lecture 11
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inputs/lecture_10.tex
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inputs/lecture_10.tex
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\lecture{10}{}{} % Mirko
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Applications of induction and recursion:
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\begin{fact}
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For every set $x$ there is a transitive set $t$
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such that $x \in t$.
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\end{fact}
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\begin{proof}
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Take $R = \in $.
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We want a function $F$ with domain $\omega$
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such that $F(0) = \{x\}$
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and $F(n+1) = \bigcup F(n)$.
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Once we have such a function,
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$\{x\} \cup \bigcup \ran(F)$
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is a set as desired.
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\todo{insert formal application of recursion theorem}
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\end{proof}
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\begin{notation}
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Let $\OR$ denote the class of all ordinals
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and $V$ the class of all sets.
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\end{notation}
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\begin{lemma}
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There is a function $F\colon \OR \to V$
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such that $F(\alpha) = \bigcup \{\cP(F(\beta)): \beta < \alpha\}$.
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\end{lemma}
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\begin{proof}
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\todo{TODO}
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\end{proof}
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\begin{notation}
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Usually, one write $V_\alpha$ for $F(\alpha)$.
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They are called the \vocab{rank initial segments} of $V$.
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\end{notation}
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\begin{lemma}
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If $x$ is any set, then there is some $\alpha \in \OR$
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such that $x \in V_\alpha$,
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i.e.~$V = \bigcup \{V_{\alpha} : \alpha \in \OR\}$.
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\end{lemma}
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inputs/lecture_11.tex
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\lecture{11}{2023-11-23}{Cardinals}
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\subsection{Cardinals}
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Consequence of the Mostowski collapse:
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If $<$ is a well-order on a set $a$
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then there is some transitive $b$
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with $(b, \in\defon{b}) \cong (a, <)$.
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\begin{definition}
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Let $a$ be any set.
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The \vocab{cardinality} of $a$
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denoted by $\overline{\overline{a}}$, $|a|$ or $\card(a)$,
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is the smallest ordinal $\alpha$
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such that there is some bijection $f\colon \alpha \to a$.
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An ordinal $\alpha$ is called a \vocab{cardinal},
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iff there is some set $a$ with $|a| = \alpha$
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(equivalently, $|\alpha| = \alpha$).
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\end{definition}
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We often write $\kappa, \lambda, \ldots$ for cardinals.
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\begin{lemma}
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For every cardinal $\kappa$,
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there is come cardinal $\lambda > \kappa$.
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\end{lemma}
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\begin{proof}
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Consider the powerset of $\kappa$.
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We know that there is no surjection $\kappa \twoheadrightarrow \cP(\kappa)$.
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Hence $\kappa < |2^{\kappa}|$.
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\end{proof}
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\begin{definition}
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For each cardinal $\kappa$,
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$\kappa^+$ denotes the least cardinal $\lambda > \kappa$.
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\end{definition}
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\begin{warning}
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This has nothing to do with the ordinal successor of $\kappa$.
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\end{warning}
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\begin{lemma}
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Let $X$ be any set of cardinals.
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Then $\sup X$ is a cardinal.
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\end{lemma}
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\begin{proof}
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If there is some $\kappa \in X$ with $\lambda \le \kappa$
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for all $\lambda \in X$,
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then $\kappa = \sup(X)$ is a cardinal.
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Let us now assume that for all $\kappa \in X$
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there is some $\lambda \in X$
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with $\lambda > \kappa$.
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Suppose that $\sup(X)$ is no a cardinal
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and write $\mu = |\sup(X)|$.
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Then $\mu \in \sup(X)$,
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since $\sup(X)$ is an ordinal.
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However $\sup(X)$ is the least ordinal
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larger than all $\alpha \in X$,
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so there is $\lambda \in X$
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with $\lambda > \mu$.
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However, there exists $\mu \twoheadrightarrow \sup(X)$,
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hence also $\mu \twoheadrightarrow \lambda$
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(which is in contradiction to $\lambda$ being a cardinal).
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\end{proof}
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We may now use the recursion theorem
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to define a sequence $\langle \aleph_\alpha : \alpha \in \OR \rangle$
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with the following properties:
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\begin{IEEEeqnarray*}{rCl}
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\aleph_0 &=& \omega,\\
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\aleph_{\alpha+1} &=& (\aleph_\alpha)^+,\\
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\aleph_{\lambda} &=& \sup \{\aleph_\alpha : \alpha < \lambda\}.
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\end{IEEEeqnarray*}
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Each $\aleph_\alpha$ is a cardinal.
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Also, a trivial induction show that $\alpha \le \aleph_\alpha$.
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In particular $|\alpha| \le \aleph_{\alpha}$.
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Therefore the $\aleph_\alpha$ are all the infinite cardinals:
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If $a$ is any infinite set, then $|a| \le \aleph_{|a|}$,
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so $|a| = \aleph_\beta$ for some $\beta \le |\alpha|$.
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\begin{notation}
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Sometimes we write $\omega_\alpha$ for $\aleph_\alpha$
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(when viewing it as an ordinal).
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\end{notation}
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\begin{notation}
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Let ${}^a b \coloneqq \{f : f \text{ is a function}, \dom(f) = \alpha, \ran(f) \subseteq b\}$.
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\end{notation}
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\begin{definition}[Cardinal arithmetic]
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Let $\kappa$, $\lambda$ be cardinals.
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Define
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\begin{IEEEeqnarray*}{rCl}
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\kappa + \lambda &\coloneqq & |\{0\} \times \kappa \cup \{1\} \times \lambda|,\\
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\kappa \cdot \lambda &\coloneqq & |\kappa \times \lambda|,\\
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\kappa^{\lambda} &\coloneqq & |{}^{\lambda}\kappa|.
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\end{IEEEeqnarray*}
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\end{definition}
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\begin{warning}
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This is very different from ordinal arithmetic!
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\end{warning}
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\begin{theorem}[Hessenberg]
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\label{thm:hessenberg}
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For all $\alpha$ we have
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\[
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\aleph_\alpha \cdot \aleph_\alpha = \aleph_\alpha.
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\]
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\end{theorem}
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\begin{corollary}
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For all $\alpha, \beta$ it is
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\[
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\aleph_\alpha + \aleph_\beta = \aleph_\alpha \cdot \aleph_\beta = \max \{\aleph_\alpha, \aleph_\beta\}.
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\]
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\end{corollary}
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\begin{proof}
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Wlog.~$\alpha \le \beta$.
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Trivially $\aleph_\alpha \le \aleph_\beta$.
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It is also clear that
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\[
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\aleph_{\beta} \le \aleph_\alpha + \aleph_{\beta} \le \aleph_\alpha \cdot \aleph_\beta \le \aleph_\beta \cdot \aleph_\beta = \aleph_\beta.
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\]
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\end{proof}
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\begin{refproof}{thm:hessenberg}
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Define a well-order $<^\ast$ on $\OR \times \OR$ by setting
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\[
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(\alpha,\beta) <^\ast (\gamma,\delta)
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\]
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iff
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\begin{itemize}
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\item $\max(\alpha,\beta) < \max(\gamma,\delta)$ or
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\item $\max(\alpha,\beta) = \max(\gamma,\delta)$ and $\alpha < \gamma$ or
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\item $\max(\alpha,\beta) = \max(\gamma,\delta)$ and $\alpha = \gamma$ and $\beta < \delta$.
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\end{itemize}
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It is clear that this is a well-order.
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There is an isomorphism
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\[
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(\OR, <) \cong^{\Gamma^{-1}} (\OR \times \OR, <^\ast).
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\]
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$\Gamma$ is called the \vocab{Gödel pairing function}.
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\begin{claim}
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For all $\alpha$ it is
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$\ran(\Gamma\defon{\aleph_\alpha \times \aleph_\alpha})) = \aleph_\alpha$,
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i.e.
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\[
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\aleph_\alpha = \{\xi: \exists \eta, \eta' < \aleph_\alpha.~\xi = \Gamma((\eta,\eta'))\}.
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\]
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\end{claim}
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\begin{subproof}
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We use induction of $\alpha$.
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The claim is trivial for $\alpha = 0$.
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Now let $\alpha > 0$ and suppose the claim
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to be true for all $\beta < \alpha$.
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It is easy to see that
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\[
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\ran(\Gamma\defon{\aleph_\alpha \times \aleph_\alpha}) \supseteq \aleph_\alpha,
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\]
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as otherwise $\Gamma\defon{\aleph_\alpha \times \alepha_\alpha}: \alepha_{\alpha} \times \alepha_\alpha \to \eta$
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would be a bijection for some $\eta < \alepha_\alpha$,
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but $\alepha_\alpha$ is a cardinal.
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Suppose that $\ran(\Gamma\defon{\aleph_\alpha \times \alepha_\alpha}) \supsetneq \aleph_\alpha$.
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Then there exist $\eta, \eta' < \aleph_\alpha$ with
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\[
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\Gamma((\eta, \eta')) = \alepha_\alpha.
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\]
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So $\Gamma\defon{\{(\gamma,\delta) : (\gamma,\delta) <^\ast (\eta, \eta'\}}$
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is bijective onto $\alepha_\alpha$.
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If $(\gamma,\delta) <^\ast (\eta, \eta')$,
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then $\max \{\gamma,\delta\} \le \max \{\eta, \eta'\}$.
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Say $\eta \le \eta' < \alepha_\alpha$
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and let $\aleph_\alpha = |\eta'|$.
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There is a surjection
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\[f\colon \underbrace{(\eta +1)}_{ \le \aleph_\beta} \times \underbrace{(\eta' + 1)}_{\sim \aleph_\beta} \twoheadrightarrow \aleph_\alpha.\]
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This gives rise to a surjection $f^\ast \colon \aleph_\beta \times \aleph_\beta \to \aleph_\alpha$.
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The inductive hypothesis then produces a surjection
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$f^\ast\colon \aleph_\beta \to \aleph_\alpha \lightning$.
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\end{subproof}
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\end{refproof}
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However, exponentiation of cardinals is far from trivial:
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\begin{observe}
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$2^{\kappa} = |\cP(\kappa)|$,
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since ${}^{\kappa} \{0,1\} \leftrightarrow\cP(\kappa)$.
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Hence by Cantor $2^{\kappa}\ge \kappa^+$.
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\end{observe}
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This is basically all we can say.
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The \vocab{continuum hypothesis} states that $2^{\aleph_0} = \aleph_1$.
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