502 – The constructible universe

December 9, 2009

In this set of notes I want to sketch Gödel’s proof that {{\sf CH}} is consistent with the other axioms of set theory. Gödel’s argument goes well beyond this result; his identification of the class {L} of constructible sets eventually led to the development of inner model theory, one of the main areas of active research within set theory nowadays.

A good additional reference for the material in these notes is Constructibility by Keith Devlin.

1. Definability

The idea behind the constructible universe is to only allow those sets that one must necessarily include. In effect, we are trying to find the smallest possible transitive class model of set theory.

{L} is defined as

\displaystyle L=\bigcup_{\alpha\in{\sf ORD}} L_\alpha,

where {L_0=\emptyset,} {L_\lambda=\bigcup_{\alpha<\lambda}L_\alpha} for {\lambda} limit, and {L_{\alpha+1}={\rm D{}ef}(L_\alpha),} where

\displaystyle \begin{array}{rcl} {\rm D{}ef}(X)=\{a\subseteq X&\mid&\exists \varphi\,\exists\vec b\in X\\ && a=\{c\in X\mid(X,\in)\models\varphi(\vec b,c)\}\}. \end{array}

The first question that comes to mind is whether this definition even makes sense. In order to formalize this, we need to begin by coding a bit of logic inside set theory. The recursive constructions that we did at the beginning of the term now prove useful.

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580 -Cardinal arithmetic (10)

March 9, 2009

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.

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580 -Some choiceless results (5)

February 2, 2009

[This lecture was covered by Marion Scheepers. Many thanks! The notes below also cover lecture 6.]

We want to prove the following result and a few related facts.

Theorem. (Specker). {\rm CH}({\mathfrak m}) and {\rm CH}(2^{\mathfrak m}) imply 2^{\mathfrak m}=\aleph({\mathfrak m}).

It follows immediately from the theorem that {\sf GCH} implies {\sf AC}, since the result gives that any (infinite) {\mathfrak m} embeds into \aleph({\mathfrak m}).

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580 -Some choiceless results (4)

January 29, 2009

Let me begin with a remark related to the question of whether \aleph(X)\preceq {\mathcal P}^2(X). We showed that this is the case if X\sim Y^2 for some Y, or if X is Dedekind-finite.

Theorem. The axiom of choice is equivalent to the statement that any Dedekind-infinite cardinal is a square.

Proof. Let X be a set. Assuming that every D-infinite cardinal is a square, we need to show that X is well-orderable. We may assume that \omega\preceq X. Otherwise, replace X with X\cup\omega. Let \kappa=\aleph(X). Assume that X\sqcup\kappa is a square, say X\sqcup\kappa\sim Y^2. Then \kappa\preceq Y^2. By Homework problem 2, \kappa\preceq Y, so Y\sim \kappa\sqcup Z for some Z, and X\sqcup \kappa\sim Y^2\sim\kappa^2\sqcup 2\times\kappa\times Z\sqcup Z^2\succeq\kappa\times Z.

Lemma. Suppose A,B,C are D-infinite sets and \lambda is an (infinite) initial ordinal. If \lambda\times A\preceq B\cup C then either \lambda\preceq B or A\preceq C.

Proof. Let f:\lambda\times A\to B\sqcup C be an injection. If there is some a\in A such that f(\cdot,a):\lambda\to B we are done, so we may assume that for all a\in A there is some \alpha\in\lambda such that f(\alpha,a)\in C. Letting \alpha_a be the least such \alpha, the map a\mapsto f(\alpha_a,a) is an injection of A into C. {\sf QED}

By the lemma, it must be that either \kappa\preceq X or else Z\preceq\kappa. The former is impossible since \kappa=\aleph(X), so Z is well-orderable, and thus so is Y, and since Y\sim Y^2\succeq X, then X is well-orderable as well. {\sf QED}

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