Lectures 2024-03-01

iii/forcing/03_forcing.tex

@@ -851,7 +851,7 @@ \subsection{\texorpdfstring{\( \mathsf{ZF} \)}{ZF} in forcing extensions}
     By separation, it suffices to show that if \( a \in M[G] \), then
     \[ \mathcal P(a) \cap M[G] = \qty{x \in M[G] \mid x \subseteq a} \subseteq b \]
     for some set \( b \in M[G] \).
-    Fix \( a \in M[G] \) with name \( \dot x \in \M^{\mathbb P} \), and define
+    Fix \( a \in M[G] \) with name \( \dot x \in M^{\mathbb P} \), and define
     \[ S = \qty{\dot x \in M^{\mathbb P} \mid \ran \dot x \subseteq \ran \dot a} = \mathcal P(\mathbb P \times \ran \dot a)^M \]
     and let
     \[ \dot b = \qty{\langle \Bbbone, \dot x \rangle \mid x \in S} \in M^{\mathbb P} \]
@@ -898,3 +898,21 @@ \subsection{\texorpdfstring{\( \mathsf{ZF} \)}{ZF} in forcing extensions}
         \item The relativisation \( (p \Vdash \varphi)^M \) will be dropped when clear in subsequent sections.
     \end{enumerate}
 \end{remark}
+\begin{lemma}
+    Let \( M \) be a countable transitive model of \( \mathsf{ZFC} \) and let \( \mathbb P \in M \) be a forcing poset.
+    Let \( \varphi, \psi \) be \( \mathcal{FL}_{\mathbb P} \)-formulas.
+    Then, for any \( p \in \mathbb P \) and \( \dot x \in M^{\mathbb P} \),
+    \begin{enumerate}
+        \item if \( \mathsf{ZFC} \vdash \forall v.\, \varphi(v) \to \psi(v) \) then \( (p \Vdash \varphi(\dot x))^M \to (p \Vdash \psi(\dot x))^M \); and
+        \item if \( \mathsf{ZFC} \vdash \forall v.\, \varphi(v) \leftrightarrow \psi(v) \) then \( (p \Vdash \varphi(\dot x))^M \leftrightarrow (p \Vdash \psi(\dot x))^M \).
+    \end{enumerate}
+\end{lemma}
+Informally, forcing is closed under logical equivalence.
+\begin{proof}
+    Clearly (ii) follows from (i).
+    Suppose that \( \mathsf{ZFC} \vdash \forall v.\, \varphi(v) \to \psi(v) \) and \( (p \Vdash \varphi(\dot x))^M \).
+    Since \( M \) is countable, we can let \( G \) be a \( \mathbb P \)-generic filter over \( M \) such that \( p \in G \).
+    By the forcing theorem, \( M[G] \vDash \varphi(\dot x^G) \).
+    Since \( M[G] \vDash \mathsf{ZFC} \), we have \( M[G] \vDash \psi(\dot x^G) \).
+    Hence, by the forcing theorem in the reverse direction, as this is true for all generics containing \( p \) we have \( (p \Vdash \psi(\dot x))^M \).
+\end{proof}

iii/forcing/04_forcing_and_independence.tex

@@ -0,0 +1,97 @@
+\subsection{Independence of the constructible universe}
+In this subsection, we show \( \Con(\mathsf{ZFC} + \mathrm{V} \neq \mathrm{L}) \), and thus \( \mathrm{V} \neq \mathrm{L} \) is independent of the axioms of \( \mathsf{ZFC} \).
+\begin{theorem}
+    Let \( M \) be a countable transitive model of \( \mathsf{ZFC} \).
+    Then there is a countable transitive model \( N \supseteq M \) such that \( N \vDash \mathsf{ZFC} + \mathrm{V} \neq \mathrm{L} \).
+\end{theorem}
+\begin{proof}
+    Let \( M \) be a countable transitive model of \( \mathsf{ZFC} \), and let \( \mathbb P \in M \) be any atomless forcing poset (that is, it has no minimal elements), for example \( \operatorname{Fn}(\omega, 2) \).
+    Since \( M \) is countable, we can let \( G \) be a \( \mathbb P \)-generic filter over \( M \).
+    As \( \mathbb P \) is atomless, \( G \notin M \).
+    Hence \( M \subsetneq M[G] \vDash \mathsf{ZFC} \).
+
+    We show that \( M[G] \vDash \mathrm{V} \neq \mathrm{L} \).
+    Therefore,
+    \[ \mathrm{L}_{\Ord \cap M} = \mathrm{L}^M \subseteq M \subsetneq M[G] \]
+    By the generic model theorem, \( \mathrm{Ord} \cap M = \mathrm{Ord} \cap M[G] \), so \( M[G] \neq \mathrm{L}_{\Ord \cap M[G]} = \mathrm{L}^{M[G]} \).
+    In particular, we have \( (\mathrm{V} \neq \mathrm{L})^{M[G]} \).
+\end{proof}
+We will now discuss how to remove the assumption that we have a countable transitive model of \( \mathsf{ZFC} \).
+\begin{theorem}
+    If \( \Con(\mathsf{ZFC}) \), then \( \Con(\mathsf{ZFC} + \mathrm{V} \neq \mathrm{L}) \).
+    Hence, \( \mathsf{ZFC} \nvdash \mathrm{V} = \mathrm{L} \).
+\end{theorem}
+\begin{proof}
+    Suppose that \( \mathsf{ZFC} + \mathrm{V} \neq \mathrm{L} \) gives rise to a contradiction.
+    Then, from a finite set of axioms \( \Gamma \subseteq \mathsf{ZFC} + \mathrm{V} \neq \mathrm{L} \), we can find \( \psi \) such that \( \Gamma \vdash \psi \wedge \neg \psi \).
+    By following the previous proofs, there is a finite set of axioms \( \Lambda \subseteq \mathsf{ZFC} \) such that \( \mathsf{ZFC} \) proves that if there is a countable transitive model of \( \Lambda \), then there is a countable transitive model of \( \Gamma \).
+    This set \( \Lambda \) should be sufficient to do the following:
+    \begin{enumerate}
+        \item to prove basic properties of forcing and constructibility;
+        \item to prove the necessary facts about absoluteness, such as absoluteness of finiteness, partial orders and so on;
+        \item to prove facts about forcing, including the forcing theorem; and
+        \item if \( M \) is a countable transitive model of \( \Lambda \) with \( \mathbb P \in M \) and \( G \) is \( \mathbb P \)-generic over \( M \), then \( \Lambda \) proves that \( M[G] \vDash \Gamma \).
+    \end{enumerate}
+    As \( \Lambda \) is finite and a subset of the axioms of \( \mathsf{ZFC} \), then by the reflection theorem there is a countable transitive model of \( \Lambda \).
+    Hence, there is a countable transitive model \( N \) of \( \Gamma \).
+    But \( \Gamma \vdash \psi \wedge \neg\psi \), so \( N \vDash \psi \wedge \neg\psi \).
+    Hence \( (\psi \wedge \neg\psi)^N \), so in \( \mathsf{ZFC} \) we can prove \( \psi^N \wedge \neg\psi^N \), so \( \mathsf{ZFC} \) is inconsistent.
+\end{proof}
+\begin{remark}
+    Gunther, Pagano, S\'anchez Terraf, and Steinberg recently completed a formalisation of the countable transitive model approach to forcing in the interactive theorem prover Isabelle.
+    To obtain \( \Con(\mathsf{ZFC}) \to \Con(\mathsf{ZFC} + \neg\mathsf{CH}) \), they used \( \mathsf{ZC} \) together with 21 instances of replacement, which are explicitly enumerated in the paper.
+\end{remark}
+
+\subsection{Cohen forcing}
+Fix a countable transitive model \( M \) of \( \mathsf{ZFC} \).
+Recall that for \( I, J \in M \),
+\begin{enumerate}
+    \item \( \operatorname{Fn}(I, J) = \qty{p \mid p \text{ is a finite partial function } I \to J} \), together with \( \supseteq \) and \( \varnothing \), has the structure of a forcing poset.
+    \item \( \operatorname{Fn}(I, J) \) is always a set in \( M \).
+    \item \( \operatorname{Fn}(I, J) \) has the countable chain condition if and only if \( I \) is empty or \( J \) is countable.
+    \item The sets \( D_i = \qty{q \in \operatorname{Fn}(I, J) \mid i \in \dom q} \) and \( R_j = \qty{q \in \operatorname{Fn}(I, J) \mid i \in \ran q} \) are dense for all \( i \in I \) and \( j \in J \).
+\end{enumerate}
+Now, suppose that \( G \subseteq \operatorname{Fn}(I, J) \) is generic over \( M \).
+Since \( G \) is a filter, if \( p, q \in G \) then \( p \cap q \in G \).
+Hence, if \( p, q \in G \), then \( p, q \) agree on the intersection of their domains.
+Let \( f_G = \bigcup G \).
+Then \( f_G \) is a function domain contained in \( I \) and range contained in \( J \).
+Note that this function has name
+\[ \dot f = \qty{\langle p, \operatorname{op}(\check \imath, \check \jmath) \rangle \mid p \in \mathbb P, \langle i, j \rangle \in p} \]
+Since \( D_i, R_j \) are dense, we obtain \( G \cap D_i \neq \varnothing \), so we must have \( i \in \dom f_G \).
+Similarly, \( j \in \ran f_G \).
+We therefore obtain the following.
+\begin{proposition}
+    Let \( G \subseteq \operatorname{Fn}(I, J) \) be a generic filter over \( M \), and suppose \( I, J \) are nonempty.
+    Then \( M[G] \vDash f_G : I \to J \) is a surjection.
+\end{proposition}
+\begin{proposition}
+    Suppose that \( I, J \) are nonempty sets, at least one of which is infinite.
+    Then
+    \[ \abs{\operatorname{Fn}(I, J)} = \max(\abs{I}, \abs{J}) \]
+\end{proposition}
+In particular, \( \abs{\operatorname{Fn}(\omega, 2)} = \aleph_0 \).
+\begin{proof}
+    Each condition \( p \in \operatorname{Fn}(I, J) \) is a finite function, so from this it follows that
+    \[ \operatorname{Fn}(I, J) \subseteq (I \times J)^{<\omega} \]
+    Hence
+    \[ \operatorname{Fn}(I, J) \subseteq \abs{(I \times J)^{<\omega}} = \abs{I \times J} = \max(\abs{I}, \abs{J}) \]
+    For the reverse direction, if we fix \( i_0 \in I \) and \( j_0 \in J \), then
+    \[ \abs{\langle i_0, j \rangle \mid j \in J} \cup \qty{\langle i, j_0 \rangle \mid i \in I} \]
+    is a collection of \( \abs{I \cup J} \)-many distinct elements of \( \operatorname{Fn}(I, J) \).
+    Thus
+    \[ \max(\abs{I}, \abs{J}) = \abs{I \cup J} \leq \operatorname{Fn}(I, J) \]
+    as required.
+\end{proof}
+We aim to provide a model in which \( \mathsf{CH} \) fails.
+To do this, we will consider the forcing poset \( \operatorname{Fn}(\omega_2^M \times \omega, 2) \).
+We may consider \( f_G : \omega_2^M \times \omega \to 2 \), and let \( g_\alpha : \omega \to 2 \) be the function defined by \( g_\alpha(n) = f_G(\alpha, n) \).
+This provides \( \omega_2^M \)-many reals in \( M[G] \).
+To show that \( M[G] \vDash \mathsf{ZFC} + \neg\mathsf{CH} \), we must show that all of the \( g_\alpha \) are distinct, and that
+\[ \omega_1^{M[G]} = \omega_1^M;\quad \omega_2^{M[G]} = \omega_2^M \]
+It will turn out that the countable chain condition guarantees that all cardinals in \( M \) remain cardinals in \( M[G] \).
+\begin{example}
+    Let \( \kappa \) be an uncountable cardinal in \( M \), and consider \( \operatorname{Fn}(\omega, \kappa) \), which does not satisfy the countable chain condition.
+    Then in \( M[G] \), the function \( f_G : \omega \to \kappa \) is a surjection.
+    Hence, \( \kappa \) has been collapsed into a countable ordinal in \( M[G] \).
+\end{example}

iii/forcing/main.tex

@@ -16,5 +16,7 @@ \section{Constructibility}
 \input{02_constructibility.tex}
 \section{Forcing}
 \input{03_forcing.tex}
+\section{Forcing and independence results}
+\input{04_forcing_and_independence.tex}
 
 \end{document}

iii/lc/02_measurable_cardinals.tex

@@ -707,7 +707,7 @@ \subsection{Ultrapowers of the universe}
     Then \( \lambda_1 \) is inaccessible, so \( \mathrm{V}_{\lambda_1} \vDash \mathsf{ZFC} + \mathsf{M}(\lambda_0) \) by 2-stability of measurability and the fact that \( \mathrm{V}_{\lambda_0 + 2} \subseteq \mathrm{V}_{\lambda_1} \).
 \end{proof}
 
-\subsection{Fundamental theorem on measurable cardinals}
+\subsection{The fundamental theorem on measurable cardinals}
 \begin{theorem}
     Suppose \( \lambda \) is inaccessible and \( \kappa < \lambda \).
     Then the following are equivalent.
@@ -749,3 +749,101 @@ \subsection{Fundamental theorem on measurable cardinals}
         giving \( \kappa \)-completeness as required.
     \end{itemize}
 \end{proof}
+\begin{remark}
+    Given a sequence \( \vb A \) of subsets of \( \kappa \) of length \( \gamma \), then \( j(\vb A) \) is a sequence of subsets of \( j(\kappa) \) of length \( j(\gamma) \).
+    Moreover, if \( A_\alpha \) is the \( \alpha \)th element of \( \vb A \), then \( j(A_\alpha) \) is the \( j(\alpha) \)th element of \( j(\vb A) \).
+    In the situation above, \( \gamma < \kappa \), so \( j(\gamma) = \gamma \) and \( j(\alpha) = \alpha \), so \( j(\vb A) = \qty{j(A_\alpha) \mid \alpha < \gamma} \).
+    If, for example, \( \gamma = \kappa \), then \( j(\vb A) \) is a sequence of length \( j(\kappa) \), which is strictly longer.
+    Despite this, the first \( \kappa \)-many elements of the sequence are still \( j(A_\alpha) \) for \( \alpha < \kappa \).
+    Beyond \( \kappa \), we do not know what the elements of \( j(\vb A) \) look like.
+    This remark suffices for the following result.
+\end{remark}
+\begin{proposition}
+    For arbitrary embeddings \( j \) with critical point \( \kappa \), the ultrafilter \( U_j \) constructed above is normal.
+\end{proposition}
+\begin{proof}
+    Suppose \( A_\alpha \in U_j \) for each \( \alpha < \kappa \), or equivalently, \( \kappa \in j(A_\alpha) \).
+    We must show \( \kappa \in j\qty(\operatorname*{\scalerel*{\mupDelta}{\textstyle\sum}}_{\alpha < \kappa}(A_\alpha)) \).
+    We have
+    \begin{align*}
+        \xi \in \operatorname*{\scalerel*{\mupDelta}{\textstyle\sum}}_{\alpha < \kappa}(A_\alpha) &\leftrightarrow \xi \in \bigcap_{\alpha < \xi} A_\alpha \\
+        &\leftrightarrow \forall \alpha < \xi.\, \xi \in A_\alpha \\
+        \xi \in j\qty(\operatorname*{\scalerel*{\mupDelta}{\textstyle\sum}}_{\alpha < \kappa}(A_\alpha)) &\leftrightarrow \forall \alpha < \xi.\, \xi \in j(\vb A)_{j(\alpha)}
+    \end{align*}
+    Substitute \( \kappa \) for \( \xi \) and obtain
+    \begin{align*}
+        \kappa \in j\qty(\operatorname*{\scalerel*{\mupDelta}{\textstyle\sum}}_{\alpha < \kappa}(A_\alpha)) &\leftrightarrow \forall \alpha < \kappa.\, \kappa \in j(\vb A)_{j(\alpha)} \\
+        &\leftrightarrow \forall \alpha < \kappa.\, \kappa \in j(\vb A)_\alpha \\
+        &\leftrightarrow \forall \alpha < \kappa.\, \kappa \in j(A_\alpha)
+    \end{align*}
+    which holds by assumption.
+\end{proof}
+\begin{remark}
+    \begin{enumerate}
+        \item This gives an alternative proof of the existence of a normal ultrafilter on a measurable cardinal.
+        \item The operations \( U \mapsto j_U \) and \( j \mapsto U_j \) are not inverses in general.
+        In particular, if \( U \) is not normal, \( U_{j_U} \neq U \).
+    \end{enumerate}
+\end{remark}
+\begin{proposition}
+    Let \( U \) be a \( \kappa \)-complete nonprincipal ultrafilter on \( \kappa \).
+    Then the following are equivalent.
+    \begin{enumerate}
+        \item \( U \) is normal;
+        \item \( (\id) = \kappa \).
+    \end{enumerate}
+\end{proposition}
+This proposition provides an alternative view of reflection.
+Suppose that the ultrafilter \( U \) on \( \kappa \) is normal.
+If \( M \vDash \Phi(\kappa) \), then \( M \vDash \Phi((id)) \).
+By \L{}o\'s' theorem,
+\[ \qty{\alpha < \kappa \mid \Phi(\id(\alpha))} \in U \]
+So \( \Phi \) reflects not only on a set of size \( \kappa \), but on an ultrafilter set.
+In particular, if \( \Phi = \mathsf{M} \) and \( M \vDash \mathsf{M}(\kappa) \), so if \( \kappa \) is surviving, then the set of \( \alpha \) that are measurable is in \( U \).
+Using this result, we can characterise the surviving cardinals in a more elegant way.
+\begin{theorem}
+    \( \kappa \) is surviving if and only if there is a normal ultrafilter on \( \kappa \) such that \( \qty{\alpha < \kappa \mid \mathsf{M}(\alpha)} \in U \).
+\end{theorem}
+\begin{proof}
+    We have just shown one direction.
+    For the converse, suppose the set \( C = \qty{\alpha < \kappa \mid \mathsf{M}(\alpha)} \) is in \( U \).
+    Then for each \( \alpha \in C \), one can find an \( \alpha \)-complete nonprincipal ultrafilter on \( \alpha \) called \( U_\alpha \).
+    Define
+    \[ f(\alpha) = \begin{cases}
+        U_\alpha & \text{if } \alpha \in C
+        \varnothing & \text{if } \alpha \notin C
+    \end{cases} \]
+    Thus the set of \( \alpha \) such that \( f(\alpha) \) is an \( \alpha \)-complete nonprincipal ultrafilter on \( \alpha \) is \( C \), so in \( U \).
+    Equivalently, the set of \( \alpha \) such that \( f(\alpha) \) is an \( \id(\alpha) \)-complete nonprincipal ultrafilter on \( \id(\alpha) \) is in \( U \).
+    So by \L{}o\'s' theorem, \( M \) believes that \( (f) \) is an \( (id) \)-complete nonprincipal ultrafilter on \( (id) \).
+    So \( (f) \) witnesses that \( \kappa \) is measurable in \( M \).
+\end{proof}
+This shows that whether a cardinal \( \kappa \) is surviving depends only on \( \mathrm{V}_{\kappa + 2} \), and is therefore a 2-stable property.
+\begin{definition}
+    If \( U, U' \) are normal ultrafilters on \( \kappa \), we write \( U <_M U' \) if
+    \[ C = \qty{\alpha \mid \mathsf{M}(\alpha)} \in U \]
+    and there is a sequence of ultrafilters \( U_\alpha \) on \( \alpha \in C \) such that
+    \[ A \in U' \leftrightarrow \qty{\alpha \mid A \cap \alpha \in U_\alpha} \in U \]
+    This is known as the \emph{Mitchell order}.
+\end{definition}
+Then \( \kappa \) is surviving if and only if there are \( U, U' \) on \( \kappa \) such that \( U <_M U' \), because of the fact that if \( h(\alpha) = A \cap \alpha \) then \( (h) = A \).
+Note that talking about sequences of Mitchell-ordered ultrafilters is also 2-stable.
+
+% \section{Large large cardinals}
+\subsection{???}
+\begin{definition}
+    A large cardinal axiom \( \Phi\mathsf{C} \) is called an \emph{embedding axiom} if \( \Phi(\kappa) \) holds if and only if there is a transitive model \( M \) and elementary embedding \( j : \mathrm{V}_\lambda \to M \) with critical point \( \kappa \) with certain additional properties.
+\end{definition}
+\( \mathsf{M}(\kappa) \) is the simplest embedding axiom.
+The remaining large cardinal axioms in this course will take the form of embedding axioms.
+\begin{definition}
+    An embedding \( j : \mathrm{V}_\lambda \to M \) with critical point \( \kappa \) is called \emph{\( \beta \)-strong} if \( \mathrm{V}_{\kappa + \beta} \subseteq M \).
+    A cardinal \( \kappa \) is called \emph{\( \beta \)-strong} if there is a \( \beta \)-strong embedding with critical point \( \kappa \).
+\end{definition}
+\( \beta \)-stable properties are preserved by \( \beta \)-strong embeddings.
+In particular, by the reflection argument, if \( \Phi \) is \( \beta \)-stable and \( \kappa \) is \( \beta \)-strong with \( \Phi(\kappa) \), then \( \kappa \) is the \( \kappa \)th cardinal with property \( \Phi \).
+
+Note that \( \kappa \) is measurable if and only if \( \kappa \) is 1-strong, and if \( \kappa \) is 2-strong then \( \qty{\alpha < \kappa \mid \mathsf{M}(\alpha)} \) and \( \qty{\alpha < \kappa \mid \mathsf{Surv}(\alpha)} \) are of size \( \kappa \).
+If we write \( \beta\mathsf{-S}(\kappa) \) to denote that \( \kappa \) is \( \beta \)-strong, then
+\[ \mathsf{SurvC} <_{\Con} 2\mathsf{-S}(\kappa) \]
+This also gives an example of \( j_{U_j} \neq j \), as the ultrapower embedding of any ultrafilter is never 2-strong.