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diff --git a/library/set/powerset.tex b/library/set/powerset.tex new file mode 100644 index 0000000..80da4cb --- /dev/null +++ b/library/set/powerset.tex @@ -0,0 +1,107 @@ +\import{set.tex} + +\subsection{Powerset} + +\begin{abbreviation}\label{powerset_of} + The powerset of $X$ denotes $\pow{X}$. +\end{abbreviation} + +\begin{axiom}% +\label{pow_iff} + $B\in\pow{A}$ iff $B\subseteq A$. +\end{axiom} + +\begin{proposition}\label{powerset_intro} + Suppose $A\subseteq B$. + Then $A\in\pow{B}$. +\end{proposition} + +\begin{proposition}\label{powerset_elim} + Let $A\in\pow{B}$. + Then $A\subseteq B$. +\end{proposition} + +\begin{proposition}\label{powerset_bottom} + $\emptyset\in\pow{A}$. +\end{proposition} + +\begin{proposition}\label{powerset_top} + $A\in\pow{A}$. +\end{proposition} + +\begin{proposition}\label{unions_subseteq_of_powerset_is_subseteq} + Let $A$ be a set. + Let $B$ be a subset of $\pow{A}$. + Then $\unions{B}\subseteq A$. +\end{proposition} +\begin{proof} + Follows by \cref{subseteq,powerset_elim,unions_iff}. +\end{proof} + +\begin{corollary}\label{powerset_closed_unions} + Let $A$ be a set. + Let $B$ be a subset of $\pow{A}$. + Then $\unions{B}\in\pow{A}$. +\end{corollary} +\begin{proof} + Follows by \cref{pow_iff,unions_subseteq_of_powerset_is_subseteq}. +\end{proof} + +\begin{proposition}\label{unions_powerset} + $\unions{\pow{A}} = A$. +\end{proposition} +\begin{proof} + Follows by set extensionality. +\end{proof} + +\begin{proposition}\label{inters_powerset} + $\inters{\pow{A}} = \emptyset$. +\end{proposition} +\begin{proof} + Follows by set extensionality. +\end{proof} + +\begin{proposition}\label{union_powersets_subseteq} + $\pow{A}\union\pow{B} \subseteq \pow{A\union B}$. +\end{proposition} +\begin{proof} + $\pow{A}\subseteq \pow{A}\union\pow{B}$ by \cref{union_upper_left}. + $\pow{B}\subseteq \pow{A}\union\pow{B}$ by \cref{union_upper_right}. + Follows by \cref{subseteq,pow_iff,powerset_elim,union_iff,subset_transitive}. +\end{proof} + +\begin{proposition}\label{powerset_emptyset} + $\pow{\emptyset} = \{\emptyset\}$. +\end{proposition} + +% LATER +%\begin{proposition}\label{powerset_cons} +% Then $\pow{\cons{b}{A}} = \pow{A}\union \{\cons{b}{B}\mid B\in\pow{A}\}$. +%\end{proposition} + +\begin{proposition}\label{powerset_union_subseteq} + $\pow{A}\union\pow{B}\subseteq \pow{A\union B}$. +\end{proposition} + +\begin{proposition}\label{subseteq_pow_unions} + $A\subseteq\pow{\unions{A}}$. +\end{proposition} +\begin{proof} + Follows by \cref{subseteq,pow_iff,unions_intro}. +\end{proof} + +\begin{proposition}\label{unions_pow} + $\unions{\pow{A}} = A$. +\end{proposition} + + +\begin{proposition}\label{unions_elem_pow_iff} + $\unions{A}\in\pow{B}$ iff $A\in\pow{\pow{B}}$. +\end{proposition} + +\begin{proposition}\label{pow_inter} + $\pow{A\inter B} = \pow{A}\inter\pow{B}$. +\end{proposition} +\begin{proof} + Follows by \cref{setext,inter,pow_iff,subseteq_inter_iff}. +\end{proof} |
