Enderton. Infinite cartesian products.

finite-set-exercises
Joshua Potter 2023-06-29 14:05:08 -06:00
parent f885f6e334
commit 92b52ef1b8
2 changed files with 22 additions and 1 deletions

View File

@ -31,6 +31,12 @@
For any relation $R$ there is a function $H \subseteq R$ with For any relation $R$ there is a function $H \subseteq R$ with
$\dom{H} = \dom{R}$. $\dom{H} = \dom{R}$.
\section{\partial{Axiom of Choice, Second Form}}%
\label{ref:axiom-of-choice-2}
For any set $I$ and any function $H$ with domain $I$, if $H(i) \neq \emptyset$
for all $i \in I$, then $$\bigtimes_{i \in I} H(i) \neq \emptyset.$$
\section{\defined{Composition}}% \section{\defined{Composition}}%
\label{ref:composition} \label{ref:composition}
@ -3929,4 +3935,19 @@ Define
\end{proof} \end{proof}
\section{Exercise 9}%
\label{sec:exercise-9}
\subsection{\unverified{Exercise 9.31}}%
\label{sub:exercise-9.31}
Show that from the first form of the axiom of choice we can prove the second
form, and conversely.
\begin{proof}
TODO
\end{proof}
\end{document} \end{document}

View File

@ -6,7 +6,7 @@
\usepackage{environ} \usepackage{environ}
\usepackage{fancybox} \usepackage{fancybox}
\usepackage{fontawesome5} \usepackage{fontawesome5}
\usepackage{mathrsfs} \usepackage{mathabx, mathrsfs}
\usepackage{soul} \usepackage{soul}
\usepackage{stmaryrd} \usepackage{stmaryrd}
\usepackage[usenames,dvipsnames]{xcolor} \usepackage[usenames,dvipsnames]{xcolor}