Enderton. Chapter 2 exercises, 11-25.
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\documentclass{report}
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\usepackage{graphicx}
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\graphicspath{{./Set/images/}}
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\input{../../preamble}
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\makeleancommands{../..}
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@ -1720,7 +1723,7 @@ Show that $A + (B + C) = (A + B) + C$.
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\end{proof}
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\subsection{\unverified{Exercise 4.16}}%
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\subsection{\verified{Exercise 4.16}}%
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\label{sub:exercise-4.16}
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Simplify:
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@ -1728,11 +1731,26 @@ Simplify:
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\begin{proof}
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TODO
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\lean{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.exercise\_4\_16}
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Let $A$, $B$, and $C$ be arbitrary sets.
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Then
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\begin{align*}
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[(A \cup B \cup C) \cap (A \cup B)] & - [(A \cup (B - C)) \cap A] \\
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& = [A \cup B] - [A] \\
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& = \{ x \mid x \in (A \cup B) \land x \not\in A \} \\
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& = \{ x \mid x \in \{ y \mid y \in A \lor y \in B \} \land x \not\in A \} \\
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& = \{ x \mid (x \in A \lor x \in B) \land x \not\in A \} \\\
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& = \{ x \mid (x \in A \land x \not\in A) \lor (x \in B \land x \not\in A) \} \\
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& = \{ x \mid F \lor (X \in B \land x \not\in A) \} \\
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& = \{ x \mid x \in B \land x \not\in A \} \\
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& = B - A.
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\end{align*}
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\end{proof}
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\subsection{\unverified{Exercise 4.17}}%
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\subsection{\verified{Exercise 4.17}}%
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\label{sub:exercise-4.17}
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Show that the following four conditions are equivalent.
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@ -1746,11 +1764,58 @@ Show that the following four conditions are equivalent.
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\begin{proof}
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TODO
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\statementpadding
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\lean*{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.exercise\_4\_17\_i}
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\lean*{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.exercise\_4\_17\_ii}
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\lean*{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.exercise\_4\_17\_iii}
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\lean{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.exercise\_4\_17\_iv}
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Let $A$ and $B$ be arbitrary sets.
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We show that (i) $(a) \Rightarrow (b)$, (ii) $(b) \Rightarrow (c)$, (iii)
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$(c) \Rightarrow (d)$, and (iv) $(d) \Rightarrow (a)$.
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\paragraph{(i)}%
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Suppose $A \subseteq B$.
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That is, $\forall t, t \in A \Rightarrow t \in B$.
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Then there is no element such that $t \in A$ and $t \not\in B$.
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By definition of the relative complement, this immediately implies
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$A - B = \emptyset$.
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\paragraph{(ii)}%
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Suppose $A - B = \emptyset$.
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By definition of the relative complement,
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$$A - B = \emptyset = \{ x \mid x \in A \land x \not\in B \}.$$
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Then, for all $t$,
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$\neg(t \in A \land t \not\in B) = t \not\in A \lor t \in B$.
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This implies, by definition of the subset, that $A \subseteq B$.
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It then immediately follows that $A \cup B = B$.
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\paragraph{(iii)}%
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Suppose $A \cup B = B$.
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Then there is no member of $A$ that is not a member of $B$.
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In other words, $A \subseteq B$.
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This immediately implies $A \cap B = A$.
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\paragraph{(iv)}%
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Suppose $A \cap B = A$.
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Then every member of $A$ is a member of $B$.
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This immediately implies $A \subseteq B$.
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\end{proof}
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\subsection{\unverified{Exercise 4.18}}%
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\subsection{\partial{Exercise 4.18}}%
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\label{sub:exercise-4.18}
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Assume that $A$ and $B$ are subsets of $S$.
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@ -1759,11 +1824,30 @@ List all of the different sets that can be made from these three by use of the
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\begin{proof}
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TODO
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We can reason about this diagrammatically:
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\begin{figure}[ht]
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\includegraphics[width=0.6\textwidth]{venn-diagram}
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\centering
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\end{figure}
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In the above diagram, we assume the left circle corresponds to set $A$ and the
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right circle corresponds to $B$.
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The the possible sets we can make via the specified operators are:
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\begin{itemize}
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\item $A - B$, the left circle excluding the overlapping region.
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\item $A \cap B$, the overlapping region.
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\item $B - A$, the right circle excluding the overlapping region.
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\item $(A \cup B) \cap A$, the left circle.
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\item $(A \cup B) \cap B$, the right circle.
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\item $(A - B) \cup (B - A)$, the symmetric difference.
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\item $A \cup B$, the entire diagram.
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\end{itemize}
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\end{proof}
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\subsection{\unverified{Exercise 4.19}}%
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\subsection{\verified{Exercise 4.19}}%
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\label{sub:exercise-4.19}
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Is $\powerset{(A - B)}$ always equal to $\powerset{A} - \powerset{B}$?
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@ -1771,11 +1855,35 @@ Is it ever equal to $\powerset{A} - \powerset{B}$?
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\begin{proof}
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TODO
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\lean{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.exercise\_4\_19}
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Let $A$ and $B$ be arbitrary sets.
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We show (i) that $\emptyset \in \powerset{(A - B})$ and (ii)
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$\emptyset \not\in \powerset{A} - \powerset{B}$.
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\paragraph{(i)}%
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\label{par:exercise-4.19-i}
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By definition of the \nameref{ref:power-set},
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$$\powerset{(A - B)} = \{ x \mid x \subseteq A - B \}.$$
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But $\emptyset$ is a subset of \textit{every} set.
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Thus $\emptyset \in \powerset{(A - B)}$.
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\paragraph{(ii)}%
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By the same reasoning found in \nameref{par:exercise-4.19-i},
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$\emptyset \in \powerset{A}$ and $\emptyset \in \powerset{B}$.
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But then, by definition of the relative complement,
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$\emptyset \not\in \powerset{A} - \powerset{B}$.
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\paragraph{Conclusion}%
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By the \nameref{ref:extensionality-axiom}, the two sets are never equal.
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\end{proof}
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\subsection{\unverified{Exercise 4.20}}%
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\subsection{\verified{Exercise 4.20}}%
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\label{sub:exercise-4.20}
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Let $A$, $B$, and $C$ be sets such that $A \cup B = A \cup C$ and
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@ -1784,22 +1892,95 @@ Show that $B = C$.
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\begin{proof}
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TODO
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\lean{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.exercise\_4\_20}
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Let $A$, $B$, and $C$ be arbitrary sets.
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By the \nameref{ref:extensionality-axiom}, $B = C$ if and only if for all sets
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$x$, $x \in B \iff x \in C$.
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We prove both directions of this biconditional.
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\paragraph{($\Rightarrow$)}%
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Suppose $x \in B$.
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Then there are two cases to consider:
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\subparagraph{Case 1}%
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Assume $x \in A$.
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Then $x \in A \cap B$.
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By hypothesis, $A \cap B = A \cap C$.
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Thus $x \in A \cap C$ immediately implying $x \in C$.
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\subparagraph{Case 2}%
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Assume $x \not\in A$.
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Then $x \in A \cup B$.
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By hypothesis, $A \cup B = A \cup C$.
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Thus $x \in A \cup C$.
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Since $x \not\in A$, it follows $x \in C$.
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\paragraph{($\Leftarrow$)}%
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Suppose $x \in C$.
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Then there are two cases to consider:
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\subparagraph{Case 1}%
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Assume $x \in A$.
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Then $x \in A \cap C$.
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By hypothesis, $A \cap B = A \cap C$.
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Thus $x \in A \cap B$, immediately implying $x \in B$.
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\subparagraph{Case 2}%
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Assume $x \not\in A$.
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Then $x \in A \cup C$.
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By hypothesis, $A \cup B = A \cup C$.
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Thus $x \in A \cup B$.
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Since $x \not\in A$, it follows $x \in B$.
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\end{proof}
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\subsection{\unverified{Exercise 4.21}}%
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\subsection{\verified{Exercise 4.21}}%
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\label{sub:exercise-4.21}
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Show that $\bigcup (A \cup B) = \bigcup A \cup \bigcup B$.
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\begin{proof}
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TODO
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\lean{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.exercise\_4\_21}
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Let $A$ and $B$ be arbitrary sets.
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By the \nameref{ref:extensionality-axiom}, the specified equality holds if and
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only if for all sets $x$,
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$$x \in \bigcup (A \cup B) \iff x \in \bigcup A \cup \bigcup B.$$
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We prove both directions of this biconditional.
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\paragraph{($\Rightarrow$)}%
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Suppose $x \in \bigcup (A \cup B)$.
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By definition of the union of sets, there exists some $b \in A \cup B$ such
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that $x \in b$.
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If $b \in A$, then $x \in \bigcup A$ and $x \in \bigcup A \cup \bigcup B$.
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Alternatively, if $b \in B$, then $x \in \bigcup B$ and
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$x \in \bigcup A \cup \bigcup B$.
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Regardless, $x$ is in the target set.
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\paragraph{($\Leftarrow$)}%
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Suppose $x \in \bigcup A \cup \bigcup B$.
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Then $x \in \bigcup A$ or $x \in \bigcup B$.
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WLOG, suppose $x \in \bigcup A$.
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By definition of the union of sets, there exists some $b \in A$ such that
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$x \in b$.
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But then $b \in A \cup B$ meaning $x$ is also a member of
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$\bigcup (A \cup B)$.
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\end{proof}
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\subsection{\unverified{Exercise 4.22}}%
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\subsection{\verified{Exercise 4.22}}%
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\label{sub:exercise-4.22}
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Show that if $A$ and $B$ are nonempty sets, then
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@ -1807,11 +1988,39 @@ Show that if $A$ and $B$ are nonempty sets, then
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\begin{proof}
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TODO
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\lean{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.exercise\_4\_22}
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Let $A$ and $B$ be arbitrary, nonempty sets.
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By the \nameref{ref:extensionality-axiom}, the specified equality holds if and
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only if for all sets $x$,
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\begin{equation}
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\label{sub:exercise-4.22-eq1}
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x \in \bigcap (A \cup B) \iff x \in \bigcap A \cap \bigcap B.
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\end{equation}
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We prove both directions of this biconditional.
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\paragraph{($\Rightarrow$)}%
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Suppose $x \in \bigcap (A \cup B)$.
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Then for all $b \in A \cup B$, $x \in B$.
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In other words, for every member $b_1$ of $A$ and every member $b_2$ of $B$,
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$x$ is a member of both $b_1$ and $b_2$.
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But that implies $x \in \bigcap A$ and $x \in \bigcap B$.
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\paragraph{($\Leftarrow$)}%
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Suppose $x \in \bigcap A \cap \bigcap B$.
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That is, $x \in \bigcap A$ and $x \in \bigcap B$.
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By definition of the intersection of sets, forall sets $b$, if $b \in A$,
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then $x \in b$.
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Likewise, if $b \in B$, then $x \in b$.
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In other words, if $b$ is a member of either $A$ or $B$, $x \in b$.
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That immediately implies $x \in \bigcap (A \cup B$.
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\end{proof}
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\subsection{\unverified{Exercise 4.23}}%
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\subsection{\partial{Exercise 4.23}}%
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\label{sub:exercise-4.23}
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Show that if $\mathscr{B}$ is nonempty, then
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@ -1819,37 +2028,113 @@ Show that if $\mathscr{B}$ is nonempty, then
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\begin{proof}
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TODO
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Refer to \nameref{sub:general-distributive-laws}.
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\end{proof}
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\subsection{\unverified{Exercise 4.24a}}%
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\subsection{\verified{Exercise 4.24a}}%
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\label{sub:exercise-4.24a}
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Show that if $\mathscr{A}$ is nonempty, then
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$\powerset{\bigcap A} = \bigcap\; \{\powerset{X} \mid X \in \mathscr{A} \}$.
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$\powerset{\bigcap\mathscr{A}} =
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\bigcap\; \{\powerset{X} \mid X \in \mathscr{A} \}$.
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\begin{proof}
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TODO
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\lean{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.exercise\_4\_24a}
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Suppose $\mathscr{A}$ is a nonempty set.
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Then $\bigcap \mathscr{A}$ is well-defined.
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Therefore
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\begin{align*}
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\powerset{\bigcap\mathscr{A}}
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& = \{ x \mid x \subseteq \bigcap \mathscr{A} \}
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& \textref{ref:power-set} \\
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& = \{ x \mid x \subseteq
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\{ y \mid \forall X \in \mathscr{A}, y \in X \} \}
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& \text{def'n intersection} \\
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& = \{ x \mid \forall t \in x,
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t \in \{ y \mid \forall X \in \mathscr{A}, y \in X \} \}
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& \text{def'n subset} \\
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& = \{ x \mid \forall t \in x,
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(\forall X \in \mathscr{A}, t \in X) \} \\
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& = \{ x \mid \forall X \in \mathscr{A},
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(\forall t \in x, t \in X) \} \\
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& = \{ x \mid \forall X \in \mathscr{A}, x \subseteq X \} \\
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& = \{ x \mid \forall X \in \mathscr{A}, x \in \powerset{X} \}
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& \textref{ref:power-set-axiom} \\
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& = \{ x \mid
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\forall t \in \{ \powerset{X} \mid X \in \mathscr{A} \}, x \in t \} \\
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& = \bigcap\; \{\powerset{X} \mid X \in \mathscr{A}\}.
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\end{align*}
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\end{proof}
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\subsection{\unverified{Exercise 4.24b}}%
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\subsection{\verified{Exercise 4.24b}}%
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\label{sub:exercise-4.24b}
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Show that
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$$\bigcup\; \{ \powerset{X} \mid X \in \mathscr{A} \} \subseteq
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\powerset{\bigcup A}.$$
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\begin{equation}
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\label{sub:exercise-4.24b-eq1}
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\bigcup\; \{ \powerset{X} \mid X \in \mathscr{A} \} \subseteq
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\powerset{\bigcup\mathscr{A}}.
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\end{equation}
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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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\lean{Bookshelf/Enderton/Set/Chapter\_2}
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{Enderton.Set.Chapter\_2.exercise\_4\_24b}
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We first prove \eqref{sub:exercise-4.24b-eq1}.
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Let $x \in \bigcup\; \{ \powerset{X} \mid X \in \mathscr{A} \}$.
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By definition of the union of sets,
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$(\exists X \in \mathscr{A}), x \in \powerset{X}$.
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By definition of the \nameref{ref:power-set}, $x \subseteq X$.
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By \nameref{sub:exercise-3.3}, $X \subseteq \bigcup \mathscr{A}$.
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Therefore $x \subseteq \bigcup \mathscr{A}$, proving
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$x \in \powerset{\mathscr{A}}$ as expected.
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\suitdivider
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\noindent
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We show $\powerset{\bigcup A} \subseteq
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\bigcup\;\{ \powerset{X} \mid X \in \mathscr{A} \}$ if and only if
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$\bigcup\mathscr{A} \in \mathscr{A}$.
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\paragraph{($\Rightarrow$)}%
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Suppose $\powerset{\bigcup\mathscr{A}} \subseteq
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\bigcup\;\{ \powerset{X} \mid X \in \mathscr{A} \}$.
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By definition of the \nameref{ref:power-set},
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$\bigcup\mathscr{A} \in \powerset{\bigcup\mathscr{A}}$.
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By hypothesis, $\bigcup\mathscr{A} \in
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\bigcup\;\{ \powerset{X} \mid X \in \mathscr{A} \}$.
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By definition of the union of sets, there exists some $X \in \mathscr{A}$
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such that $\bigcup\mathscr{A} \in \powerset{X}$.
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That is, $\bigcup\mathscr{A} \subseteq X$.
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But $\bigcup\mathscr{A}$ cannot be a proper subset of $X$ since
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$X \in \mathscr{A}$.
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Thus $\bigcup\mathscr{A} = X$.
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This proves $\bigcup\mathscr{A} \in
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\bigcup\;\{ \powerset{X} \mid X \in \mathscr{A} \}$.
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\paragraph{($\Leftarrow$)}%
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Suppose $\bigcup\mathscr{A} \in A$.
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Let $x \in \powerset{\bigcup\mathscr{A}}$.
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Since $\bigcup\mathscr{A} \in \mathscr{A}$, it immediately follows that
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$x \in \{\powerset{X} \mid X \in \mathscr{A}\}$.
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\paragraph{Conclusion}%
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Equality follows immediately from this fact in conjunction with the proof
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of \eqref{sub:exercise-4.24b-eq1}.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\subsection{\unverified{Exercise 4.25}}%
|
||||
\subsection{\verified{Exercise 4.25}}%
|
||||
\label{sub:exercise-4.25}
|
||||
|
||||
Is $A \cup \bigcup \mathscr{B}$ always the same as
|
||||
|
|
@ -1858,7 +2143,71 @@ If not, then under what conditions does equality hold?
|
|||
|
||||
\begin{proof}
|
||||
|
||||
TODO
|
||||
\lean{Bookshelf/Enderton/Set/Chapter\_2}
|
||||
{Enderton.Set.Chapter\_2.exercise\_4\_25}
|
||||
|
||||
We prove that
|
||||
\begin{equation}
|
||||
\label{sub:exercise-4.25-eq1}
|
||||
A \cup \bigcup \mathscr{B} =
|
||||
\bigcup\;\{ A \cup X \mid X \in \mathscr{B} \}
|
||||
\end{equation}
|
||||
if and only if $A = \emptyset$ or $\mathscr{B} \neq \emptyset$.
|
||||
We prove both directions of this biconditional.
|
||||
|
||||
\paragraph{($\Rightarrow$)}%
|
||||
|
||||
Suppose \eqref{sub:exercise-4.25-eq1} holds true.
|
||||
There are two cases to consider:
|
||||
|
||||
\subparagraph{Case 1}%
|
||||
|
||||
Suppose $B \neq \emptyset$.
|
||||
Then $A = \emptyset \lor \mathscr{B} \neq \emptyset$ holds trivially.
|
||||
|
||||
\subparagraph{Case 2}%
|
||||
|
||||
Suppose $B = \emptyset$.
|
||||
Then $$A \cup \bigcup \mathscr{B} = A \cup \bigcup \emptyset = A$$ and
|
||||
$$
|
||||
\bigcup\;\{ A \cup X \mid X \in \mathscr{B} \}
|
||||
= \bigcup \emptyset \\
|
||||
= \emptyset.
|
||||
$$
|
||||
Then by hypothesis \eqref{sub:exercise-4.25-eq1}, $A = \emptyset$.
|
||||
Then $A = \emptyset \lor \mathscr{B} \neq \emptyset$ holds trivially.
|
||||
|
||||
\paragraph{($\Leftarrow$)}%
|
||||
|
||||
Suppose $A = \emptyset$ or $\mathscr{B} \neq \emptyset$.
|
||||
There are two cases to consider:
|
||||
|
||||
\paragraph{Case 1}%
|
||||
|
||||
Suppose $A = \emptyset$.
|
||||
Then $A \cup \bigcup \mathscr{B} = \bigcup{\mathscr{B}}$.
|
||||
Likewise,
|
||||
$$
|
||||
\bigcup \{ A \cup X \mid X \in \mathscr{B} \}
|
||||
= \bigcup \{ X \mid X \in \mathscr{B} \} \\
|
||||
= \bigcup \mathscr{B}.
|
||||
$$
|
||||
Therefore \eqref{sub:exercise-4.25-eq1} holds.
|
||||
|
||||
\paragraph{Case 2}%
|
||||
|
||||
Suppose $B \neq \emptyset$.
|
||||
Then
|
||||
\begin{align*}
|
||||
A \cup \bigcup\mathscr{B}
|
||||
& = \{ x \mid x \in A \lor x \in \bigcup\mathscr{B} \} \\
|
||||
& = \{ x \mid x \in A \lor (\exists b \in \mathscr{B}) x \in b \} \\
|
||||
& = \{ x \mid (\exists b \in \mathscr{B}) x \in A \lor x \in b \} \\
|
||||
& = \{ x \mid (\exists b \in \mathscr{B}) x \in A \cup b \} \\
|
||||
& = \{ x \mid x \in \bigcup \{ A \cup X \mid X \in \mathscr{B} \} \\
|
||||
& = \bigcup \{ A \cup X \mid X \in \mathscr{B} \}.
|
||||
\end{align*}
|
||||
Therefore \eqref{sub:exercise-4.25-eq1} holds.
|
||||
|
||||
\end{proof}
|
||||
|
||||
|
|
|
|||
|
|
@ -105,8 +105,6 @@ theorem exercise_3_4 (h : A ⊆ B) : ⋃₀ A ⊆ ⋃₀ B := by
|
|||
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
|
||||
|
|
@ -117,9 +115,8 @@ theorem exercise_3_5 (h : ∀ x ∈ 𝓐, x ⊆ B) : ⋃₀ 𝓐 ⊆ B := by
|
|||
|
||||
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
|
||||
theorem exercise_3_6a : ⋃₀ (𝒫 A) = A := by
|
||||
show { a | ∃ t, t ∈ { t | t ⊆ A } ∧ a ∈ t } = A
|
||||
ext x
|
||||
apply Iff.intro
|
||||
· intro hx
|
||||
|
|
@ -136,11 +133,10 @@ theorem exercise_3_6a : ⋃₀ (Set.powerset A) = A := by
|
|||
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
|
||||
: A ⊆ 𝒫 (⋃₀ A)
|
||||
∧ (A = 𝒫 (⋃₀ A) ↔ ∃ B, A = 𝒫 B) := by
|
||||
apply And.intro
|
||||
· unfold Set.powerset
|
||||
show ∀ x ∈ A, x ∈ { t | t ⊆ ⋃₀ A }
|
||||
· show ∀ x ∈ A, x ∈ { t | t ⊆ ⋃₀ A }
|
||||
intro x hx
|
||||
rw [Set.mem_setOf]
|
||||
exact exercise_3_3 x hx
|
||||
|
|
@ -156,10 +152,8 @@ theorem exercise_3_6b
|
|||
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
|
||||
: 𝒫 A ∩ 𝒫 B = 𝒫 (A ∩ B) := by
|
||||
suffices 𝒫 A ∩ 𝒫 B ⊆ 𝒫 (A ∩ B) ∧ 𝒫 (A ∩ B) ⊆ 𝒫 A ∩ 𝒫 B from
|
||||
subset_antisymm this.left this.right
|
||||
apply And.intro
|
||||
· unfold Set.powerset
|
||||
|
|
@ -178,7 +172,7 @@ theorem exercise_3_7A
|
|||
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
|
||||
: 𝒫 A ∪ 𝒫 B ⊆ 𝒫 (A ∪ B) := by
|
||||
unfold Set.powerset
|
||||
intro x hx
|
||||
simp at hx
|
||||
|
|
@ -195,7 +189,7 @@ theorem exercise_3_7b_i
|
|||
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
|
||||
: 𝒫 A ∪ 𝒫 B = 𝒫 (A ∪ B) ↔ A ⊆ B ∨ B ⊆ A := by
|
||||
unfold Set.powerset
|
||||
apply Iff.intro
|
||||
· intro h
|
||||
|
|
@ -247,15 +241,15 @@ theorem exercise_3_7b_ii
|
|||
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
|
||||
: a ∈ B ∧ 𝒫 a ∉ 𝒫 B := by
|
||||
apply And.intro
|
||||
· rw [ha, hB]
|
||||
simp
|
||||
· intro h
|
||||
have h₁ : Set.powerset a = {∅, {1}} := by
|
||||
have h₁ : 𝒫 a = {∅, {1}} := by
|
||||
rw [ha]
|
||||
exact Set.powerset_singleton 1
|
||||
have h₂ : Set.powerset B = {∅, {{1}}} := by
|
||||
have h₂ : 𝒫 B = {∅, {{1}}} := by
|
||||
rw [hB]
|
||||
exact Set.powerset_singleton {1}
|
||||
rw [h₁, h₂] at h
|
||||
|
|
@ -275,7 +269,7 @@ theorem exercise_3_9 (ha : a = {1}) (hB : B = {{1}})
|
|||
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
|
||||
: 𝒫 a ∈ 𝒫 (𝒫 (⋃₀ B)) := by
|
||||
have h₁ := exercise_3_3 a ha
|
||||
have h₂ := Chapter_1.exercise_1_3 h₁
|
||||
generalize hb : 𝒫 (⋃₀ B) = b
|
||||
|
|
@ -289,11 +283,7 @@ Show that for any sets `A` and `B`, `A = (A ∩ B) ∪ (A - B)`.
|
|||
-/
|
||||
theorem exercise_4_11_i {A B : Set α}
|
||||
: A = (A ∩ B) ∪ (A \ B) := by
|
||||
unfold Union.union Set.instUnionSet Set.union
|
||||
unfold SDiff.sdiff Set.instSDiffSet Set.diff
|
||||
unfold Inter.inter Set.instInterSet Set.inter
|
||||
unfold Membership.mem Set.instMembershipSet Set.Mem setOf
|
||||
simp only
|
||||
show A = fun a => A a ∧ B a ∨ A a ∧ ¬B a
|
||||
suffices ∀ x, (A x ∧ (B x ∨ ¬B x)) = A x by
|
||||
conv => rhs; ext x; rw [← and_or_left, this]
|
||||
intro x
|
||||
|
|
@ -310,10 +300,7 @@ Show that for any sets `A` and `B`, `A ∪ (B - A) = A ∪ B`.
|
|||
-/
|
||||
theorem exercise_4_11_ii {A B : Set α}
|
||||
: A ∪ (B \ A) = A ∪ B := by
|
||||
unfold Union.union Set.instUnionSet Set.union
|
||||
unfold SDiff.sdiff Set.instSDiffSet Set.diff
|
||||
unfold Membership.mem Set.instMembershipSet Set.Mem setOf
|
||||
simp only
|
||||
show (fun a => A a ∨ B a ∧ ¬A a) = fun a => A a ∨ B a
|
||||
suffices ∀ x, ((A x ∨ B x) ∧ (A x ∨ ¬A x)) = (A x ∨ B x) by
|
||||
conv => lhs; ext x; rw [or_and_left, this x]
|
||||
intro x
|
||||
|
|
@ -418,4 +405,249 @@ theorem exercise_4_14 : A \ (B \ C) ≠ (A \ B) \ C := by
|
|||
|
||||
end
|
||||
|
||||
/-- ### Exercise 4.16
|
||||
|
||||
Simplify:
|
||||
`[(A ∪ B ∪ C) ∩ (A ∪ B)] - [(A ∪ (B - C)) ∩ A]`
|
||||
-/
|
||||
theorem exercise_4_16 {A B C : Set α}
|
||||
: ((A ∪ B ∪ C) ∩ (A ∪ B)) \ ((A ∪ (B \ C)) ∩ A) = B \ A := by
|
||||
calc ((A ∪ B ∪ C) ∩ (A ∪ B)) \ ((A ∪ (B \ C)) ∩ A)
|
||||
_ = (A ∪ B) \ ((A ∪ (B \ C)) ∩ A) := by rw [Set.union_inter_cancel_left]
|
||||
_ = (A ∪ B) \ A := by rw [Set.union_inter_cancel_left]
|
||||
_ = B \ A := by rw [Set.union_diff_left]
|
||||
|
||||
/-! ### Exercise 4.17
|
||||
|
||||
Show that the following four conditions are equivalent.
|
||||
|
||||
(a) `A ⊆ B`
|
||||
(b) `A - B = ∅`
|
||||
(c) `A ∪ B = B`
|
||||
(d) `A ∩ B = A`
|
||||
-/
|
||||
|
||||
theorem exercise_4_17_i {A B : Set α} (h : A ⊆ B)
|
||||
: A \ B = ∅ := by
|
||||
ext x
|
||||
apply Iff.intro
|
||||
· intro hx
|
||||
exact absurd (h hx.left) hx.right
|
||||
· intro hx
|
||||
exact False.elim hx
|
||||
|
||||
theorem exercise_4_17_ii {A B : Set α} (h : A \ B = ∅)
|
||||
: A ∪ B = B := by
|
||||
suffices A ⊆ B from Set.left_subset_union_eq_self this
|
||||
show ∀ t, t ∈ A → t ∈ B
|
||||
intro t ht
|
||||
rw [Set.ext_iff] at h
|
||||
by_contra nt
|
||||
exact (h t).mp ⟨ht, nt⟩
|
||||
|
||||
theorem exercise_4_17_iii {A B : Set α} (h : A ∪ B = B)
|
||||
: A ∩ B = A := by
|
||||
suffices A ⊆ B from Set.inter_eq_left_iff_subset.mpr this
|
||||
exact Set.union_eq_right_iff_subset.mp h
|
||||
|
||||
theorem exercise_4_17_iv {A B : Set α} (h : A ∩ B = A)
|
||||
: A ⊆ B := Set.inter_eq_left_iff_subset.mp h
|
||||
|
||||
/-- ### Exercise 4.19
|
||||
|
||||
Is `𝒫 (A - B)` always equal to `𝒫 A - 𝒫 B`? Is it ever equal to `𝒫 A - 𝒫 B`?
|
||||
-/
|
||||
theorem exercise_4_19 {A B : Set α}
|
||||
: 𝒫 (A \ B) ≠ (𝒫 A) \ (𝒫 B) := by
|
||||
intro h
|
||||
have he : ∅ ∈ 𝒫 (A \ B) := by simp
|
||||
have ne : ∅ ∉ (𝒫 A) \ (𝒫 B) := by simp
|
||||
rw [Set.ext_iff] at h
|
||||
have := h ∅
|
||||
exact absurd (this.mp he) ne
|
||||
|
||||
/-- ### Exercise 4.20
|
||||
|
||||
Let `A`, `B`, and `C` be sets such that `A ∪ B = A ∪ C` and `A ∩ B = A ∩ C`.
|
||||
Show that `B = C`.
|
||||
-/
|
||||
theorem exercise_4_20 {A B C : Set α}
|
||||
(hu : A ∪ B = A ∪ C) (hi : A ∩ B = A ∩ C) : B = C := by
|
||||
ext x
|
||||
apply Iff.intro
|
||||
· intro hB
|
||||
by_cases hA : x ∈ A
|
||||
· have : x ∈ A ∩ B := Set.mem_inter hA hB
|
||||
rw [hi] at this
|
||||
exact this.right
|
||||
· have : x ∈ A ∪ B := Set.mem_union_right A hB
|
||||
rw [hu] at this
|
||||
exact Or.elim this (absurd · hA) (by simp)
|
||||
· intro hC
|
||||
by_cases hA : x ∈ A
|
||||
· have : x ∈ A ∩ C := Set.mem_inter hA hC
|
||||
rw [← hi] at this
|
||||
exact this.right
|
||||
· have : x ∈ A ∪ C := Set.mem_union_right A hC
|
||||
rw [← hu] at this
|
||||
exact Or.elim this (absurd · hA) (by simp)
|
||||
|
||||
/-- ### Exercise 4.21
|
||||
|
||||
Show that `⋃ (A ∪ B) = (⋃ A) ∪ (⋃ B)`.
|
||||
-/
|
||||
theorem exercise_4_21 {A B : Set (Set α)}
|
||||
: ⋃₀ (A ∪ B) = (⋃₀ A) ∪ (⋃₀ B) := by
|
||||
ext x
|
||||
apply Iff.intro
|
||||
· intro hx
|
||||
have ⟨t, ht⟩ : ∃ t, t ∈ A ∪ B ∧ x ∈ t := hx
|
||||
apply Or.elim ht.left
|
||||
· intro hA
|
||||
exact Or.inl ⟨t, ⟨hA, ht.right⟩⟩
|
||||
· intro hB
|
||||
exact Or.inr ⟨t, ⟨hB, ht.right⟩⟩
|
||||
· intro hx
|
||||
apply Or.elim hx
|
||||
· intro hA
|
||||
have ⟨t, ht⟩ : ∃ t, t ∈ A ∧ x ∈ t := hA
|
||||
exact ⟨t, ⟨Set.mem_union_left B ht.left, ht.right⟩⟩
|
||||
· intro hB
|
||||
have ⟨t, ht⟩ : ∃ t, t ∈ B ∧ x ∈ t := hB
|
||||
exact ⟨t, ⟨Set.mem_union_right A ht.left, ht.right⟩⟩
|
||||
|
||||
/-- ### Exercise 4.22
|
||||
|
||||
Show that if `A` and `B` are nonempty sets, then `⋂ (A ∪ B) = ⋂ A ∩ ⋂ B`.
|
||||
-/
|
||||
theorem exercise_4_22 {A B : Set (Set α)}
|
||||
: ⋂₀ (A ∪ B) = ⋂₀ A ∩ ⋂₀ B := by
|
||||
ext x
|
||||
apply Iff.intro
|
||||
· intro hx
|
||||
have : ∀ t : Set α, t ∈ A ∪ B → x ∈ t := hx
|
||||
show (∀ t : Set α, t ∈ A → x ∈ t) ∧ (∀ t : Set α, t ∈ B → x ∈ t)
|
||||
rw [← forall_and]
|
||||
intro t
|
||||
exact ⟨
|
||||
fun ht => this t (Set.mem_union_left B ht),
|
||||
fun ht => this t (Set.mem_union_right A ht)
|
||||
⟩
|
||||
· intro hx
|
||||
have : ∀ t : Set α, (t ∈ A → x ∈ t) ∧ (t ∈ B → x ∈ t) := by
|
||||
have : (∀ t : Set α, t ∈ A → x ∈ t) ∧ (∀ t : Set α, t ∈ B → x ∈ t) := hx
|
||||
rwa [← forall_and] at this
|
||||
show ∀ (t : Set α), t ∈ A ∪ B → x ∈ t
|
||||
intro t ht
|
||||
apply Or.elim ht
|
||||
· intro hA
|
||||
exact (this t).left hA
|
||||
· intro hB
|
||||
exact (this t).right hB
|
||||
|
||||
/-- ### Exercise 4.24a
|
||||
|
||||
Show that is `𝓐` is nonempty, then `𝒫 (⋂ 𝓐) = ⋂ { 𝒫 X | X ∈ 𝓐 }`.
|
||||
-/
|
||||
theorem exercise_4_24a {𝓐 : Set (Set α)}
|
||||
: 𝒫 (⋂₀ 𝓐) = ⋂₀ { 𝒫 X | X ∈ 𝓐 } := by
|
||||
calc 𝒫 (⋂₀ 𝓐)
|
||||
_ = { x | x ⊆ ⋂₀ 𝓐 } := rfl
|
||||
_ = { x | x ⊆ { y | ∀ X, X ∈ 𝓐 → y ∈ X } } := rfl
|
||||
_ = { x | ∀ t ∈ x, t ∈ { y | ∀ X, X ∈ 𝓐 → y ∈ X } } := rfl
|
||||
_ = { x | ∀ t ∈ x, (∀ X, X ∈ 𝓐 → t ∈ X) } := rfl
|
||||
_ = { x | ∀ X ∈ 𝓐, (∀ t, t ∈ x → t ∈ X) } := by
|
||||
ext
|
||||
rw [Set.mem_setOf, Set.mem_setOf, forall_mem_comm (· ∈ ·)]
|
||||
_ = { x | ∀ X ∈ 𝓐, x ⊆ X} := rfl
|
||||
_ = { x | ∀ X ∈ 𝓐, x ∈ 𝒫 X } := rfl
|
||||
_ = { x | ∀ t ∈ { 𝒫 X | X ∈ 𝓐 }, x ∈ t} := by simp
|
||||
_ = ⋂₀ { 𝒫 X | X ∈ 𝓐 } := rfl
|
||||
|
||||
/-- ### Exercise 4.24b
|
||||
|
||||
Show that
|
||||
```
|
||||
⋃ {𝒫 X | X ∈ 𝓐} ⊆ 𝒫 ⋃ 𝓐.
|
||||
```
|
||||
Under what conditions does equality hold?
|
||||
-/
|
||||
theorem exercise_4_24b {𝓐 : Set (Set α)}
|
||||
: (⋃₀ { 𝒫 X | X ∈ 𝓐 } ⊆ 𝒫 ⋃₀ 𝓐)
|
||||
∧ ((⋃₀ { 𝒫 X | X ∈ 𝓐 } = 𝒫 ⋃₀ 𝓐) ↔ (⋃₀ 𝓐 ∈ 𝓐)) := by
|
||||
have hS : (⋃₀ { 𝒫 X | X ∈ 𝓐 } ⊆ 𝒫 ⋃₀ 𝓐) := by
|
||||
simp
|
||||
exact exercise_3_3
|
||||
refine ⟨hS, ?_⟩
|
||||
apply Iff.intro
|
||||
· intro rS
|
||||
have rS : 𝒫 ⋃₀ 𝓐 ⊆ ⋃₀ { 𝒫 X | X ∈ 𝓐 } :=
|
||||
(Set.Subset.antisymm_iff.mp rS).right
|
||||
have hA : ⋃₀ 𝓐 ∈ ⋃₀ { 𝒫 X | X ∈ 𝓐 } :=
|
||||
rS Set.self_mem_powerset_self
|
||||
conv at hA =>
|
||||
rhs
|
||||
unfold Set.sUnion sSup Set.instSupSetSet
|
||||
simp only
|
||||
have ⟨X, ⟨⟨x, hx⟩, ht⟩⟩ := Set.mem_setOf.mp hA
|
||||
have : ⋃₀ 𝓐 = x := by
|
||||
rw [← hx.right] at ht
|
||||
have hl : ⋃₀ 𝓐 ⊆ x := ht
|
||||
have hr : x ⊆ ⋃₀ 𝓐 := exercise_3_3 x hx.left
|
||||
exact Set.Subset.antisymm hl hr
|
||||
rw [← this] at hx
|
||||
exact hx.left
|
||||
· intro hA
|
||||
suffices 𝒫 ⋃₀ 𝓐 ⊆ ⋃₀ { 𝒫 X | X ∈ 𝓐 } from
|
||||
subset_antisymm hS this
|
||||
show ∀ x, x ∈ 𝒫 ⋃₀ 𝓐 → x ∈ ⋃₀ { x | ∃ X, X ∈ 𝓐 ∧ 𝒫 X = x }
|
||||
intro x hx
|
||||
unfold Set.sUnion sSup Set.instSupSetSet
|
||||
simp only [Set.mem_setOf_eq, exists_exists_and_eq_and, Set.mem_powerset_iff]
|
||||
exact ⟨⋃₀ 𝓐, ⟨hA, hx⟩⟩
|
||||
|
||||
/-- ### Exercise 4.25
|
||||
|
||||
Is `A ∪ (⋃ 𝓑)` always the same as `⋃ { A ∪ X | X ∈ 𝓑 }`? If not, then under
|
||||
what conditions does equality hold?
|
||||
-/
|
||||
theorem exercise_4_25 {A : Set α} (𝓑 : Set (Set α))
|
||||
: (A ∪ (⋃₀ 𝓑) = ⋃₀ { A ∪ X | X ∈ 𝓑 }) ↔ (A = ∅ ∨ Set.Nonempty 𝓑) := by
|
||||
apply Iff.intro
|
||||
· intro h
|
||||
by_cases h𝓑 : Set.Nonempty 𝓑
|
||||
· exact Or.inr h𝓑
|
||||
· have : 𝓑 = ∅ := Set.not_nonempty_iff_eq_empty.mp h𝓑
|
||||
rw [this] at h
|
||||
simp at h
|
||||
exact Or.inl h
|
||||
· intro h
|
||||
apply Or.elim h
|
||||
· intro hA
|
||||
rw [hA]
|
||||
simp
|
||||
· intro h𝓑
|
||||
calc A ∪ (⋃₀ 𝓑)
|
||||
_ = { x | x ∈ A ∨ x ∈ ⋃₀ 𝓑} := rfl
|
||||
_ = { x | x ∈ A ∨ (∃ b ∈ 𝓑, x ∈ b) } := rfl
|
||||
_ = { x | ∃ b ∈ 𝓑, x ∈ A ∨ x ∈ b } := by
|
||||
ext x
|
||||
rw [Set.mem_setOf, Set.mem_setOf]
|
||||
apply Iff.intro
|
||||
· intro hx
|
||||
apply Or.elim hx
|
||||
· intro hA
|
||||
have ⟨b, hb⟩ := Set.nonempty_def.mp h𝓑
|
||||
exact ⟨b, ⟨hb, Or.inl hA⟩⟩
|
||||
· intro ⟨b, hb⟩
|
||||
exact ⟨b, ⟨hb.left, Or.inr hb.right⟩⟩
|
||||
· intro ⟨b, ⟨hb, hx⟩⟩
|
||||
apply Or.elim hx
|
||||
· exact (Or.inl ·)
|
||||
· intro h
|
||||
exact Or.inr ⟨b, ⟨hb, h⟩⟩
|
||||
_ = { x | ∃ b ∈ 𝓑, x ∈ A ∪ b } := rfl
|
||||
_ = { x | ∃ t, t ∈ { y | ∃ X, X ∈ 𝓑 ∧ A ∪ X = y } ∧ x ∈ t } := by simp
|
||||
_ = ⋃₀ { A ∪ X | X ∈ 𝓑 } := rfl
|
||||
|
||||
end Enderton.Set.Chapter_2
|
||||
Binary file not shown.
|
After Width: | Height: | Size: 51 KiB |
|
|
@ -9,10 +9,23 @@ Additional theorems and definitions related to basic logic.
|
|||
/--
|
||||
The de Morgan law that distributes negation across a conjunction.
|
||||
-/
|
||||
theorem not_and_de_morgan {a b : Prop} : (¬(a ∧ b)) ↔ (¬ a ∨ ¬ b) := by
|
||||
theorem not_and_de_morgan : (¬(p ∧ q)) ↔ (¬p ∨ ¬q) := by
|
||||
tauto
|
||||
|
||||
/--
|
||||
Renaming of `not_or` to indicate its relationship to de Morgan's laws.
|
||||
-/
|
||||
theorem not_or_de_morgan : ¬(p ∨ q) ↔ ¬p ∧ ¬q := not_or
|
||||
|
||||
/--
|
||||
Universal quantification across nested set memberships can be commuted in either
|
||||
order.
|
||||
-/
|
||||
theorem forall_mem_comm {X : Set α} {Y : Set β} (p : α → β → Prop)
|
||||
: (∀ u ∈ X, (∀ v, v ∈ Y → p u v)) = (∀ v ∈ Y, (∀ u, u ∈ X → p u v)) := by
|
||||
refine propext ?_
|
||||
apply Iff.intro
|
||||
· intro h v hv u hu
|
||||
exact h u hu v hv
|
||||
· intro h u hu v hv
|
||||
exact h v hv u hu
|
||||
|
|
@ -89,6 +89,14 @@ theorem mem_mem_imp_pair_subset {x y : α}
|
|||
· intro hy'
|
||||
rwa [hy']
|
||||
|
||||
/-! ## Powerset -/
|
||||
|
||||
/--
|
||||
Every `Set` is a member of its own powerset.
|
||||
-/
|
||||
theorem self_mem_powerset_self {A : Set α}
|
||||
: A ∈ 𝒫 A := subset_self A
|
||||
|
||||
/-! ## Symmetric Difference -/
|
||||
|
||||
/--
|
||||
|
|
@ -116,8 +124,7 @@ This is the contraposition of `mem_symm_diff_iff_exclusive_mem`.
|
|||
-/
|
||||
theorem not_mem_symm_diff_inter_or_not_union {A B : Set α}
|
||||
: x ∉ (A ∆ B) ↔ (x ∈ A ∩ B) ∨ (x ∉ A ∪ B) := by
|
||||
unfold symmDiff
|
||||
simp
|
||||
show ¬(x ∈ A ∧ ¬x ∈ B ∨ x ∈ B ∧ ¬x ∈ A) ↔ x ∈ A ∧ x ∈ B ∨ ¬(x ∈ A ∨ x ∈ B)
|
||||
rw [
|
||||
not_or_de_morgan,
|
||||
not_and_de_morgan, not_and_de_morgan,
|
||||
|
|
@ -125,16 +132,12 @@ theorem not_mem_symm_diff_inter_or_not_union {A B : Set α}
|
|||
not_or_de_morgan
|
||||
]
|
||||
apply Iff.intro
|
||||
· intro hx
|
||||
apply Or.elim hx.left
|
||||
· intro nx
|
||||
apply Or.elim nx.left
|
||||
· intro nA
|
||||
exact Or.elim hx.right
|
||||
(fun nB => Or.inr ⟨nA, nB⟩)
|
||||
(fun hA => absurd hA nA)
|
||||
exact Or.elim nx.right (Or.inr ⟨nA, ·⟩) (absurd · nA)
|
||||
· intro hB
|
||||
apply Or.elim hx.right
|
||||
(fun nB => absurd hB nB)
|
||||
(fun hA => Or.inl ⟨hA, hB⟩)
|
||||
exact Or.elim nx.right (absurd hB ·) (Or.inl ⟨·, hB⟩)
|
||||
· intro hx
|
||||
apply Or.elim hx
|
||||
· intro hy
|
||||
|
|
|
|||
Loading…
Reference in New Issue