r
/
bookshelf
1
Fork 0
Code Issues Pull Requests Packages Projects Releases Wiki Activity

Enderton. Continue proving laws in "Axioms and Operations."

finite-set-exercises
Joshua Potter 2023-05-22 16:33:57 -06:00
parent 995e0ce327
commit 6deeb57409
1 changed files with 176 additions and 19 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

195
Bookshelf/Enderton/Set.tex
View File

@ -897,49 +897,80 @@ For any sets $A$ and $B$,
\paragraph{(i)}% \paragraph{(i)}%
By the definition of the union of sets, By the definition of the union of sets,
$A \cup B = \{ x \mid x \in A \lor x \in B \}$.
Since $\lor$ is commutative, it follows
\begin{align*} \begin{align*}
A \cup B A \cup B
& = \{ x \mid x \in A \lor x \in B \} \\ & = \{ x \mid x \in A \lor x \in B \} \\
& = \{ x \mid x \in B \lor x \in A \} \\ & = \{ x \mid x \in B \lor x \in A \} \\
& = B \cup A, & = B \cup A.
\end{align*} \end{align*}
where the last equality follows again from the definition of the union of
sets.
\paragraph{(ii)}% \paragraph{(ii)}%
By the definition of the intersection of sets, By the definition of the intersection of sets,
$A \cap B = \{ x \mid x \in A \land x \in B \}$.
Since $\land$ is commutative, it follows
\begin{align*} \begin{align*}
A \cap B A \cap B
& = \{ x \mid x \in A \land x \in B \} \\ & = \{ x \mid x \in A \land x \in B \} \\
& = \{ x \mid x \in B \land x \in A \} \\ & = \{ x \mid x \in B \land x \in A \} \\
& = B \land A, & = B \land A.
\end{align*} \end{align*}
where the last equality follows again from the definition of the
intersection of sets.
\end{proof} \end{proof}
\subsection{\unverified{Associative Laws}}% \subsection{\verified{Associative Laws}}%
\label{sub:associative-laws} \label{sub:associative-laws}
For any sets $A$, $B$ and $C$, For any sets $A$, $B$ and $C$,
\begin{align*} \begin{align*}
A \cup (B \cup C) & = (A \cup B) \cup C \\ A \cup (B \cup C) & = (A \cup B) \cup C \\
(A \cup B) \cup C & = A \cup (B \cup C) A \cap (B \cap C) & = (A \cap B) \cap C
\end{align*} \end{align*}
\begin{proof} \begin{proof}
TODO \statementpadding
\lean*{Mathlib/Data/Set/Basic}{Set.union\_assoc}
\lean{Mathlib/Data/Set/Basic}{Set.inter\_assoc}
Let $A$, $B$, and $C$ be sets.
We show (i) $A \cup (B \cup C) = (A \cup B) \cup C$ and then (ii)
$A \cap (B \cap C) = (A \cap B) \cap C$.
\paragraph{(i)}%
By the definition of the union of sets,
\begin{align*}
A \cup (B \cup C)
& = \{ x \mid x \in A \lor x \in (B \cup C) \} \\
& = \{ x \mid x \in A \lor x \in \{ y \mid y \in B \lor C \}\} \\
& = \{ x \mid x \in A \lor (x \in B \lor x \in C) \} \\
& = \{ x \mid (x \in A \lor x \in B) \lor x \in C \} \\
& = \{ x \mid x \in \{ y \mid y \in A \lor y \in B \} \lor
x \in C \} \\
& = \{ x \mid x \in (A \cup B) \lor x \in C \} \\
& = (A \cup B) \cup C.
\end{align*}
\paragraph{(ii)}%
By the definition of the intersection of sets,
\begin{align*}
A \cap (B \cap C)
& = \{ x \mid x \in A \land x \in (B \cap C) \} \\
& = \{ x \mid x \in A \land
x \in \{ y \mid y \in B \land y \in C \}\} \\
& = \{ x \mid x \in A \land (x \in B \land x \in C) \} \\
& = \{ x \mid (x \in A \land x \in B) \land x \in C \} \\
& = \{ x \mid x \in \{ y \mid y \in A \land y \in B \} \land
x \in C \} \\
& = \{ x \mid x \in (A \cap B) \land x \in C \} \\
& = (A \cap B) \cap C.
\end{align*}
\end{proof} \end{proof}
\subsection{\unverified{Distributive Laws}}% \subsection{\verified{Distributive Laws}}%
\label{sub:distributive-laws} \label{sub:distributive-laws}
For any sets $A$, $B$, and $C$, For any sets $A$, $B$, and $C$,
@ -950,11 +981,49 @@ For any sets $A$, $B$, and $C$,
\begin{proof} \begin{proof}
TODO \statementpadding
\lean*{Mathlib/Data/Set/Basic}{Set.inter\_distrib\_left}
\lean{Mathlib/Data/Set/Basic}{Set.union\_distrib\_left}
Let $A$, $B$, and $C$ be sets.
We show (i) $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ and then (ii)
$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.
\paragraph{(i)}%
By the definition of the union and intersection of sets,
\begin{align*}
A \cap (B \cup C)
& = \{ x \mid x \in A \land x \in B \cup C \} \\
& = \{ x \mid x \in A \land
x \in \{ y \mid y \in B \lor y \in C \}\} \\
& = \{ x \mid x \in A \land (x \in B \lor x \in C) \} \\
& = \{ x \mid (x \in A \land x \in B) \lor
(x \in A \land x \in C) \} \\
& = \{ x \mid x \in A \cap B \lor x \in A \cap C \} \\
& = (A \cap B) \cup (A \cap C).
\end{align*}
\paragraph{(ii)}%
By the definition of the union and intersection of sets,
\begin{align*}
A \cup (B \cap C)
& = \{ x \mid x \in A \lor x \in B \cap C \} \\
& = \{ x \mid x \in A \lor
x \in \{ y \mid y \in B \land y \in C \}\} \\
& = \{ x \mid x \in A \lor (x \in B \land x \in C) \} \\
& = \{ x \mid (x \in A \lor x \in B) \land
(x \in A \lor x \in C) \} \\
& = \{ x \mid x \in A \cup B \land x \in A \cup C \} \\
& = (A \cup B) \cap (A \cup C).
\end{align*}
\end{proof} \end{proof}
\subsection{\unverified{De Morgan's Laws}}% \subsection{\verified{De Morgan's Laws}}%
\label{sub:de-morgans-laws} \label{sub:de-morgans-laws}
For any sets $A$, $B$, and $C$, For any sets $A$, $B$, and $C$,
@ -965,11 +1034,51 @@ For any sets $A$, $B$, and $C$,
\begin{proof} \begin{proof}
TODO \statementpadding
\lean*{Mathlib/Data/Set/Basic}{Set.diff\_inter\_diff}
\lean{Mathlib/Data/Set/Basic}{Set.diff\_inter}
Let $A$, $B$, and $C$ be sets.
We show (i) $C - (A \cup B) = (C - A) \cap (C - B)$ and then (ii)
$C - (A \cap B) = (C - A) \cup (C - B)$.
\paragraph{(i)}%
By definition of the union, intersection, and relative complements of sets,
\begin{align*}
C - (A \cup B)
& = \{ x \mid x \in C \land x \not\in A \cup B \} \\
& = \{ x \mid x \in C \land
x \not\in \{ y \mid y \in A \lor y \in B \}\} \\
& = \{ x \mid x \in C \land \neg(x \in A \lor x \in B) \} \\
& = \{ x \mid x \in C \land (x \not\in A \land x \not\in B) \} \\
& = \{ x \mid (x \in C \land x \not\in A) \land
(x \in C \land x \not\in B) \} \\
& = \{ x \mid x \in (C - A) \land x \in (C - B) \} \\
& = (C - A) \cap (C - B).
\end{align*}
\paragraph{(ii)}%
By definition of the union, intersection, and relative complements of sets,
\begin{align*}
C - (A \cap B)
& = \{ x \mid x \in C \land x \not\in A \cap B \} \\
& = \{ x \mid x \in C \land
x \not\in \{ y \mid y \in A \land y \in B \}\} \\
& = \{ x \mid x \in C \land \neg(x \in A \land x \in B) \} \\
& = \{ x \mid x \in C \land (x \not\in A \lor x \not\in B) \} \\
& = \{ x \mid (x \in C \land x \not\in A) \lor
(x \in C \land x \not\in B) \} \\
& = \{ x \mid x \in (C - A) \lor x \in (C - B) \} \\
& = (C - A) \cup (C - B).
\end{align*}
\end{proof} \end{proof}
\subsection{\unverified{ \subsection{\verified{%
Identities Involving \texorpdfstring{$\emptyset$}{the Empty Set}}}% Identities Involving \texorpdfstring{$\emptyset$}{the Empty Set}}}%
\label{sub:identitives-involving-empty-set} \label{sub:identitives-involving-empty-set}
@ -982,7 +1091,55 @@ For any set $A$,
\begin{proof} \begin{proof}
TODO \statementpadding
\lean*{Mathlib/Data/Set/Basic}{Set.union\_empty}
\lean*{Mathlib/Data/Set/Basic}{Set.inter\_empty}
\lean{Mathlib/Data/Set/Basic}{Set.inter\_diff\_self}
Let $A$ be an arbitrary set. We prove (i) that $A \cup \emptyset = A$, (ii)
$A \cap \emptyset = \emptyset$, and (iii) $A \cap (C - A) = \emptyset$.
\paragraph{(i)}%
By definition of the emptyset and union of sets,
\begin{align*}
A \cup \emptyset
& = \{ x \mid x \in A \lor x \in \emptyset \} \\
& = \{ x \mid x \in A \lor F \} \\
& = \{ x \mid x \in A \} \\
& = A.
\end{align*}
\paragraph{(ii)}%
By definition of the emptyset and intersection of sets,
\begin{align*}
A \cap \emptyset
& = \{ x \mid x \in A \land x \in \emptyset \} \\
& = \{ x \mid x \in A \land F \} \\
& = \{ x \mid F \} \\
& = \{ x \mid x \neq x \} \\
& = \emptyset.
\end{align*}
\paragraph{(iii)}%
By definition of the emptyset, and the intersection and relative complement
of sets,
\begin{align*}
A \cap (C - A)
& = \{ x \mid x \in A \land x \in C - A \} \\
& = \{ x \mid x \in A \land
x \in \{ y \mid y \in C \land y \not\in A \}\} \\
& = \{ x \mid x \in A \land (x \in C \land x \not\in A) \} \\
& = \{ x \mid x \in C \land F \} \\
& = \{ x \mid F \} \\
& = \{ x \mid x \neq x \} \\
& = \emptyset.
\end{align*}
\end{proof} \end{proof}