131 lines
4.5 KiB
TeX
131 lines
4.5 KiB
TeX
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\lecture{23}{2024-01-25}{Negation of CH is consistent to ZFC}
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\begin{goal}
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We want to construct a model of $\ZFC$
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such that $2^{\aleph_0} \ge \aleph_2$.
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% = ?
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\end{goal}
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Let $M$ be a countable transitive model of $\ZFC$.
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Suppose that $M \models \CH$
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(otherwise we are done).
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Let $\alpha = \omega_2^M$.
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Let $\bC(\alpha) \coloneqq \{p : p\colon \alpha \to \bC \text{ is a function such that } \{\xi < \alpha : p(\xi) \neq \emptyset\} \text{ is finite} \}$,
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ordered by $p \le_{\bC(\alpha)} q$ iff
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$p(\xi) \le_{\bC} q(\xi)$ for all $\xi < \alpha$.
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Recall that $\bC$ is the set of finite sequences of natural numbers
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ordered by $p \le_{\bC} q$ iff $p \supseteq q$.
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Let $g $ be $\bC(\alpha)$-generic over $M$.
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For $\xi < \alpha$ let $x_\xi = \bigcup \{ p(\xi) : p \in g\}$.
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We have already seen that $x_\xi\colon \omega \to \omega$
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is a function
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and $x_\xi \neq x_\eta$ for $\xi \neq \eta$.
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We have $M[g] \models\ZFC$.\footnote{We only handwaved this step.}
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As $g \in M[g]$,
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we have $\langle x_\xi : \xi < \alpha\rangle \in M[g]$.
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Therefore $M[g] \models \text{``$2^{\aleph_0} \ge \alpha$''}$.
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Also $\alpha = \omega_2^M$.
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However the proof is not finished yet,
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since we need to make sure,
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that $M[g]$ does not collapse cardinals.
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We only have $M[g] \models 2^{\aleph_0} \ge \aleph_2^M$,
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i.e.~we need to see $\aleph_2^{M[g]} = \aleph_2^M$.
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\begin{claim}
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Every cardinal of $M$ is still
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a cardinal of $M[g]$.
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\end{claim}
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This suffices, because
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then $\aleph_0^M = \aleph_0^{M[g]}$,
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$\aleph_1^M = \aleph_1^{M[g]}$,
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$\aleph_2^{M}= \aleph_2^{M[g]}$, $\ldots$
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\begin{definition}
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Let $(\bP, \le )$ be a partial order.
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We say that $\bP$ has the
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\vocab{countable chain condition}
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(\vocab{c.c.c.})%
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\footnote{it should really be the ``countable antichain condition''}
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iff there is no uncountable
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antichain,
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i.e.~every uncountable $V \subseteq \bP$
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contains compatible $p \neq q$.
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\end{definition}
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We shall prove:
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\begin{claim}
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\label{l23:c:1}
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For all $\beta$, $\bC(\beta)$ has the c.c.c.
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\end{claim}
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\begin{claim}
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\label{l23:c:2}
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If $\bP \in M$ and
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$M \models \text{`` $\bP$ has the c.c.c.''}$
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and $h$ is generic over $M$,
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then all $M$-cardinals are still
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$M[h]$ cardinals.
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\footnote{Being a cardinal is $\Pi_1$,
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so $M[h]$ cardinals are always $M$ cardinals.}
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\end{claim}
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\begin{refproof}{l23:c:2}
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Suppose not.
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Let $\kappa$ be minimal such that $M \models \text{``$\kappa$ is a cardinal''}$,
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but $M[h] \models \text{``$\kappa$ is not a cardinal''}$.
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Then $\kappa = (\lambda^+)^M$ for some unique $M$-cardinal $\lambda < \kappa$.
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By minimality, $\lambda$ is also an $M[h]$-cardinal.
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Let $f \in M[h]$ be such that $M[h] \models \text{``$f$ is a surjection from $\lambda$ onto $\kappa$''}$.
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There is a name $\tau \in M^{\bP}$ with $\tau^h = f$.
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We then have some $p \in h$
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with $p \Vdash_M^\bP \text{``$\tau$ is a surjection from $\check{\lambda}$ onto $\check{\kappa}$''}$.
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Let $\xi < \lambda$.
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Consider $X_\xi \coloneqq \{\eta < \kappa: \exists q \le_{\bP} p .~q \Vdash \tau(\check{\xi}) = \check{\eta}\} \in M$.
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$X_\xi$ is countable in $M$ by the following argument (in $M$):
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For every $\eta \in X_\xi$,
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let $q_\eta \le p$ be such that $q_\xi \Vdash^{\bP}_M \tau(\check{\xi}) = \check{\eta}$.
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The set $\{q_\eta : \eta \in X_\xi\}$ is an antichain
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as for $\eta_1 \neq \eta_2$ we have that $q_{\eta_i} \Vdash \tau(\check{\xi}) = \check{\eta_i}$,
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so they are not compatible.
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So $\{q_\eta : \eta \in X_\xi\}$ is countable by the c.c.c.
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Thus $X_\xi$ is countable.
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Therefore we may define a function in $M$
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\begin{IEEEeqnarray*}{rCl}
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F\colon \lambda \times \omega &\longrightarrow & \kappa
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\end{IEEEeqnarray*}
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such that for all $\xi < \lambda$
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\[
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\{F(\xi,n) : n < \omega\} = X_\xi.
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\]
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$F$ is surjective since $f$ is surjective:
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For $\eta < \kappa$,
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there is some $\xi < \lambda$ such that $M[h] \models \text{``$f(\xi) = \eta$''}$,
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there is some $\overline{q} \in h$ with $\overline{q} \Vdash ^{\bP}_M \tau(\check{\xi}) = \check{\eta}$.
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Pick $q \le \overline{q},p$.
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This shows $\eta \in X_\xi$
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hence $\eta = F(\xi, n)$ for some $n$.
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But $|\lambda \times \omega| = |\lambda| = \lambda$,
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so in $M$ there is a surjection $F' \colon \lambda \to \kappa$,
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but $\kappa$ is a cardinal in $M$ $\lightning$.
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\end{refproof}
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\begin{refproof}{l23:c:1}
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Omitted.
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% TODO combinatorial argument
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\end{refproof}
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