Enderton. Infinite cartesian products.
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For any relation $R$ there is a function $H \subseteq R$ with
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$\dom{H} = \dom{R}$.
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\section{\partial{Axiom of Choice, Second Form}}%
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\label{ref:axiom-of-choice-2}
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For any set $I$ and any function $H$ with domain $I$, if $H(i) \neq \emptyset$
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for all $i \in I$, then $$\bigtimes_{i \in I} H(i) \neq \emptyset.$$
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\section{\defined{Composition}}%
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\label{ref:composition}
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@ -3929,4 +3935,19 @@ Define
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\end{proof}
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\section{Exercise 9}%
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\label{sec:exercise-9}
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\subsection{\unverified{Exercise 9.31}}%
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\label{sub:exercise-9.31}
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Show that from the first form of the axiom of choice we can prove the second
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form, and conversely.
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\begin{proof}
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TODO
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\end{proof}
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\end{document}
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@ -6,7 +6,7 @@
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\usepackage{environ}
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\usepackage{fancybox}
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\usepackage{fontawesome5}
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\usepackage{mathrsfs}
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\usepackage{mathabx, mathrsfs}
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\usepackage{soul}
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\usepackage{stmaryrd}
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\usepackage[usenames,dvipsnames]{xcolor}
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