,}$ i.e., we have violated ${{\sf SCH}.}$

Moreover, the consistency strength of both assumptions (the existence of a measurable where ${{\sf GCH}}$ fails and the failure of ${{\sf SCH}}$ ) is the same: Both statements are equiconsistent over ${{\sf ZFC}}$ to the existence of a cardinal ${\kappa}$ of Mitchell order ${\kappa^{++}.}$ This is a notion significantly stronger than measurability, but weaker than ${2}$ -strength.

Definition 18 The Mitchell order of a measurable cardinal ${\kappa}$ is defined as follows: First, for normal ultrafilters ${{\mathcal U}}$ and ${{\mathcal V}}$ over ${\kappa,}$ set ${{\mathcal U}\lhd{\mathcal V}}$ iff ${{\mathcal U}\in{\rm Ult}(V,{\mathcal V}).}$ It is easy to see that this implies that ${i_{\mathcal U}(\kappa)<i_{\mathcal V}(\kappa),}$ and therefore ${\lhd}$ is well-founded.

We define the Mitchell order of ${\kappa}$ by ${o(\kappa)=\sup\{\|{\mathcal U}\|_{\lhd}+1:{\mathcal U}}$ is a normal nonprincipal ${\kappa}$ -complete ultrafilter over ${\kappa\}.}$

It follows that ${o(\kappa)\ge1}$ iff ${\kappa}$ is measurable. The assumption ${o(\kappa)=2}$ implies that the measurable cardinals below ${\kappa}$ form a stationary set, but is stronger than this.

Homework problem 12. Assume that ${\kappa}$ is ${2}$ -strong and determine ${o(\kappa).}$

One final remark is in order: Through this lecture I have used proper classes freely, without worrying about whether this can be formalized in ${{\sf ZFC}.}$ This is indeed the case, since just as the existence of embeddings corresponds to the existence of ${\sigma}$ -complete nonprincipal ultrafilters, the embeddings discussed here can be coded via sequences of ultrafilters (and so, any mention of proper classes can be eliminated by referring to the (definable) ultrapowers from which they arise). Nevertheless, the right setting for Kunen’s theorem is a set theory that allows classes, so it is perhaps more reasonable to think that we are arguing in either ${{\sf GBC}}$ or (even better) ${{\sf MKC}.}$ My personal opinion is that these matters are not mathematical in nature and should therefore not worry us.

Through this lecture, I have made good use of Akihiro Kanamori, The Higher Infinite, Springer (1994). For an introduction to the Rudin-Keisler order, see Wistar Comfort, Stylianos Negrepontis, The theory of ultrafilters, Springer (1974).

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6 Responses to 580 -Cardinal arithmetic (10)

580 -Cardinal arithmetic (11) « Teaching page says:

March 12, 2009 at 11:13 am

[…] sets for ), is not -complete. It follows that From, say, Theorem 6 and preceeding remarks from last lecture, it follows that By choice, this implies […]

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580 -Partition calculus (3) « Teaching page says:

April 6, 2009 at 12:20 am

[…] gives us another proof of Kunen’s inconsistency Theorem 13 from lectures II.10 and II.12. Corollary 9 (Kunen) If is elementary, […]

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580 -Partition calculus (5) « Teaching page says:

April 21, 2009 at 11:33 pm

[…] showed that there are no embeddings All the known proofs of this result (see for example lectures II.10, II.12, and III.3) use the axiom of choice in an essential way. Theorem 27 (Sargsyan) In assume […]

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580 -Partition calculus (5) « A kind of library says:

September 10, 2012 at 11:49 am

[…] showed that there are no embeddings All the known proofs of this result (see for example lectures II.10, II.12, and III.3) use the axiom of choice in an essential way. Theorem 27 (Sargsyan) In assume […]

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580 -Partition calculus (3) « A kind of library says:

January 20, 2013 at 9:30 am

[…] gives us another proof of Kunen’s inconsistency Theorem 13 from lectures II.10 and […]

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580 -Cardinal arithmetic (11) | A kind of library says:

August 5, 2020 at 12:18 pm

[…] sets for ), is not -complete. It follows that From, say, Theorem 6 and preceding remarks from last lecture, it follows that By choice, this implies […]

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