Enderton, basic axioms/unions exercises.
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@ -432,7 +432,7 @@ List all the members of $V_4$.
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\section{Exercises 3}%
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\label{sec:exercises-3}
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\subsection{\unverified{Exercise 3.1}}%
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\subsection{\verified{Exercise 3.1}}%
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\label{sub:exercise-3.1}
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Assume that $A$ is the set of integers divisible by $4$.
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@ -440,25 +440,31 @@ Similarly assume that $B$ and $C$ are the sets of integers divisible by $9$ and
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$10$, respectively.
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What is in $A \cap B \cap C$?
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\begin{proof}
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\begin{answer}
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TODO
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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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\end{proof}
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The set of integers divisible by $4$, $9$, and $10$.
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\subsection{\unverified{Exercise 3.2}}%
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\end{answer}
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\subsection{\verified{Exercise 3.2}}%
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\label{sub:exercise-3.2}
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Give an example of sets $A$ and $B$ for which $\bigcup A = \bigcup B$ but
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$A \neq B$.
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\begin{proof}
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\begin{answer}
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TODO
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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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\end{proof}
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Let $A = \{\{1\}, \{2\}\}$ and $B = \{\{1, 2\}\}$.
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\subsection{\unverified{Exercise 3.3}}%
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\end{answer}
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\subsection{\verified{Exercise 3.3}}%
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\label{sub:exercise-3.3}
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Show that every member of a set $A$ is a subset of $\bigcup A$.
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@ -466,22 +472,38 @@ Show that every member of a set $A$ is a subset of $\bigcup A$.
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\begin{proof}
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TODO
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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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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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Then $\{ y \mid y \in x\} \subseteq \bigcup A$.
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But $\{ y \mid y \in x\} = x$.
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Thus $x \subseteq \bigcup A$.
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\end{proof}
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\subsection{\unverified{Exercise 3.4}}%
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\subsection{\verified{Exercise 3.4}}%
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\label{sub:exercise-3.4}
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Show that if $A \subseteq B$, then $\bigcup A \subseteq \bigcup B$.
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\begin{proof}
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TODO
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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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Let $A$ and $B$ be sets such that $A \subseteq B$.
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Let $x \in \bigcup A$.
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By definition of the union, there exists some $b \in A$ such that $x \in b$.
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By definition of the subset, $b \in B$.
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This immediatley implies $x \in \bigcup B$.
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Since this holds for all $x \in \bigcup A$, it follows
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$\bigcup A \subseteq \bigcup B$.
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\end{proof}
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\subsection{\unverified{Exercise 3.5}}%
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\subsection{\verified{Exercise 3.5}}%
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\label{sub:exercise-3.5}
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Assume that every member of $\mathscr{A}$ is a subset of $B$.
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@ -489,22 +511,61 @@ Show that $\bigcup \mathscr{A} \subseteq B$.
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\begin{proof}
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TODO
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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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Let $x \in \bigcup \mathscr{A}$.
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By definition,
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$$\bigcup \mathscr{A} = \{ y \mid (\exists b \in A)y \in b \}.$$
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Then there exists some $b \in A$ such that $x \in b$.
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By hypothesis, $b \subseteq B$.
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Thus $x$ must also be a member of $B$.
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Since this holds for all $x \in \bigcup \mathscr{A}$, it follows
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$\bigcup \mathscr{A} \subseteq B$.
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\end{proof}
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\subsection{\unverified{Exercise 3.6a}}%
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\subsection{\verified{Exercise 3.6a}}%
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\label{sub:exercise-3.6a}
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Show that for any set $A$, $\bigcup \powerset{A} = A$.
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\begin{proof}
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TODO
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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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We prove that (i) $\bigcup \powerset{A} \subseteq A$ and (ii)
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$A \subseteq \bigcup \powerset{A}$.
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\paragraph{(i)}%
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\label{par:exercise-3.6a-i}
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By definition, the \nameref{ref:power-set} of $A$ is the set of all subsets
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of $A$.
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In other words, every member of $\powerset{A}$ is a subset of $A$.
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By \nameref{sub:exercise-3.5}, $\bigcup \powerset{A} \subseteq A$.
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\paragraph{(ii)}%
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\label{par:exercise-3.6a-ii}
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Let $x \in A$.
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By definition of the power set of $A$, $\{x\} \in \powerset{A}$.
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By definition of the union,
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$$\bigcup \powerset{A} =
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\{ y \mid (\exists b \in \powerset{A}), y \in b).$$
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Since $x \in \{x\}$ and $\{x\} \in \powerset{A}$, it follows
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$x \in \bigcup \powerset{A}$.
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Thus $A \subseteq \bigcup \powerset{A}$.
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\paragraph{Conclusion}%
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By \nameref{par:exercise-3.6a-i} and \nameref{par:exercise-3.6a-ii},
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$\bigcup \powerset{A} = A$.
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\end{proof}
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\subsection{\unverified{Exercise 3.6b}}%
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\subsection{\verified{Exercise 3.6b}}%
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\label{sub:exercise-3.6b}
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Show that $A \subseteq \powerset{\bigcup A}$.
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@ -512,35 +573,176 @@ Under what conditions does equality hold?
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\begin{proof}
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TODO
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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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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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By the definition of the \nameref{ref:power-set},
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$$\powerset{\bigcup A} = \{ y \mid y \subseteq \bigcup A \}.$$
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Therefore $x \in \powerset{\bigcup A}$.
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Since this holds for all $x \in A$, $A \subseteq \powerset{\bigcup A}$.
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\suitdivider
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We show equality holds if and only if there exists some set $B$ such that
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$A = \powerset{B}$.
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\paragraph{($\Rightarrow$)}%
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\label{par:exercise-3.6b-right}
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Suppose $A = \powerset{\bigcup A}$.
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Then our statement immediately follows by settings $B = \bigcup A$.
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\paragraph{($\Leftarrow$)}%
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\label{par:exercise-3.6b-left}
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Suppose there exists some set $B$ such that $A = \powerset{B}$.
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Therefore
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\begin{align*}
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\powerset{\bigcup A}
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& = \powerset{\left(\bigcup {\powerset {B}}\right)} \\
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& = \powerset{B} & \textref{sub:exercise-3.6a} \\
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& = A.
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\end{align*}
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\paragraph{Conclusion}%
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By \nameref{par:exercise-3.6b-right} and \nameref{par:exercise-3.6b-left},
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$A = \powerset{\bigcup A}$ if and only if there exists some set $B$ such
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that $A = \powerset{B}$.
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\end{proof}
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\subsection{\unverified{Exercise 3.7a}}%
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\subsection{\verified{Exercise 3.7a}}%
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\label{sub:exercise-3.7a}
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Show that for any sets $A$ and $B$,
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$$\powerset{A} \cap \powerset{B} = \powerset(A \cap B).$$
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$$\powerset{A} \cap \powerset{B} = \powerset{(A \cap B)}.$$
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\begin{proof}
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TODO
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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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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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show that $\powerset{A} \cap \powerset{B} \supseteq \powerset{(A \cap B)}$.
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\paragraph{($\subseteq$)}%
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Let $x \in \powerset{A} \cap \powerset{B}$.
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That is, $x \in \powerset{A}$ and $x \in \powerset{B}$.
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By the definition of the \nameref{ref:power-set},
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\begin{align*}
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\powerset{A} & = \{ y \mid y \subseteq A \} \\
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\powerset{B} & = \{ y \mid y \subseteq B \}
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\end{align*}
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Thus $x \subseteq A$ and $x \subseteq B$, meaning $x \subseteq A \cap B$.
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But then $x \in \powerset{(A \cap B)}$, the set of all subsets of
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$A \cap B$.
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Since this holds for all $x \in \powerset{A} \cap \powerset{B}$, it follows
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$$\powerset{A} \cap \powerset{B} \subseteq \powerset{(A \cap B)}.$$
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\paragraph{($\supseteq$)}%
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Let $x \in \powerset{(A \cap B)}$.
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By the definition of the \nameref{ref:power-set},
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$$\powerset{(A \cap B)} = \{ y \mid y \subseteq A \cap B \}.$$
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Thus $x \subseteq A \cap B$, meaning $x \subseteq A$ and $x \subseteq B$.
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But this implies $x \in \powerset{A}$, the set of all subsets of $A$.
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Likewise $x \in \powerset{B}$, the set of all subsets of $B$.
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Thus $x \in \powerset{A} \cap \powerset{B}$.
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Since this holds for all $x \in \powerset{(A \cap B)}$, it follows
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$$\powerset{(A \cap B)} \subseteq \powerset{A} \cap \powerset{B}.$$
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\paragraph{Conclusion}%
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Since each side of our identity is a subset of the other,
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$$\powerset{(A \cap B)} = \powerset{A} \cap \powerset{B}.$$
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\end{proof}
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\subsection{\unverified{Exercise 3.7b}}%
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\subsection{\verified{Exercise 3.7b}}%
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\label{sub:exercise-3.7b}
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Show that $\powerset{A} \cup \powerset{B} \subseteq \powerset(A \cup B)$.
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Show that $\powerset{A} \cup \powerset{B} \subseteq \powerset{(A \cup B)}$.
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Under what conditions does equality hold?
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\begin{proof}
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TODO
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\ % Add space.
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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\_1}
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{Enderton.Set.Chapter\_1.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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By the definition of the \nameref{ref:power-set},
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\begin{align*}
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\powerset{A} &= \{ y \mid y \subseteq A \} \\
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\powerset{B} &= \{ y \mid y \subseteq B \}.
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\end{align*}
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Thus $x \subseteq A$ or $x \subseteq B$.
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Therefore $x \subseteq A \cup B$.
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But then $x \in \powerset{(A \cup B)}$, the set of all subsets of $A \cup B$.
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\suitdivider
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We show equality holds if and only if one of $A$ or $B$ is a subset of the
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other.
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\paragraph{($\Rightarrow$)}%
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\label{par:exercise-3.7b-right}
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Suppose
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\begin{equation}
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\label{sub:exercise-3.7b-eq1}
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\powerset{A} \cup \powerset{B} = \powerset{(A \cup B)}.
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\end{equation}
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By the definition of the \nameref{ref:power-set},
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$A \cup B \in \powerset{(A \cup B)}$.
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Then \eqref{sub:exercise-3.7b-eq1} implies
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$A \cup B \in \powerset{A} \cup \powerset{B}$.
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That is, $A \cup B \in \powerset{A}$ or $A \cup B \in \powerset{B}$ (or
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both).
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For the sake of contradiction, suppose $A \not\subseteq B$ and
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$B \not\subseteq A$.
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Then there exists an element $x \in A$ such that $x \not\in B$ and there
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exists an element $y \in B$ such that $y \not\in A$.
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But then $A \cup B \not\in \powerset{A}$ since $y$ cannot be a member of a
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member of $\powerset{A}$.
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Likewise, $A \cup B \not\in \powerset{B}$ since $x$ cannot be a member of a
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member of $\powerset{B}$.
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Therefore our assumption is incorrect.
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In other words, $A \subseteq B$ or $B \subseteq A$.
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\paragraph{($\Leftarrow$)}%
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\label{par:exercise-3.7b-left}
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WLOG, suppose $A \subseteq B$.
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Then, by \nameref{sub:exercise-1.3}, $\powerset{A} \subseteq \powerset{B}$.
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Thus
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\begin{align*}
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\powerset{A} \cup \powerset{B}
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& = \powerset{B} \\
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& = \powerset{A \cup B}.
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\end{align*}
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\paragraph{Conclusion}%
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By \nameref{par:exercise-3.7b-right} and \nameref{par:exercise-3.7b-left},
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it follows
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$\powerset{A} \cup \powerset{B} \subseteq \powerset{(A \cup B)}$ if and
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only if $A \subseteq B$ or $B \subseteq A$.
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\end{proof}
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\subsection{\unverified{Exercise 3.8}}%
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\subsection{\partial{Exercise 3.8}}%
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\label{sub:exercise-3.8}
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Show that there is no set to which every singleton (that is, every set of the
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@ -550,23 +752,37 @@ Show that there is no set to which every singleton (that is, every set of the
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\begin{proof}
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TODO
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We proceed by contradiction.
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Suppose there existed a set $A$ consisting of every singleton.
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Then the \nameref{ref:union-axiom} suggests $\bigcup A$ is a set.
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But this set is precisely the class of all sets, which is \textit{not} a set.
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Thus our original assumption was incorrect.
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That is, there is no set to which every singleton belongs.
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\end{proof}
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\subsection{\unverified{Exercise 3.9}}%
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\subsection{\verified{Exercise 3.9}}%
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\label{sub:exercise-3.9}
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Give an example of sets $a$ and $B$ for which $a \in B$ but
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$\powerset{A} \not\in \powerset{B}$.
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$\powerset{a} \not\in \powerset{B}$.
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\begin{proof}
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\begin{answer}
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TODO
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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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\end{proof}
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Let $a = \{1\}$ and $B = \{\{1\}\}$.
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Then
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\begin{align*}
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\powerset{a} & = \{\emptyset, \{1\}\} \\
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\powerset{B} & = \{\emptyset, \{\{1\}\}\}.
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\end{align*}
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It immediately follows that $\powerset{a} \not\in \powerset{B}$.
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\subsection{\unverified{Exercise 3.10}}%
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\end{answer}
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\subsection{\verified{Exercise 3.10}}%
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\label{sub:exercise-3.10}
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Show that if $a \in B$, then $\powerset{a} \in \powerset{\powerset{\bigcup B}}$.
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@ -574,7 +790,16 @@ Show that if $a \in B$, then $\powerset{a} \in \powerset{\powerset{\bigcup B}}$.
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\begin{proof}
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TODO
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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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Suppose $a \in B$.
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By \nameref{sub:exercise-3.3}, $a \subseteq \bigcup B$.
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By \nameref{sub:exercise-1.3}, $\powerset{a} \subseteq \powerset{\bigcup B}$.
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By the definition of the \nameref{ref:power-set},
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$$\powerset{\powerset{\bigcup B}} =
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\{ y \mid y \subseteq \powerset{\bigcup B} \}.$$
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Therefore $\powerset{a} \in \powerset{\powerset{\bigcup B}}$.
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\end{proof}
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@ -1,7 +1,10 @@
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import Mathlib.Init.Set
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import Mathlib.Logic.Basic
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import Mathlib.Data.Set.Basic
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import Mathlib.Data.Set.Lattice
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import Mathlib.Tactic.LibrarySearch
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import Common.Set.Basic
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/-! # Enderton.Chapter_1
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Introduction
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@ -104,8 +107,6 @@ theorem exercise_1_2
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Show that if `B ⊆ C`, then `𝓟 B ⊆ 𝓟 C`.
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-/
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theorem exercise_1_3 (h : B ⊆ C) : Set.powerset B ⊆ Set.powerset C := by
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unfold Set.powerset
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simp
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intro x hx
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exact Set.Subset.trans hx h
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@ -130,4 +131,273 @@ theorem exercise_1_4 (x y : α) (hx : x ∈ B) (hy : y ∈ B)
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(Set.singleton_subset_iff.mpr hx)
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(Set.singleton_subset_iff.mpr hy)
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/-- ### Exercise 3.1
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||||
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
|
||||
|
|
@ -7,15 +7,17 @@ Additional theorems and definitions useful in the context of `Set`s.
|
|||
|
||||
namespace Set
|
||||
|
||||
/-! ## Minkowski Sum -/
|
||||
|
||||
/-
|
||||
The Minkowski sum of two sets `s` and `t` is the set
|
||||
The Minkowski sum of two `Set`s `s` and `t` is the set
|
||||
`s + t = { a + b : a ∈ s, b ∈ t }`.
|
||||
-/
|
||||
def minkowskiSum {α : Type u} [Add α] (s t : Set α) :=
|
||||
{ x | ∃ a ∈ s, ∃ b ∈ t, x = a + b }
|
||||
|
||||
/--
|
||||
The sum of two sets is nonempty **iff** the summands are nonempty.
|
||||
The sum of two `Set`s is nonempty **iff** the summands are nonempty.
|
||||
-/
|
||||
theorem nonempty_minkowski_sum_iff_nonempty_add_nonempty {α : Type u} [Add α]
|
||||
{s t : Set α}
|
||||
|
|
@ -30,12 +32,59 @@ theorem nonempty_minkowski_sum_iff_nonempty_add_nonempty {α : Type u} [Add α]
|
|||
· intro ⟨⟨a, ha⟩, ⟨b, hb⟩⟩
|
||||
exact ⟨a + b, ⟨a, ⟨ha, ⟨b, ⟨hb, rfl⟩⟩⟩⟩⟩
|
||||
|
||||
/-! ## Characteristic Function -/
|
||||
|
||||
/--
|
||||
The characteristic function of a set `S`.
|
||||
The characteristic function of a `Set` `S`.
|
||||
|
||||
It returns `1` if the specified input belongs to `S` and `0` otherwise.
|
||||
-/
|
||||
def characteristic (S : Set α) (x : α) [Decidable (x ∈ S)] : Nat :=
|
||||
if x ∈ S then 1 else 0
|
||||
|
||||
/-! ## Subsets -/
|
||||
|
||||
/--
|
||||
Every `Set` is a subset of itself.
|
||||
-/
|
||||
theorem subset_self (S : Set α) : S ⊆ S := by
|
||||
intro _ hs
|
||||
exact hs
|
||||
|
||||
/--
|
||||
If `Set` `A` is a subset of `Set` `B`, then `A ∪ B = B`.
|
||||
-/
|
||||
theorem left_subset_union_eq_self {A B : Set α} (h : A ⊆ B)
|
||||
: A ∪ B = B := by
|
||||
rw [Set.ext_iff]
|
||||
intro x
|
||||
apply Iff.intro
|
||||
· intro hU
|
||||
apply Or.elim hU
|
||||
· intro hA
|
||||
exact h hA
|
||||
· simp
|
||||
· intro hB
|
||||
exact Or.inr hB
|
||||
|
||||
/--
|
||||
If `Set` `B` is a subset of `Set` `A`, then `A ∪ B = B`.
|
||||
-/
|
||||
theorem right_subset_union_eq_self {A B : Set α} (h : B ⊆ A)
|
||||
: A ∪ B = A := by
|
||||
rw [Set.union_comm]
|
||||
exact left_subset_union_eq_self h
|
||||
|
||||
/--
|
||||
If `x` and `y` are members of `Set` `A`, it follows `{x, y}` is a subset of `A`.
|
||||
-/
|
||||
theorem mem_mem_imp_pair_subset {x y : α}
|
||||
(hx : x ∈ A) (hy : y ∈ A) : ({x, y} : Set α) ⊆ A := by
|
||||
intro a ha
|
||||
apply Or.elim ha
|
||||
· intro hx'
|
||||
rwa [hx']
|
||||
· intro hy'
|
||||
rwa [hy']
|
||||
|
||||
end Set
|
||||
|
|
@ -29,7 +29,7 @@
|
|||
\newcommand\@linespace{\vspace{10pt}}
|
||||
\newcommand\linedivider{\@linespace\hrule\@linespace}
|
||||
\WithSuffix\newcommand\linedivider*{\@linespace\hrule}
|
||||
\newcommand\suitdivider{$$\spadesuit\spadesuit\spadesuit$$}
|
||||
\newcommand\suitdivider{$$\spadesuit\;\spadesuit\;\spadesuit$$}
|
||||
|
||||
% ========================================
|
||||
% Linking
|
||||
|
|
@ -89,6 +89,7 @@
|
|||
\newcommand\@statement[1]{%
|
||||
\linedivider*\paragraph{\normalfont\normalsize\textit{#1.}}}
|
||||
|
||||
\newenvironment{answer}{\@statement{Answer}}{\hfill$\square$}
|
||||
\newenvironment{axiom}{\@statement{Axiom}}{\hfill$\square$}
|
||||
\newenvironment{definition}{\@statement{Definition}}{\hfill$\square$}
|
||||
\renewenvironment{proof}{\@statement{Proof}}{\hfill$\square$}
|
||||
|
|
@ -131,7 +132,7 @@
|
|||
\newcommand{\ico}[2]{\left[#1, #2\right)}
|
||||
\newcommand{\ioc}[2]{\left(#1, #2\right]}
|
||||
\newcommand{\ioo}[2]{\left(#1, #2\right)}
|
||||
\newcommand{\powerset}[1]{\mathscr{P}\,#1}
|
||||
\newcommand{\powerset}[1]{\mathscr{P}#1}
|
||||
\newcommand{\ubar}[1]{\text{\b{$#1$}}}
|
||||
|
||||
\let\oldemptyset\emptyset
|
||||
|
|
|
|||
Loading…
Reference in New Issue