Enderton "Algebra of Sets" identities and create `Chapter_2.lean`.
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@ -1 +1,2 @@
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import Bookshelf.Enderton.Set.Chapter_1
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import Bookshelf.Enderton.Set.Chapter_1
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import Bookshelf.Enderton.Set.Chapter_2
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@ -510,8 +510,8 @@ What is in $A \cap B \cap C$?
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\begin{answer}
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\lean{Bookshelf/Enderton/Set/Chapter\_1}
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{Enderton.Set.Chapter\_1.exercise\_3\_1}
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\lean{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.exercise\_3\_1}
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The set of integers divisible by $4$, $9$, and $10$.
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@ -525,8 +525,8 @@ Give an example of sets $A$ and $B$ for which $\bigcup A = \bigcup B$ but
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\begin{answer}
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\lean{Bookshelf/Enderton/Set/Chapter\_1}
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{Enderton.Set.Chapter\_1.exercise\_3\_2}
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\lean{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.exercise\_3\_2}
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Let $A = \{\{1\}, \{2\}\}$ and $B = \{\{1, 2\}\}$.
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@ -540,8 +540,8 @@ Show that every member of a set $A$ is a subset of $\bigcup A$.
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\begin{proof}
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\lean{Bookshelf/Enderton/Set/Chapter\_1}
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{Enderton.Set.Chapter\_1.exercise\_3\_3}
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\lean{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.exercise\_3\_3}
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Let $x \in A$.
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By definition, $$\bigcup A = \{ y \mid (\exists b \in A) y \in b\}.$$
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@ -558,8 +558,8 @@ Show that if $A \subseteq B$, then $\bigcup A \subseteq \bigcup B$.
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\begin{proof}
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\lean{Bookshelf/Enderton/Set/Chapter\_1}
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{Enderton.Set.Chapter\_1.exercise\_3\_4}
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\lean{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.exercise\_3\_4}
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Let $A$ and $B$ be sets such that $A \subseteq B$.
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Let $x \in \bigcup A$.
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@ -579,8 +579,8 @@ Show that $\bigcup \mathscr{A} \subseteq B$.
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\begin{proof}
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\lean{Bookshelf/Enderton/Set/Chapter\_1}
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{Enderton.Set.Chapter\_1.exercise\_3\_5}
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\lean{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.exercise\_3\_5}
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Let $x \in \bigcup \mathscr{A}$.
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By definition,
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@ -600,8 +600,8 @@ Show that for any set $A$, $\bigcup \powerset{A} = A$.
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\begin{proof}
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\lean{Bookshelf/Enderton/Set/Chapter\_1}
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{Enderton.Set.Chapter\_1.exercise\_3\_6a}
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\lean{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.exercise\_3\_6a}
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We prove that (i) $\bigcup \powerset{A} \subseteq A$ and (ii)
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$A \subseteq \bigcup \powerset{A}$.
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@ -641,8 +641,8 @@ Under what conditions does equality hold?
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\begin{proof}
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\lean{Bookshelf/Enderton/Set/Chapter\_1}
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{Enderton.Set.Chapter\_1.exercise\_3\_6b}
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\lean{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.exercise\_3\_6b}
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Let $x \in A$.
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By \nameref{sub:exercise-3.3}, $x$ is a subset of $\bigcup A$.
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@ -690,8 +690,8 @@ Show that for any sets $A$ and $B$,
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\begin{proof}
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\lean{Bookshelf/Enderton/Set/Chapter\_1}
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{Enderton.Set.Chapter\_1.exercise\_3\_7a}
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\lean{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.exercise\_3\_7a}
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Let $A$ and $B$ be arbitrary sets. We show that
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$\powerset{A} \cap \powerset{B} \subseteq \powerset{(A \cap B)}$ and then
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@ -741,11 +741,11 @@ Under what conditions does equality hold?
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\statementpadding
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\lean*{Bookshelf/Enderton/Set/Chapter\_1}
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{Enderton.Set.Chapter\_1.exercise\_3\_7b\_i}
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\lean*{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.exercise\_3\_7b\_i}
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\lean{Bookshelf/Enderton/Set/Chapter\_1}
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{Enderton.Set.Chapter\_1.exercise\_3\_7b\_ii}
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\lean{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.exercise\_3\_7b\_ii}
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Let $x \in \powerset{A} \cup \powerset{B}$.
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By definition, $x \in \powerset{A}$ or $x \in \powerset{B}$ (or both).
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@ -837,8 +837,8 @@ Give an example of sets $a$ and $B$ for which $a \in B$ but
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\begin{answer}
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\lean{Bookshelf/Enderton/Set/Chapter\_1}
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{Enderton.Set.Chapter\_1.exercise\_3\_9}
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\lean{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.exercise\_3\_9}
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Let $a = \{1\}$ and $B = \{\{1\}\}$.
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Then
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@ -858,8 +858,8 @@ Show that if $a \in B$, then $\powerset{a} \in \powerset{\powerset{\bigcup B}}$.
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\begin{proof}
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\lean{Bookshelf/Enderton/Set/Chapter\_1}
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{Enderton.Set.Chapter\_1.exercise\_3\_10}
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\lean{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.exercise\_3\_10}
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Suppose $a \in B$.
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By \nameref{sub:exercise-3.3}, $a \subseteq \bigcup B$.
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@ -891,8 +891,12 @@ For any sets $A$ and $B$,
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\lean{Mathlib/Data/Set/Basic}{Set.inter\_comm}
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Let $A$ and $B$ be sets.
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We show (i) $A \cup B = B \cup A$ and then (ii) $A \cap B = B \cap A$.
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\noindent Let $A$ and $B$ be sets.
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We prove that
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\begin{enumerate}[(i)]
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\item $A \cup B = B \cup A$
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\item $A \cap B = B \cap A$.
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\end{enumerate}
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\paragraph{(i)}%
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@ -933,9 +937,12 @@ For any sets $A$, $B$ and $C$,
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\lean{Mathlib/Data/Set/Basic}{Set.inter\_assoc}
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Let $A$, $B$, and $C$ be sets.
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We show (i) $A \cup (B \cup C) = (A \cup B) \cup C$ and then (ii)
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$A \cap (B \cap C) = (A \cap B) \cap C$.
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\noindent Let $A$, $B$, and $C$ be sets.
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We prove that
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\begin{enumerate}[(i)]
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\item $A \cup (B \cup C) = (A \cup B) \cup C$
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\item $A \cap (B \cap C) = (A \cap B) \cap C$
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\end{enumerate}
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\paragraph{(i)}%
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@ -987,9 +994,12 @@ For any sets $A$, $B$, and $C$,
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\lean{Mathlib/Data/Set/Basic}{Set.union\_distrib\_left}
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Let $A$, $B$, and $C$ be sets.
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We show (i) $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ and then (ii)
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$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.
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\noindent Let $A$, $B$, and $C$ be sets.
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We prove that
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\begin{enumerate}[(i)]
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\item $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$
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\item $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$
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\end{enumerate}
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\paragraph{(i)}%
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@ -1040,9 +1050,12 @@ For any sets $A$, $B$, and $C$,
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\lean{Mathlib/Data/Set/Basic}{Set.diff\_inter}
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Let $A$, $B$, and $C$ be sets.
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We show (i) $C - (A \cup B) = (C - A) \cap (C - B)$ and then (ii)
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$C - (A \cap B) = (C - A) \cup (C - B)$.
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\noindent Let $A$, $B$, and $C$ be sets.
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We prove that
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\begin{enumerate}[(i)]
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\item $C - (A \cup B) = (C - A) \cap (C - B)$
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\item $C - (A \cap B) = (C - A) \cup (C - B)$
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\end{enumerate}
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\paragraph{(i)}%
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@ -1099,8 +1112,13 @@ For any set $A$,
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\lean{Mathlib/Data/Set/Basic}{Set.inter\_diff\_self}
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Let $A$ be an arbitrary set. We prove (i) that $A \cup \emptyset = A$, (ii)
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$A \cap \emptyset = \emptyset$, and (iii) $A \cap (C - A) = \emptyset$.
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\noindent Let $A$ be an arbitrary set.
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We prove that
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\begin{enumerate}[(i)]
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\item $A \cup \emptyset = A$
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\item $A \cap \emptyset = \emptyset$
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\item $A \cap (C - A) = \emptyset$
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\end{enumerate}
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\paragraph{(i)}%
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@ -1143,7 +1161,7 @@ For any set $A$,
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\end{proof}
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\subsection{\unverified{Monotonicity}}%
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\subsection{\verified{Monotonicity}}%
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\label{sub:monotonicity}
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For any sets $A$, $B$, and $C$,
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@ -1155,11 +1173,68 @@ For any sets $A$, $B$, and $C$,
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\begin{proof}
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TODO
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\statementpadding
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\lean*{Mathlib/Data/Set/Basic}{Set.union\_subset\_union\_left}
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\lean*{Mathlib/Data/Set/Basic}{Set.inter\_subset\_inter\_left}
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\lean{Mathlib/Data/Set/Lattice}{Set.sUnion\_mono}
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\noindent Let $A$, $B$, and $C$ be arbitrary sets.
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We prove that
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\begin{enumerate}[(i)]
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\item $A \subseteq B \Rightarrow A \cup C \subseteq B \cup C$
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\item $A \subseteq B \Rightarrow A \cap C \subseteq B \cap C$
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\item $A \subseteq B \Rightarrow \bigcup A \subseteq \bigcup B$
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\end{enumerate}
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\paragraph{(i)}%
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Suppose $A \subseteq B$.
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Let $x \in A \cup C$.
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There are two cases to consider.
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\subparagraph{Case 1}%
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Suppose $x \in A$.
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Then, by definition of the subset, $x \in B$.
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Therefore $x \in B \cup C$.
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\subparagraph{Case 2}%
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Suppose $x \in C$.
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Then $x$ is trivially a member of $B \cup C$.
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\subparagraph{Conclusion}%
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Since these cases are exhaustive and both imply $x \in B \cup C$, it
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follows $A \cup C \subseteq B \cup C$.
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\paragraph{(ii)}%
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Suppose $A \subseteq B$.
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Let $x \in A \cap C$.
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Then, by definition of the intersection of sets, $x \in A$ and $x \in C$.
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By definition of the subset, $x \in A$ implies $x \in B$.
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Therefore $x \in B$ and $x \in C$.
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That is, $x \in B \cap C$.
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Since this holds for arbitrary $x \in A \cap C$, it follows
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$A \cap C \subseteq B \cap C$.
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\paragraph{(iii)}%
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Suppose $A \subseteq B$.
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Let $x \in \bigcup A$.
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Then, by definition of the union of sets, there exists some $b \in A$ such
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that $x \in b$.
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By definition of the subset, $b \in B$ as well.
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Another application of the definition of the union of sets immediately
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implies that $x$ is a member of $\bigcup B$.
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\end{proof}
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\subsection{\unverified{Anti-monotonicity}}%
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\subsection{\verified{Anti-monotonicity}}%
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\label{sub:anti-monotonicity}
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For any sets $A$, $B$, and $C$,
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@ -1170,11 +1245,46 @@ For any sets $A$, $B$, and $C$,
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\begin{proof}
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TODO
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\statementpadding
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\lean*{Mathlib/Data/Set/Basic}{Set.diff\_subset\_diff\_right}
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\lean{Mathlib/Data/Set/Lattice}{Set.sInter\_subset\_sInter}
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\noindent Let $A$, $B$, and $C$ be arbitrary sets.
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We prove that
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\begin{enumerate}[(i)]
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\item $A \subseteq B \Rightarrow C - B \subseteq C - A$
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\item $\emptyset \neq A \subseteq B \Rightarrow
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\bigcap B \subseteq \bigcap A$
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\end{enumerate}
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\paragraph{(i)}%
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Suppose $A \subseteq B$.
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Let $x \in C - B$.
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By definition of the relative complement, $x \in C$ and $x \not\in B$.
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Then $x$ cannot be a member of $A$, since otherwise this would contradict
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our subset hypothesis.
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That is, $x \in C$ and $x \not\in A$.
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Therefore $x \in C - A$.
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Since this holds for arbitrary $x \in C - B$, it follows that
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$C - B \subseteq C - A$.
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\paragraph{(ii)}%
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Suppose $A \neq \emptyset$ and $A \subseteq B$.
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Then $B \neq \emptyset$.
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Let $x \in \bigcap B$.
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By definition of the intersection of sets, for all $b \in B$, $x \in b$.
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But then, by definition of the subset, for all $a \in A$, $x \in a$.
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Therefore $x \in \bigcap A$.
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Since this holds for arbitrary $x \in \bigcap B$, it follows that
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$\bigcap B \subseteq \bigcap A$.
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\end{proof}
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\subsection{\unverified{General Distributive Laws}}%
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\subsection{\partial{General Distributive Laws}}%
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\label{sub:general-distributive-laws}
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For any sets $A$ and $\mathscr{B}$,
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\begin{proof}
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TODO
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Let $A$ and $\mathscr{B}$ be sets.
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We prove that
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\begin{enumerate}[(i)]
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\item For $\mathscr{B} \neq \emptyset$,
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$A \cup \bigcap \mathscr{B} =
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\bigcap\; \{ A \cup X \mid X \in \mathscr{B} \}$.
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\item $A \cap \bigcup \mathscr{B} =
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\bigcup\; \{ A \cap X \mid X \in \mathscr{B} \}$
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\end{enumerate}
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\paragraph{(i)}%
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Suppose $\mathscr{B}$ is nonempty.
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Then $\bigcap \mathscr{B}$ is defined.
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By definition of the union and intersection of sets,
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\begin{align*}
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A \cup \bigcap \mathscr{B}
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& = \{ x \mid x \in A \lor x \in \bigcap \mathscr{B} \} \\
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& = \{ x \mid x \in A \lor
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x \in \{ y \mid (\forall b \in \mathscr{B}), y \in b \}\} \\
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& = \{ x \mid x \in A \lor (\forall b \in \mathscr{B}), x \in b \} \\
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& = \{ x \mid \forall b \in \mathscr{B}, x \in A \lor x \in b \} \\
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& = \{ x \mid \forall b \in \mathscr{B}, x \in A \cup b \} \\
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& = \{ x \mid
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x \in \bigcap\; \{ A \cup X \mid X \in \mathscr{B} \}\} \\
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& = \bigcap\; \{ A \cup X \mid X \in \mathscr{B} \}.
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\end{align*}
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\paragraph{(ii)}%
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By definition of the intersection and union of sets,
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\begin{align*}
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A \cap \bigcup \mathscr{B}
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& = \{ x \mid x \in A \land x \in \bigcup \mathscr{B} \} \\
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& = \{ x \mid x \in A \land
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x \in \{ y \mid (\exists b \in \mathscr{B}), y \in b \}\} \\
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& = \{ x \mid x \in A \land (\exists b \in \mathscr{B}), x \in b \} \\
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& = \{ x \mid \exists b \in \mathscr{B}, x \in A \land x \in b \} \\
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& = \{ x \mid \exists b \in \mathscr{B} x \in A \cap b \} \\
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& = \{ x \mid
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x \in \bigcup\; \{ A \cap X \mid X \in \mathscr{B} \}\} \\
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& = \bigcup\; \{ A \cap X \mid X \in \mathscr{B} \}.
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\end{align*}
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\end{proof}
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\subsection{\unverified{General De Morgan's Laws}}%
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\subsection{\partial{General De Morgan's Laws}}%
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\label{sub:general-de-morgans-laws}
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For any set $C$ and $\mathscr{A} \neq \emptyset$,
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@ -1203,7 +1355,52 @@ For any set $C$ and $\mathscr{A} \neq \emptyset$,
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\begin{proof}
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TODO
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Let $C$ and $\mathscr{A}$ be sets such that $\mathscr{A} \neq \emptyset$.
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We prove that
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\begin{enumerate}[(i)]
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\item $C - \bigcup \mathscr{A} =
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\bigcap\; \{ C - X \mid X \in \mathscr{A} \}$
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\item $C - \bigcap \mathscr{A} =
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\bigcup\; \{ C - X \mid X \in \mathscr{A} \}$
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\end{enumerate}
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\paragraph{(i)}%
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By definition of the relative complement, union, and intersection of sets,
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\begin{align*}
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C - \bigcup \mathscr{A}
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& = \{ x \mid x \in C \land x \not\in \bigcup \mathscr{A} \} \\
|
||||
& = \{ x \mid x \in C \land
|
||||
x \not\in \{ y \mid (\exists b \in \mathscr{A}) y \in b \}\} \\
|
||||
& = \{ x \mid x \in C \land
|
||||
\neg(\exists b \in \mathscr{A}, x \in b) \} \\
|
||||
& = \{ x \mid x \in C \land
|
||||
(\forall b \in \mathscr{A}, x \not\in b) \} \\
|
||||
& = \{ x \mid
|
||||
\forall b \in \mathscr{A}, x \in C \land x \not\in b \} \\
|
||||
& = \{ x \mid \forall b \in \mathscr{A}, x \in C - b \} \\
|
||||
& = \{ x \mid x \in \bigcap\; \{ C - X \mid X \in \mathscr{A} \} \\
|
||||
& = \bigcap\; \{ C - X \mid X \in \mathscr{A} \}.
|
||||
\end{align*}
|
||||
|
||||
\paragraph{(ii)}%
|
||||
|
||||
By definition of the relative complement, union, and intersection of sets,
|
||||
\begin{align*}
|
||||
C - \bigcap \mathscr{A}
|
||||
& = \{ x \mid x \in C \land x \not\in \bigcap \mathscr{A} \} \\
|
||||
& = \{ x \mid x \in C \land
|
||||
x \not\in \{ y \mid (\forall b \in \mathscr{A}) y \in b \}\} \\
|
||||
& = \{ x \mid x \in C \land
|
||||
\neg(\forall b \in \mathscr{A}, x \in b) \} \\
|
||||
& = \{ x \mid x \in C \land
|
||||
\exists b \in \mathscr{A}, x \not\in b \} \\
|
||||
& = \{ x \mid
|
||||
\exists b \in \mathscr{A}, x \in C \land x \not\in b \} \\
|
||||
& = \{ x \mid \exists b \in \mathscr{A}, x \in C - b \} \\
|
||||
& = \{ x \mid x \in \bigcup\; \{ C - X \mid X \in \mathscr{A} \} \} \\
|
||||
& = \bigcup\; \{ C - X \mid X \in \mathscr{A} \}.
|
||||
\end{align*}
|
||||
|
||||
\end{proof}
|
||||
|
||||
|
|
|
|||
|
|
@ -1,9 +1,4 @@
|
|||
import Mathlib.Logic.Basic
|
||||
import Mathlib.Data.Set.Basic
|
||||
import Mathlib.Data.Set.Lattice
|
||||
import Mathlib.Tactic.LibrarySearch
|
||||
|
||||
import Common.Set.Basic
|
||||
|
||||
/-! # Enderton.Chapter_1
|
||||
|
||||
|
|
@ -131,273 +126,4 @@ theorem exercise_1_4 (x y : α) (hx : x ∈ B) (hy : y ∈ B)
|
|||
(Set.singleton_subset_iff.mpr hx)
|
||||
(Set.singleton_subset_iff.mpr hy)
|
||||
|
||||
/-- ### Exercise 3.1
|
||||
|
||||
Assume that `A` is the set of integers divisible by `4`. Similarly assume that
|
||||
`B` and `C` are the sets of integers divisible by `9` and `10`, respectively.
|
||||
What is in `A ∩ B ∩ C`?
|
||||
-/
|
||||
theorem exercise_3_1 {A B C : Set ℤ}
|
||||
(hA : A = { x | Dvd.dvd 4 x })
|
||||
(hB : B = { x | Dvd.dvd 9 x })
|
||||
(hC : C = { x | Dvd.dvd 10 x })
|
||||
: ∀ x ∈ (A ∩ B ∩ C), (4 ∣ x) ∧ (9 ∣ x) ∧ (10 ∣ x) := by
|
||||
intro x h
|
||||
rw [Set.mem_inter_iff] at h
|
||||
have ⟨⟨ha, hb⟩, hc⟩ := h
|
||||
refine ⟨?_, ⟨?_, ?_⟩⟩
|
||||
· rw [hA] at ha
|
||||
exact Set.mem_setOf.mp ha
|
||||
· rw [hB] at hb
|
||||
exact Set.mem_setOf.mp hb
|
||||
· rw [hC] at hc
|
||||
exact Set.mem_setOf.mp hc
|
||||
|
||||
/-- ### Exercise 3.2
|
||||
|
||||
Give an example of sets `A` and `B` for which `⋃ A = ⋃ B` but `A ≠ B`.
|
||||
-/
|
||||
theorem exercise_3_2 {A B : Set (Set ℕ)}
|
||||
(hA : A = {{1}, {2}}) (hB : B = {{1, 2}})
|
||||
: Set.sUnion A = Set.sUnion B ∧ A ≠ B := by
|
||||
apply And.intro
|
||||
· show ⋃₀ A = ⋃₀ B
|
||||
ext x
|
||||
show (∃ t, t ∈ A ∧ x ∈ t) ↔ ∃ t, t ∈ B ∧ x ∈ t
|
||||
apply Iff.intro
|
||||
· intro ⟨t, ⟨ht, hx⟩⟩
|
||||
rw [hA] at ht
|
||||
refine ⟨{1, 2}, ⟨by rw [hB]; simp, ?_⟩⟩
|
||||
apply Or.elim ht <;>
|
||||
· intro ht'
|
||||
rw [ht'] at hx
|
||||
rw [hx]
|
||||
simp
|
||||
· intro ⟨t, ⟨ht, hx⟩⟩
|
||||
rw [hB] at ht
|
||||
rw [ht] at hx
|
||||
apply Or.elim hx
|
||||
· intro hx'
|
||||
exact ⟨{1}, ⟨by rw [hA]; simp, by rw [hx']; simp⟩⟩
|
||||
· intro hx'
|
||||
exact ⟨{2}, ⟨by rw [hA]; simp, by rw [hx']; simp⟩⟩
|
||||
· show A ≠ B
|
||||
-- Find an element that exists in `B` but not in `A`. Extensionality
|
||||
-- concludes the proof.
|
||||
intro h
|
||||
rw [hA, hB, Set.ext_iff] at h
|
||||
have h₁ := h {1, 2}
|
||||
simp at h₁
|
||||
rw [Set.ext_iff] at h₁
|
||||
have h₂ := h₁ 2
|
||||
simp at h₂
|
||||
|
||||
/-- ### Exercise 3.3
|
||||
|
||||
Show that every member of a set `A` is a subset of `U A`. (This was stated as an
|
||||
example in this section.)
|
||||
-/
|
||||
theorem exercise_3_3 {A : Set (Set α)}
|
||||
: ∀ x ∈ A, x ⊆ Set.sUnion A := by
|
||||
intro x hx
|
||||
show ∀ y ∈ x, y ∈ { a | ∃ t, t ∈ A ∧ a ∈ t }
|
||||
intro y hy
|
||||
rw [Set.mem_setOf_eq]
|
||||
exact ⟨x, ⟨hx, hy⟩⟩
|
||||
|
||||
/-- ### Exercise 3.4
|
||||
|
||||
Show that if `A ⊆ B`, then `⋃ A ⊆ ⋃ B`.
|
||||
-/
|
||||
theorem exercise_3_4 (h : A ⊆ B) : ⋃₀ A ⊆ ⋃₀ B := by
|
||||
show ∀ x ∈ { a | ∃ t, t ∈ A ∧ a ∈ t }, x ∈ { a | ∃ t, t ∈ B ∧ a ∈ t }
|
||||
intro x hx
|
||||
rw [Set.mem_setOf_eq] at hx
|
||||
have ⟨t, ⟨ht, hxt⟩⟩ := hx
|
||||
rw [Set.mem_setOf_eq]
|
||||
exact ⟨t, ⟨h ht, hxt⟩⟩
|
||||
|
||||
/-- ### Exercise 3.5
|
||||
|
||||
Assume that every member of `𝓐` is a subset of `B`. Show that `⋃ 𝓐 ⊆ B`.
|
||||
-/
|
||||
theorem exercise_3_5 (h : ∀ x ∈ 𝓐, x ⊆ B) : ⋃₀ 𝓐 ⊆ B := by
|
||||
unfold Set.sUnion sSup Set.instSupSetSet
|
||||
simp only
|
||||
show ∀ y ∈ { a | ∃ t, t ∈ 𝓐 ∧ a ∈ t }, y ∈ B
|
||||
intro y hy
|
||||
rw [Set.mem_setOf_eq] at hy
|
||||
have ⟨t, ⟨ht𝓐, hyt⟩⟩ := hy
|
||||
exact (h t ht𝓐) hyt
|
||||
|
||||
/-- ### Exercise 3.6a
|
||||
|
||||
Show that for any set `A`, `⋃ 𝓟 A = A`.
|
||||
-/
|
||||
theorem exercise_3_6a : ⋃₀ (Set.powerset A) = A := by
|
||||
unfold Set.sUnion sSup Set.instSupSetSet Set.powerset
|
||||
simp only
|
||||
ext x
|
||||
apply Iff.intro
|
||||
· intro hx
|
||||
rw [Set.mem_setOf_eq] at hx
|
||||
have ⟨t, ⟨htl, htr⟩⟩ := hx
|
||||
rw [Set.mem_setOf_eq] at htl
|
||||
exact htl htr
|
||||
· intro hx
|
||||
rw [Set.mem_setOf_eq]
|
||||
exact ⟨A, ⟨by rw [Set.mem_setOf_eq], hx⟩⟩
|
||||
|
||||
/-- ### Exercise 3.6b
|
||||
|
||||
Show that `A ⊆ 𝓟 ⋃ A`. Under what conditions does equality hold?
|
||||
-/
|
||||
theorem exercise_3_6b
|
||||
: A ⊆ Set.powerset (⋃₀ A)
|
||||
∧ (A = Set.powerset (⋃₀ A) ↔ ∃ B, A = Set.powerset B) := by
|
||||
apply And.intro
|
||||
· unfold Set.powerset
|
||||
show ∀ x ∈ A, x ∈ { t | t ⊆ ⋃₀ A }
|
||||
intro x hx
|
||||
rw [Set.mem_setOf]
|
||||
exact exercise_3_3 x hx
|
||||
· apply Iff.intro
|
||||
· intro hA
|
||||
exact ⟨⋃₀ A, hA⟩
|
||||
· intro ⟨B, hB⟩
|
||||
conv => rhs; rw [hB, exercise_3_6a]
|
||||
exact hB
|
||||
|
||||
/-- ### Exercise 3.7a
|
||||
|
||||
Show that for any sets `A` and `B`, `𝓟 A ∩ 𝓟 B = 𝓟 (A ∩ B)`.
|
||||
-/
|
||||
theorem exercise_3_7A
|
||||
: Set.powerset A ∩ Set.powerset B = Set.powerset (A ∩ B) := by
|
||||
suffices
|
||||
Set.powerset A ∩ Set.powerset B ⊆ Set.powerset (A ∩ B) ∧
|
||||
Set.powerset (A ∩ B) ⊆ Set.powerset A ∩ Set.powerset B from
|
||||
subset_antisymm this.left this.right
|
||||
apply And.intro
|
||||
· unfold Set.powerset
|
||||
intro x hx
|
||||
simp only [Set.mem_inter_iff, Set.mem_setOf_eq] at hx
|
||||
rwa [Set.mem_setOf, Set.subset_inter_iff]
|
||||
· unfold Set.powerset
|
||||
simp
|
||||
intro x hA _
|
||||
exact hA
|
||||
|
||||
-- theorem false_of_false_iff_true : (false ↔ true) → false := by simp
|
||||
|
||||
/-- ### Exercise 3.7b (i)
|
||||
|
||||
Show that `𝓟 A ∪ 𝓟 B ⊆ 𝓟 (A ∪ B)`.
|
||||
-/
|
||||
theorem exercise_3_7b_i
|
||||
: Set.powerset A ∪ Set.powerset B ⊆ Set.powerset (A ∪ B) := by
|
||||
unfold Set.powerset
|
||||
intro x hx
|
||||
simp at hx
|
||||
apply Or.elim hx
|
||||
· intro hA
|
||||
rw [Set.mem_setOf_eq]
|
||||
exact Set.subset_union_of_subset_left hA B
|
||||
· intro hB
|
||||
rw [Set.mem_setOf_eq]
|
||||
exact Set.subset_union_of_subset_right hB A
|
||||
|
||||
/-- ### Exercise 3.7b (ii)
|
||||
|
||||
Under what conditions does `𝓟 A ∪ 𝓟 B = 𝓟 (A ∪ B)`.?
|
||||
-/
|
||||
theorem exercise_3_7b_ii
|
||||
: Set.powerset A ∪ Set.powerset B = Set.powerset (A ∪ B) ↔ A ⊆ B ∨ B ⊆ A := by
|
||||
unfold Set.powerset
|
||||
apply Iff.intro
|
||||
· intro h
|
||||
by_contra nh
|
||||
rw [not_or] at nh
|
||||
have ⟨a, hA⟩ := Set.not_subset.mp nh.left
|
||||
have ⟨b, hB⟩ := Set.not_subset.mp nh.right
|
||||
rw [Set.ext_iff] at h
|
||||
have hz := h {a, b}
|
||||
-- `hz` states that `{a, b} ⊆ A ∨ {a, b} ⊆ B ↔ {a, b} ⊆ A ∪ B`. We show the
|
||||
-- left-hand side is false but the right-hand side is true, yielding our
|
||||
-- contradiction.
|
||||
suffices ¬({a, b} ⊆ A ∨ {a, b} ⊆ B) by
|
||||
have hz₁ : a ∈ A ∪ B := by
|
||||
rw [Set.mem_union]
|
||||
exact Or.inl hA.left
|
||||
have hz₂ : b ∈ A ∪ B := by
|
||||
rw [Set.mem_union]
|
||||
exact Or.inr hB.left
|
||||
exact absurd (hz.mpr $ Set.mem_mem_imp_pair_subset hz₁ hz₂) this
|
||||
intro hAB
|
||||
exact Or.elim hAB
|
||||
(fun y => absurd (y $ show b ∈ {a, b} by simp) hB.right)
|
||||
(fun y => absurd (y $ show a ∈ {a, b} by simp) hA.right)
|
||||
· intro h
|
||||
ext x
|
||||
apply Or.elim h
|
||||
· intro hA
|
||||
apply Iff.intro
|
||||
· intro hx
|
||||
exact Or.elim hx
|
||||
(Set.subset_union_of_subset_left · B)
|
||||
(Set.subset_union_of_subset_right · A)
|
||||
· intro hx
|
||||
refine Or.inr (Set.Subset.trans hx ?_)
|
||||
exact subset_of_eq (Set.left_subset_union_eq_self hA)
|
||||
· intro hB
|
||||
apply Iff.intro
|
||||
· intro hx
|
||||
exact Or.elim hx
|
||||
(Set.subset_union_of_subset_left · B)
|
||||
(Set.subset_union_of_subset_right · A)
|
||||
· intro hx
|
||||
refine Or.inl (Set.Subset.trans hx ?_)
|
||||
exact subset_of_eq (Set.right_subset_union_eq_self hB)
|
||||
|
||||
/-- ### Exercise 3.9
|
||||
|
||||
Give an example of sets `a` and `B` for which `a ∈ B` but `𝓟 a ∉ 𝓟 B`.
|
||||
-/
|
||||
theorem exercise_3_9 (ha : a = {1}) (hB : B = {{1}})
|
||||
: a ∈ B ∧ Set.powerset a ∉ Set.powerset B := by
|
||||
apply And.intro
|
||||
· rw [ha, hB]
|
||||
simp
|
||||
· intro h
|
||||
have h₁ : Set.powerset a = {∅, {1}} := by
|
||||
rw [ha]
|
||||
exact Set.powerset_singleton 1
|
||||
have h₂ : Set.powerset B = {∅, {{1}}} := by
|
||||
rw [hB]
|
||||
exact Set.powerset_singleton {1}
|
||||
rw [h₁, h₂] at h
|
||||
simp at h
|
||||
apply Or.elim h
|
||||
· intro h
|
||||
rw [Set.ext_iff] at h
|
||||
have := h ∅
|
||||
simp at this
|
||||
· intro h
|
||||
rw [Set.ext_iff] at h
|
||||
have := h 1
|
||||
simp at this
|
||||
|
||||
/-- ### Exercise 3.10
|
||||
|
||||
Show that if `a ∈ B`, then `𝓟 a ∈ 𝓟 𝓟 ⋃ B`.
|
||||
-/
|
||||
theorem exercise_3_10 (ha : a ∈ B)
|
||||
: Set.powerset a ∈ Set.powerset (Set.powerset (⋃₀ B)) := by
|
||||
have h₁ := exercise_3_3 a ha
|
||||
have h₂ := exercise_1_3 h₁
|
||||
generalize hb : 𝒫 (⋃₀ B) = b
|
||||
conv => rhs; unfold Set.powerset
|
||||
rw [← hb, Set.mem_setOf_eq]
|
||||
exact h₂
|
||||
|
||||
end Enderton.Set.Chapter_1
|
||||
|
|
@ -0,0 +1,284 @@
|
|||
import Mathlib.Data.Set.Basic
|
||||
import Mathlib.Data.Set.Lattice
|
||||
|
||||
import Bookshelf.Enderton.Set.Chapter_1
|
||||
import Common.Set.Basic
|
||||
|
||||
/-! # Enderton.Chapter_2
|
||||
|
||||
Axioms and Operations
|
||||
-/
|
||||
|
||||
namespace Enderton.Set.Chapter_2
|
||||
|
||||
|
||||
/-- ### Exercise 3.1
|
||||
|
||||
Assume that `A` is the set of integers divisible by `4`. Similarly assume that
|
||||
`B` and `C` are the sets of integers divisible by `9` and `10`, respectively.
|
||||
What is in `A ∩ B ∩ C`?
|
||||
-/
|
||||
theorem exercise_3_1 {A B C : Set ℤ}
|
||||
(hA : A = { x | Dvd.dvd 4 x })
|
||||
(hB : B = { x | Dvd.dvd 9 x })
|
||||
(hC : C = { x | Dvd.dvd 10 x })
|
||||
: ∀ x ∈ (A ∩ B ∩ C), (4 ∣ x) ∧ (9 ∣ x) ∧ (10 ∣ x) := by
|
||||
intro x h
|
||||
rw [Set.mem_inter_iff] at h
|
||||
have ⟨⟨ha, hb⟩, hc⟩ := h
|
||||
refine ⟨?_, ⟨?_, ?_⟩⟩
|
||||
· rw [hA] at ha
|
||||
exact Set.mem_setOf.mp ha
|
||||
· rw [hB] at hb
|
||||
exact Set.mem_setOf.mp hb
|
||||
· rw [hC] at hc
|
||||
exact Set.mem_setOf.mp hc
|
||||
|
||||
/-- ### Exercise 3.2
|
||||
|
||||
Give an example of sets `A` and `B` for which `⋃ A = ⋃ B` but `A ≠ B`.
|
||||
-/
|
||||
theorem exercise_3_2 {A B : Set (Set ℕ)}
|
||||
(hA : A = {{1}, {2}}) (hB : B = {{1, 2}})
|
||||
: Set.sUnion A = Set.sUnion B ∧ A ≠ B := by
|
||||
apply And.intro
|
||||
· show ⋃₀ A = ⋃₀ B
|
||||
ext x
|
||||
show (∃ t, t ∈ A ∧ x ∈ t) ↔ ∃ t, t ∈ B ∧ x ∈ t
|
||||
apply Iff.intro
|
||||
· intro ⟨t, ⟨ht, hx⟩⟩
|
||||
rw [hA] at ht
|
||||
refine ⟨{1, 2}, ⟨by rw [hB]; simp, ?_⟩⟩
|
||||
apply Or.elim ht <;>
|
||||
· intro ht'
|
||||
rw [ht'] at hx
|
||||
rw [hx]
|
||||
simp
|
||||
· intro ⟨t, ⟨ht, hx⟩⟩
|
||||
rw [hB] at ht
|
||||
rw [ht] at hx
|
||||
apply Or.elim hx
|
||||
· intro hx'
|
||||
exact ⟨{1}, ⟨by rw [hA]; simp, by rw [hx']; simp⟩⟩
|
||||
· intro hx'
|
||||
exact ⟨{2}, ⟨by rw [hA]; simp, by rw [hx']; simp⟩⟩
|
||||
· show A ≠ B
|
||||
-- Find an element that exists in `B` but not in `A`. Extensionality
|
||||
-- concludes the proof.
|
||||
intro h
|
||||
rw [hA, hB, Set.ext_iff] at h
|
||||
have h₁ := h {1, 2}
|
||||
simp at h₁
|
||||
rw [Set.ext_iff] at h₁
|
||||
have h₂ := h₁ 2
|
||||
simp at h₂
|
||||
|
||||
/-- ### Exercise 3.3
|
||||
|
||||
Show that every member of a set `A` is a subset of `U A`. (This was stated as an
|
||||
example in this section.)
|
||||
-/
|
||||
theorem exercise_3_3 {A : Set (Set α)}
|
||||
: ∀ x ∈ A, x ⊆ Set.sUnion A := by
|
||||
intro x hx
|
||||
show ∀ y ∈ x, y ∈ { a | ∃ t, t ∈ A ∧ a ∈ t }
|
||||
intro y hy
|
||||
rw [Set.mem_setOf_eq]
|
||||
exact ⟨x, ⟨hx, hy⟩⟩
|
||||
|
||||
/-- ### Exercise 3.4
|
||||
|
||||
Show that if `A ⊆ B`, then `⋃ A ⊆ ⋃ B`.
|
||||
-/
|
||||
theorem exercise_3_4 (h : A ⊆ B) : ⋃₀ A ⊆ ⋃₀ B := by
|
||||
show ∀ x ∈ { a | ∃ t, t ∈ A ∧ a ∈ t }, x ∈ { a | ∃ t, t ∈ B ∧ a ∈ t }
|
||||
intro x hx
|
||||
rw [Set.mem_setOf_eq] at hx
|
||||
have ⟨t, ⟨ht, hxt⟩⟩ := hx
|
||||
rw [Set.mem_setOf_eq]
|
||||
exact ⟨t, ⟨h ht, hxt⟩⟩
|
||||
|
||||
/-- ### Exercise 3.5
|
||||
|
||||
Assume that every member of `𝓐` is a subset of `B`. Show that `⋃ 𝓐 ⊆ B`.
|
||||
-/
|
||||
theorem exercise_3_5 (h : ∀ x ∈ 𝓐, x ⊆ B) : ⋃₀ 𝓐 ⊆ B := by
|
||||
unfold Set.sUnion sSup Set.instSupSetSet
|
||||
simp only
|
||||
show ∀ y ∈ { a | ∃ t, t ∈ 𝓐 ∧ a ∈ t }, y ∈ B
|
||||
intro y hy
|
||||
rw [Set.mem_setOf_eq] at hy
|
||||
have ⟨t, ⟨ht𝓐, hyt⟩⟩ := hy
|
||||
exact (h t ht𝓐) hyt
|
||||
|
||||
/-- ### Exercise 3.6a
|
||||
|
||||
Show that for any set `A`, `⋃ 𝓟 A = A`.
|
||||
-/
|
||||
theorem exercise_3_6a : ⋃₀ (Set.powerset A) = A := by
|
||||
unfold Set.sUnion sSup Set.instSupSetSet Set.powerset
|
||||
simp only
|
||||
ext x
|
||||
apply Iff.intro
|
||||
· intro hx
|
||||
rw [Set.mem_setOf_eq] at hx
|
||||
have ⟨t, ⟨htl, htr⟩⟩ := hx
|
||||
rw [Set.mem_setOf_eq] at htl
|
||||
exact htl htr
|
||||
· intro hx
|
||||
rw [Set.mem_setOf_eq]
|
||||
exact ⟨A, ⟨by rw [Set.mem_setOf_eq], hx⟩⟩
|
||||
|
||||
/-- ### Exercise 3.6b
|
||||
|
||||
Show that `A ⊆ 𝓟 ⋃ A`. Under what conditions does equality hold?
|
||||
-/
|
||||
theorem exercise_3_6b
|
||||
: A ⊆ Set.powerset (⋃₀ A)
|
||||
∧ (A = Set.powerset (⋃₀ A) ↔ ∃ B, A = Set.powerset B) := by
|
||||
apply And.intro
|
||||
· unfold Set.powerset
|
||||
show ∀ x ∈ A, x ∈ { t | t ⊆ ⋃₀ A }
|
||||
intro x hx
|
||||
rw [Set.mem_setOf]
|
||||
exact exercise_3_3 x hx
|
||||
· apply Iff.intro
|
||||
· intro hA
|
||||
exact ⟨⋃₀ A, hA⟩
|
||||
· intro ⟨B, hB⟩
|
||||
conv => rhs; rw [hB, exercise_3_6a]
|
||||
exact hB
|
||||
|
||||
/-- ### Exercise 3.7a
|
||||
|
||||
Show that for any sets `A` and `B`, `𝓟 A ∩ 𝓟 B = 𝓟 (A ∩ B)`.
|
||||
-/
|
||||
theorem exercise_3_7A
|
||||
: Set.powerset A ∩ Set.powerset B = Set.powerset (A ∩ B) := by
|
||||
suffices
|
||||
Set.powerset A ∩ Set.powerset B ⊆ Set.powerset (A ∩ B) ∧
|
||||
Set.powerset (A ∩ B) ⊆ Set.powerset A ∩ Set.powerset B from
|
||||
subset_antisymm this.left this.right
|
||||
apply And.intro
|
||||
· unfold Set.powerset
|
||||
intro x hx
|
||||
simp only [Set.mem_inter_iff, Set.mem_setOf_eq] at hx
|
||||
rwa [Set.mem_setOf, Set.subset_inter_iff]
|
||||
· unfold Set.powerset
|
||||
simp
|
||||
intro x hA _
|
||||
exact hA
|
||||
|
||||
-- theorem false_of_false_iff_true : (false ↔ true) → false := by simp
|
||||
|
||||
/-- ### Exercise 3.7b (i)
|
||||
|
||||
Show that `𝓟 A ∪ 𝓟 B ⊆ 𝓟 (A ∪ B)`.
|
||||
-/
|
||||
theorem exercise_3_7b_i
|
||||
: Set.powerset A ∪ Set.powerset B ⊆ Set.powerset (A ∪ B) := by
|
||||
unfold Set.powerset
|
||||
intro x hx
|
||||
simp at hx
|
||||
apply Or.elim hx
|
||||
· intro hA
|
||||
rw [Set.mem_setOf_eq]
|
||||
exact Set.subset_union_of_subset_left hA B
|
||||
· intro hB
|
||||
rw [Set.mem_setOf_eq]
|
||||
exact Set.subset_union_of_subset_right hB A
|
||||
|
||||
/-- ### Exercise 3.7b (ii)
|
||||
|
||||
Under what conditions does `𝓟 A ∪ 𝓟 B = 𝓟 (A ∪ B)`.?
|
||||
-/
|
||||
theorem exercise_3_7b_ii
|
||||
: Set.powerset A ∪ Set.powerset B = Set.powerset (A ∪ B) ↔ A ⊆ B ∨ B ⊆ A := by
|
||||
unfold Set.powerset
|
||||
apply Iff.intro
|
||||
· intro h
|
||||
by_contra nh
|
||||
rw [not_or] at nh
|
||||
have ⟨a, hA⟩ := Set.not_subset.mp nh.left
|
||||
have ⟨b, hB⟩ := Set.not_subset.mp nh.right
|
||||
rw [Set.ext_iff] at h
|
||||
have hz := h {a, b}
|
||||
-- `hz` states that `{a, b} ⊆ A ∨ {a, b} ⊆ B ↔ {a, b} ⊆ A ∪ B`. We show the
|
||||
-- left-hand side is false but the right-hand side is true, yielding our
|
||||
-- contradiction.
|
||||
suffices ¬({a, b} ⊆ A ∨ {a, b} ⊆ B) by
|
||||
have hz₁ : a ∈ A ∪ B := by
|
||||
rw [Set.mem_union]
|
||||
exact Or.inl hA.left
|
||||
have hz₂ : b ∈ A ∪ B := by
|
||||
rw [Set.mem_union]
|
||||
exact Or.inr hB.left
|
||||
exact absurd (hz.mpr $ Set.mem_mem_imp_pair_subset hz₁ hz₂) this
|
||||
intro hAB
|
||||
exact Or.elim hAB
|
||||
(fun y => absurd (y $ show b ∈ {a, b} by simp) hB.right)
|
||||
(fun y => absurd (y $ show a ∈ {a, b} by simp) hA.right)
|
||||
· intro h
|
||||
ext x
|
||||
apply Or.elim h
|
||||
· intro hA
|
||||
apply Iff.intro
|
||||
· intro hx
|
||||
exact Or.elim hx
|
||||
(Set.subset_union_of_subset_left · B)
|
||||
(Set.subset_union_of_subset_right · A)
|
||||
· intro hx
|
||||
refine Or.inr (Set.Subset.trans hx ?_)
|
||||
exact subset_of_eq (Set.left_subset_union_eq_self hA)
|
||||
· intro hB
|
||||
apply Iff.intro
|
||||
· intro hx
|
||||
exact Or.elim hx
|
||||
(Set.subset_union_of_subset_left · B)
|
||||
(Set.subset_union_of_subset_right · A)
|
||||
· intro hx
|
||||
refine Or.inl (Set.Subset.trans hx ?_)
|
||||
exact subset_of_eq (Set.right_subset_union_eq_self hB)
|
||||
|
||||
/-- ### Exercise 3.9
|
||||
|
||||
Give an example of sets `a` and `B` for which `a ∈ B` but `𝓟 a ∉ 𝓟 B`.
|
||||
-/
|
||||
theorem exercise_3_9 (ha : a = {1}) (hB : B = {{1}})
|
||||
: a ∈ B ∧ Set.powerset a ∉ Set.powerset B := by
|
||||
apply And.intro
|
||||
· rw [ha, hB]
|
||||
simp
|
||||
· intro h
|
||||
have h₁ : Set.powerset a = {∅, {1}} := by
|
||||
rw [ha]
|
||||
exact Set.powerset_singleton 1
|
||||
have h₂ : Set.powerset B = {∅, {{1}}} := by
|
||||
rw [hB]
|
||||
exact Set.powerset_singleton {1}
|
||||
rw [h₁, h₂] at h
|
||||
simp at h
|
||||
apply Or.elim h
|
||||
· intro h
|
||||
rw [Set.ext_iff] at h
|
||||
have := h ∅
|
||||
simp at this
|
||||
· intro h
|
||||
rw [Set.ext_iff] at h
|
||||
have := h 1
|
||||
simp at this
|
||||
|
||||
/-- ### Exercise 3.10
|
||||
|
||||
Show that if `a ∈ B`, then `𝓟 a ∈ 𝓟 𝓟 ⋃ B`.
|
||||
-/
|
||||
theorem exercise_3_10 (ha : a ∈ B)
|
||||
: Set.powerset a ∈ Set.powerset (Set.powerset (⋃₀ B)) := by
|
||||
have h₁ := exercise_3_3 a ha
|
||||
have h₂ := Chapter_1.exercise_1_3 h₁
|
||||
generalize hb : 𝒫 (⋃₀ B) = b
|
||||
conv => rhs; unfold Set.powerset
|
||||
rw [← hb, Set.mem_setOf_eq]
|
||||
exact h₂
|
||||
|
||||
end Enderton.Set.Chapter_2
|
||||
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Reference in New Issue