From 0594f94d8dda1cc653822ce9e58b8fdf4e4a43fd Mon Sep 17 00:00:00 2001 From: tomit4 Date: Wed, 22 Jul 2026 15:40:15 -0700 Subject: [PATCH] :construction: Fin 6.3 --- chapter_6/exercises.md | 1175 +++++++++++++++++++++++++++++++++++- chapter_6/test_yourself.md | 8 + 2 files changed, 1182 insertions(+), 1 deletion(-) diff --git a/chapter_6/exercises.md b/chapter_6/exercises.md index 968443f..70ea5be 100644 --- a/chapter_6/exercises.md +++ b/chapter_6/exercises.md @@ -2719,15 +2719,119 @@ Assume all sets are subsets of a universal set $U$. $$ (A \cup B) \cap C = A \cup (B \cap C) $$ +**Disproof (by counterexample):** + +Let $A$, $B$, and $C$ be any sets where $A$ has an element not in $C$. For +example: + +$$ A = \{1, 2\}, B = \{2\}, C = \{2\} $$ + +Then, the left-hand side of the given equation becomes: + +$$ A \cup B = \{1, 2\} $$ + +$$ (A \cup B) \cap C = \{\2\} $$ + +Then, the right-hand side of the given equation becomes: + +$$ B \cap C = \{2\} $$ + +$$ A \cup (B \cap C) = \{1, 2\} $$ + +Thus we can see that: + +$$ (A \cup B) \cap C = \{2\} \neq \{1, 2\} = A \cup (B \cap C) $$ + +as $1 \in $A \cup (B \cap C)$, but $1 \notin (A \cup B) \cap C$. + +Hence it has been shown that $(A \cup B) \cap C \neq A \cup (B \cap C)$ by +definition of subset. + +Q.E.D. + 2. For all sets $A$ and $B$, $(A \cup B)^c = A^c \cup B^c$. +**Disproof (by counterexample):** + +Let $U = \{1, 2\}, A = \{1\}, B = \{2\}$. + +Then, left-hand side: + +$$ A \cup B = \{1, 2\} $$ + +$$ (A \cup B)^c = U - \{1, 2\} $$ + +$$ = \emptyset $$ + +Then, right-hand side: + +$$ A^c = U - \{1\} $$ + +$$ B^c = U - \{2\} $$ + +$$ A^c \cup B^c = (U - \{1\}) \cup (U - \{2\}) $$ + +$$ = \{2\} \cup \{1\} $$ + +$$ = \{1, 2\} $$ + +Thus the given equality does not hold. + +$$ (A \cup B)^c = \emptyset \neq \{1, 2\} = A^c \cup B^c $$ + +Therefore it has been shown that $(A \cup B)^c \neq A^c \cup B^c$ by the +definition of subset. + +Q.E.D. + 3. For all sets $A$, $B$, and $C$, if $A \nsubseteq B$ and $B \nsubseteq C$ then $A \nsubseteq C$. +**Disproof (by counterexample):** + +Suppose $A$, $B$, and $C$ are any sets such that $A \nsubseteq B$ and +$B \nsubseteq C$, but $A$ has an element that is in $C$. For example: + +Let $A = \{1\}, B = \{2\}, C = \{1, 3\}$. + +It is trivially true that $A \nsubsteq B$ and $B \nsubseteq C$ (since $B$ does +not have any elements that are in $A$), but notice that $A \subseteq C$ since +$\{1\} \in \{1, 3\}$. + +Therefore the supposition is false by the definition of subset. + +Q.E.D. + 4. For all sets $A$, $B$, and $C$, if $B \cup C \subseteq A$ then $$ (A - B) \cap (A - C) = \emptyset $$ +**Disproof (by counterexample):** + +Suppose $A$, $B$, and $C$ are any sets such that $B \cup C \subseteq A$. For +example: + +Let $A, = \{1, 2, 3\}, B = \{2\}, C =\{3\}$. + +It is trivially true that $B \cup C \subseteq A$, since +$\{2\} \cup \{3\} = \{2, 3\} \subseteq \{1, 2, 3\}$. + +Then evaluating the left-hand side of the given equality: + +$$ (A - B) \cap (A - C) = (\{1, 2, 3\} - \{2\} ) \cap (\{1, 2, 3\} - \{3\}) $$ + +$$ = \{1, 3\} \cap \{1, 2\} $$ + +$$ = \{1\} $$ + +And note by the definition of $\emptyset$ that: + +$$ \{1\} \neq \emptyset $$ + +Therefore the supposition is false since $(A - B) \cap (A - C) \neq \emptyset$. + +Q.E.D. + For each of 5-21 prove each statement that is true and find a counterexample for each statement that is false. Assume all sets are subsets of a universal set $U$. @@ -2736,86 +2840,765 @@ $U$. $$ A - (B - C) = (A - B) - C $$ +**Disproof (by counterexample):** + +Let $A = \{1, 2, 3\}, B = \{2, 3\}, C = \{3\}$. + +Then, evaluating the left-hand side of the given equality: + +$$ A - (B - C) = \{1, 2, 3\} - (\{2, 3\} - \{3\}) $$ + +$$ = \{1, 2, 3\} - \{2\} $$ + +$$ = \{1, 3\} $$ + +Then, evaluating the right-hand side of the given equality: + +$$ (A - B) - C = (\{1, 2, 3\} - \{2, 3\}) - \{3\} $$ + += \{1\} - \{3\} $$ + += \{1\} $$ + +Notice that: + +$$ A - (B - C) = \{1, 3\} \neq \{1\} = (A - B) - C $$ + +Therefore the supposition is false since $A - (B - C) \neq (A - B) - C$. + +Q.E.D. + 6. For all sets $A$ and $B$, $A \cap (A \cup B) = A$. +**Proof:** + +Let $A$ and $B$ be any sets. + +To prove $A \cap (A \cup B) = A$, it must be shown that +$A \cap (A \cup B) \subseteq A$ and also $A \subseteq A \cap (A \cup B)$. + +_Proof $A \cap (A \cup B) \subseteq A$:_ + +Suppose $x$ is any element such that $x \in A \cap (A \cup B)$. + +By the definition of $\cap$, this means that $x \in A$ and $x \in A \cup B$. +Furthermore, by the definition of $\cup$, this means that $x \in A$ and +$x \in A$ or $x \in B$. In particular, it is known in either case that +$x \in A$. + +Therefore, by definition of a subset, $A \cap (A \cup B) \subseteq A$. + +This is what was to be shown. + +_Proof $A \subseteq A \cap (A \cup B)$:_ + +Suppose $x$ is any element such that $x \in A$. By definition of $\cup$, this +means that $x \in A \cup B$. Furthermore, by definition of $\cap$, it follows +that $x \in A \cap (A \cup B)$. + +Therefore, by definition of subset, $A \subseteq A \cap (A \cup B)$. + +This is what was to be shown. + +Therefore, since both subset relations have been proved, it can be concluded +that $A \cap (A \cup B) = A$. + +Q.E.D. + 7. For all sets $A$, $B$, and $C$, $$ (A - B) \cap (C - B) = A - (B \cup C) $$ +**Disproof (by counterexample):** + +Suppose $A$, $B$, and $C$ are any sets such that they are disjoint. For example: + +Let $A = \{1\}, B = \{2\}, C = \{3\}$. + +Then evaluating the left-hand side of the given equality: + +$$ (A - B) \cap (C - B) = (\{1\} - \{2\}) \cap (\{3\} - \{2\}) $$ + +$$ = \{1\} \cap \{3\} $$ + +$$ = \emptyset $$ + +Then evaluating the right-hand side of the given equality: + +$$ A - (B \cup C) = \{1\} - (\{2\} \cup \{3\}) $$ + +$$ = \{1\} - {2, 3\} $$ + +$$ = \{1\} $$ + +Notice that: + +$$ (A - B) \cap (C - B) = \emptyset \neq \{1\} = A - (B \cup C) $$ + +Therefore, the supposition is false since +$(A - B) \cap (C - B) \neq A - (B \cup C)$. + +Q.E.D. + 8. For all sets $A$ and $B$, if $A^c \subseteq B$ then $A \cup B = U$. +**Proof:** + +Let $A$ and $B$ be any sets such that $A^c \subseteq B$. + +In order to prove that $A \cup B = U$, it must be shown that +$A \cup B \subseteq U$ and $U \subseteq A \cup B$. + +_Proof $A \cup B \substeq U$:_ + +Suppose $x$ is any element such that $x \in A \cup B$. + +By definition of $\cup$, this means that $x \in A$ or $x \in B$. Regardless of +whether $x \in A$ or $x \in B$, $x \in U$, by definition of the universal set. + +Therefore it has been shown that $A \cup B \subseteq U$. + +_Proof $U \subseteq A \cup B$:_ + +Suppose $x$ is any element. Since $x$ is any element, by definition of the +universal set $U$, $x \in U$. + +It follows then that either $x \in A$ or $x \in A^c$. + +_Case $x \in A$:_ + +Since $x \in A$, by the definition of $\cup$, $x \in A \cup B$. + +_Case $x \in A^c$:_ + +Since $x \in A^c$, by the supposition, this means that $x \in B$ since +$A^c \subseteq B$. + +Since $x \in B$, by definition of $\cup$, $x \in A \cup B$. + +In either case $x \in A \cup B$. This is what was to be shown. + +_Conclusion:_ + +Since both subset relations have been proved, it can be concluded that +$A \cup B = U$. + +Q.E.D. + 9. For all sets $A$ ,$B$, and $C$, if $A \subseteq C$ and $B \subseteq C$ then $A \cup B \subseteq C$. +**Proof:** + +Suppose $A$, $B$, and $C$ are any sets such that $A \subseteq C$ and +$B \subseteq C$. + +Let $x$ be any element such that $x \in A \cup B$. + +By definition of $\cup$, this means that $x \in A$ or $x \in B$. + +_Case $x \in A$:_ + +Since $x \in A$, $x \in C$ since $A \subseteq C$ (by the supposition and the +definition of subset). + +_Case $x \in B$:_ + +Since $x \in B$, $x \in C$ since $B \subseteq C$ (by the supposition and the +definition of subset). + +_Conclusion:_ + +In either case, $x \in C$, so by definition of subset, it can be concluded that +$A \cup B \subseteq C$. + +Q.E.D. + 10. For all sets $A$ and $B$, if $A \subseteq B$ then $A \cap B^c = \emptyset$. +**Proof (by contradiction):** + +Let $A$ and $B$ be any sets such that $A \subseteq B$ and +$A \cap B^c \neq \emptyset$. + +Let $x$ be any element such that $x \in A \cap B^c$ + +By definition of $\cap$, this means that $x \in A$ and $x \in B^c$. By +definition of complement, it follows that this means that $x \in A$ and +$x \notin B$. + +Since $x \in A$, $x \in B$ since $A \subseteq B$ (by the supposition and by the +definition of subset). + +Thus $x \in B$ and $x \notin B$. This is a contradiction. + +Hence the supposition is false, and $A \cap B^c = \emptyset$. + +Q.E.D. + 11. For all sets $A$, $B$, and $C$, if $A \subseteq B$ then $A \cap (B \cap C)^c = \emptyset$. +**Disproof (by counterexample):** + +Suppose $A$, $B$, $C$, are any sets and $U$ is the universal set such that +$A \subseteq B$ and $C \subseteq A$ and $C \subseteq B$. For example: + +Let $U = \{1, 2, 3, 4\}, A = \{1, 2\}, B = \{1, 2, 3\}, C = \{2\}$. + +Then, the left-hand side of the given equality becomes: + +$$ A \cap (B \cap C)^c = \{1, 2\} \cap (\{1, 2, 3\} \cap \{2\})^c $$ + +$$ = \{1, 2\} \cap (\{2\})^c $$ + +$$ = \{1, 2\} \cap (U - \{2\}) $$ + +$$ = \{1, 2\} \cap (\{1, 2, 3, 4\} - \{2\}) $$ + +$$ = \{1, 2\} \cap \{1, 3, 4\} $$ + +$$ = \{1\} $$ + +Thus $A \cap (B \cap C)^c \neq \emptyset$ since $\{1\} \neq \emptyset$. + +Q.E.D. + 12. For all sets $A$, $B$, and $C$, $$ A \cap (B - C) = (A \cap B) - (A \cap C) $$ +**Proof:** + +Suppose $A$, $B$, and $C$ are any sets. + +To prove $A \cap (B - C) = (A \cap B) - (A \cap C)$, it must be shown that +$A \cap (B - C) \subseteq (A \cap B) - (A \cap C)$ and that +$(A \cap B) - (A \cap C) \subseteq A \cap (B - C)$. + +_Proof $A \cap (B - C) \subseteq (A \cap B) - (A \cap C)$:_ + +Let $x$ be any element such that $x \in A \cap (B - C)$. + +By the definition of $\cap$, this means that $x \in A$ and $x \in (B - C)$. By +the definition of difference, this means $x \in A$ and $x \in B$ and +$x \notin C$. + +Since $x \in A$ and $x \in B$, it follows that $x \in A \cap B$ (by the +definition of $\cap$). + +Since $x \in A$ and $x \notin C$, it follows that $x \notin A \cap C$ (by the +definition of $\cap$.) + +Hence $x \in (A \cap B) - (A \cap C)$, by the definition of difference. + +Therefore it has been shown that +$A \cap (B - C) \subseteq (A \cap B) - (A \cap C)$, which is what was to be +shown. + +_Proof $(A \cap B) - (A \cap C) \subseteq A \cap (B - C)$:_ + +Let $x$ is any element such that $x \in (A \cap B) - (A \cap C)$. + +This can be rewritten as $x \in (A \cap B) \cap (A \cap C)^c$, by the set +difference law. + +Furthermore, this can be written as $x \in (A \cap B) \cap (A^c \cup C^c)$ by De +Morgan's Laws of sets. + +This means that $x \in A$ and $x \in B$ and $x \notin A$ or $x \notin C$. + +Since we know that $x \in A$, then $x \notin C$ must be true. + +Hence $x \in A$ and $x \in B$ and $x \notin C$, or: + +$$ x \in A \cap (B - C) $$ + +Therefore it has been shown that +$(A \cap B) - (A \cap C) \subseteq A \cap (B - C)$. + +This is what was to be shown. + +_Conclusion:_ + +Since both subset relations have been proved, it can be concluded that +$A \cap (B - C) = (A \cap B) - (A \cap C)$. + +Q.E.D. + 13. For all sets $A$, $B$, and $C$, $$ A \cup (B - C) = (A \cup B) - (A \cup C) $$ +**Disproof (by counterexample):** + +Suppose $A$, $B$, and $C$ are any sets such that they are disjoint. For example: + +Let $A = \{1}, B = \{2\}, C = \{3\}$. + +The left-hand side of the equality is: + +$$ A \cup (B - C) = \{1\} \cup (\{2\} - \{3\}) $$ + +$$ = \{1\} \cup \{2\} $$ + +$$ = \{1, 2\} $$ + +The right-hand side of the equality is: + +$$ (A \cup B) - (A \cup C) = (\{1\} \cup \{2\}) - (\{1\} \cup \{3\}) $$ + +$$ = \{1, 2\} - \{1, 3\} $$ + +$$ = \{2\} $$ + +Note that: + +$$ A \cup (B - C) = \{1, 2\} \neq \{2\} = (A \cup B) - (A \cup C) $$ + +Therefore the supposition has been disproven. + +Q.E.D. + 14. For all sets $A$, $B$, and $C$, if $A \cap C = B \cap C$ and $A \cup C = B \cup C$, then $A = B$. +**Proof:** + +Let $A$, $B$, and $C$ be any sets such that $A \cap C = B \cap C$ and +$A \cup C = B \cup C$. + +To prove $A = B$, it must be shown that $A \subseteq B$, and $B \subseteq A$. + +_Proof $A \buseteq B$:_ + +Suppose $x is any element such that $x \in A$. + +By the supposition, we know that $A \cap C \subseteq B \cap C$ since +$A \cap C = B \cap C$. + +Therefore since $x \in A$, we must determine if $x \in C$ to determine if +$A \cap C$ is true. + +_Case $x \in C$:_ + +If $x \in C$, then, since $x \in A$ also, it follows that $A \cap C$ is true, +and thus by the supposition $A \cap C = B \cap C$ is true. + +It follows that $x \in B \cap C$, and thus $x \in B$. + +_Case $x \notin C$:_ + +If $x \notin C$, then $A \cup C$ is true (by virtue of $x \in A$). Then, by +supposition, $A \cup C = B \cup C$. It follows that $x \in B \cup C$. Since +$x\notin C$, this means that $x \in B$. + +In either case $x \in B$. Therefore it has been shown that $A \subseteq B$. + +_Proof $B \buseteq A$:_ + +Suppose $x is any element such that $x \in B$. + +By the supposition, we know that $B \cap C \subseteq A \cap C$ since +$A \cap C = B \cap C$. + +Therefore since $x \in B$, we must determine if $x \in C$ to determine if +$B \cap C$ is true. + +_Case $x \in C$:_ + +If $x \in C$, then, since $x \in B$ also, it follows that $B \cap C$ is true, +and thus by the supposition $B \cap C = A \cap C$ is true. + +It follows that $x \in A \cap C$, and thus $x \in A$. + +_Case $x \notin C$:_ + +If $x \notin C$, then $B \cup C$ is true (by virtue of $x \in B$). Then, by +supposition, $B \cup C = A \cup C$. It follows that $x \in A \cup C$. Since +$x\notin C$, this means that $x \in A$. + +In either case $x \in A$. Therefore it has been shown that $B \subseteq A$. + +_Conclusion:_ + +Since both subset relations have been proved, it can be concluded that $A = B$. + +Q.E.D. + 15. For all sets $A$, $B$, and $C$, $(A - B) \cup C \subseteq A \cup (C - B)$. +**Disproof (by counterexample):** + +Suppose $A$, $B$, and $C$ are any sets such that $A \cap B \neq \emptyset$, and +$B \cap C \neq \emptyset$, but $A \cap C = \emptyset$. For example: + +Let $A = \{1, 2\}, B = \{2, 3\}, C = \{3, 4\}$. + +Then, evaluating the left-hand side of the given equality: + +$$ (A - B) \cup C = (\{1, 2\} - \{2, 3\}) \cup \{3, 4\} $$ + +$$ = \{1\} \cup \{3, 4\} $$ + +$$ = \{1, 3, 4\} $$ + +Then, evaluating the right-hand side of the given equality: + +$$ A \cup (C - B) = \{1, 2\} \cup (\{3, 4\} - \{2, 3\}) $$ + +$$ = \{1, 2\} \cup \{4\} $$ + +$$ = \{1, 2, 4\} $$ + +Note that: + +$(A - B) \cup C = \{1, 3, 4\} \nsubseteq \{1, 2, 4\} = A \cup (C - B)$ + +Hence the given supposition is false since $3 \in (A - B) \cup C$ and +$3 \notin A \cup (C - B)$. + +Q.E.D. + 16. For all sets $A$ and $B$, if $A \cap B = \emptyset$ then $A \times B = \emptyset$. +**Disproof (by counterexample):** + +Suppose $A$ and $B$ are any sets such that $A \cap B = \emptyset$. For example: + +Let $A = \{1\}, B = \{2\}$. + +Note that the supposition is true given the example since: + +$$ A \cap B = \{1\} \cap \{2\} = \emptyset $$ + +By the definition of $\cap$. + +Note, though, that: + +$$ A \times B = \{1\} \times \{2\} = (1, 2) \neq \emptyset $$ + +Hence the given supposition is false since $\{1\} \cap \{2\} = \emptyset$, but +$(1, 2) \neq \emptyset$. + +Q.E.D. + 17. For all sets $A$ and $B$, if $A \subseteq B$ then $\mathscr{P}(A) \subseteq \mathscr{P}(B)$. +**Proof:** + +Let $A$ and $B$ be any sets such that $A \subseteq B$. + +Suppose $X$ is any set such that $X \in \mathscr{P}(A)$. By the definition of +power set, it follows that $X \subseteq A$. + +Since $X \subseteq A$, $X \subseteq B$ by the supposition/definition of subset +($A \subseteq B$). + +Since $X \subseteq B$, this means that $X \in \mathscr{P}(B)$, by definition of +power set. + +Hence $\mathscr{P}(A) \subseteq \mathscr{P}(B)$. + +Q.E.D. + 18. For all sets $A$ and $B$, $\mathscr{P}(A \cup B) \subseteq \mathscr{P}(A) \cup \mathscr{P}(B)$. +**Disproof (by counterexample):** + +Let $A$ and $B$ be any two disjoint sets. For example: + +$$ A = \{1\}, B = \{2\} $$ + +Then, the left-hand side of the given subset relation is: + +$$ \mathscr{P}(A \cup B) = \mathscr{P}(\{1\} \cup \{2\}) $$ + +$$ = \mathscr{P}(\{1, 2\}) $$ + +$$ = \{\emptyset, \{1\}, \{2\}, \{1, 2\}\} $$ + +Then, the right-hand side of the given subset relation is: + +$$ \mathscr{P}(A) \cup \mathscr{P}(B) = \mathscr{P}(\{1\}) \cup \mathscr{P}(\{2\}) $$ + +$$ = \{\emptyset, \{1\}\} \cup \{\emptyset, \{2\}\} $$ + +$$ = \{\emptyset, \{1\}, \{2\}\} $$ + +Now, note that: + +$$ \mathscr{P}(A \cup B) = \{\emptyset, \{1\}, \{2\}, \{1, 2\}\} \nsubseteq \{\emptyset, \{1\}, \{2\}\} = \mathscr{P}(A) \cup \mathscr{P}(B) $$ + +Hence it has been shown that the supposition is false since $\{1, 2\}$ is not in +$\{\emptyset, \{1\}, \{2\}\}$. + +Q.E.D. + 19. For all sets $A$ and $B$, $\mathscr{P}(A) \cup \mathscr{P}(B) \subseteq \mathscr{P}(A \cup B)$. +**Proof:** + +Suppose $A$ and $B$ are any sets. + +Let $X$ be some set such that $X \in \mathscr{P}(A) \cup \mathscr{P}(B)$. + +By the definition of $\cup$, this means that $X \in \mathscr{P}(A)$ or +$X \in \mathscr{P}(B)$. + +_Case $X \in \mathscr{P}(A)$:_ + +By the definition of power set, $X \subseteq A$. + +Since $X \subseteq A$, it follows that $X \subseteq A \cup B$, by definition of +$\cup$. + +_Case $X \in \mathscr{P}(B)$:_ + +By the definition of power set, $X \subseteq B$. + +Since $X \subseteq B$, it follows that $X \subseteq A \cup B$, by definition of +$\cup$. + +In either case $X \subseteq A \cup B$. By the definition of power set, this +means that $X \in \mathscr{P}(A \cup B)$. + +This is what was to be shown. + +Q.E.D. + 20. For all sets $A$ and $B$, $\mathscr{P}(A \cap B) = \mathscr{P}(A) \cap \mathscr{P}(B)$. +**Proof:** + +Let $A$ and $B$ be any sets. + +To prove that $\mathscr{P}(A \cap B) = \mathscr{P}(A) \cap \mathscr{P}(B)$, it +must be shown that +$\mathscr{P}(A \cap B) \subseteq \mathscr{P}(A) \cap \mathscr{P}(B)$ and that +$\mathscr{P}(A) \cap \mathscr{P}(B) \subseteq \mathscr{P}(A \cap B)$. + +_Proof $\mathscr{P}(A \cap B) \subseteq \mathscr{P}(A) \cap \mathscr{P}(B)$:_ + +Suppose $X$ is some set such that $X \in \mathscr{P}(A \cap B)$. + +By the definition of power set, this means that $X \subseteq A \cap B$. + +By definition of $\cap$, this means that $X \subseteq A$ and $X \subseteq B$. + +By the definition of power set, since $X \subseteq A$, this means that +$X \in \mathscr{P}(A)$. + +By the definition of power set, since $X \subseteq B$, this means that +$X \in \mathscr{P}(B)$. + +Hence $X \in \mathscr{P}(A) \cap \mathscr{P}(B)$. + +Thus it has been shown that +$\mathscr{P}(A \cap B) \subseteq \mathscr{P}(A) \cap \mathscr{P}(B)$. + +_Proof $\mathscr{P}(A) \cap \mathscr{P}(B) \subseteq \mathscr{P}(A \cap B)$:_ + +Suppose $X$ is some set such that $X \in \mathscr{P}(A) \cap \mathscr{P}(B)$. + +By the definition of $\cap$, this means that $X \in \mathscr{P}(A)$ and +$X \in \mathscr{P}(B)$. + +Since $X \in \mathscr{P}(A)$, by the definition of power set, $X \subseteq A$. + +Since $X \in \mathscr{P}(B)$, by the definition of power set, $X \subseteq B$. + +Since $X \subseteq A$ and $X \subseteq B$, it follows that +$X \subseteq A \cap B$, by the definition of subset and $\cap$. + +By the definition of power set, since $X \subseteq A \cap B$, this means that +$X \in \mathscr{P}(A \cap B)$. + +Thus it has been shown that +$\mathscr{P}(A) \cap \mathscr{P}(B) \subseteq \mathscr{P}(A \cap B)$. + +_Conclusion:_ + +Since both subset relations have been proved, it can be concluded that +$\mathscr{P}(A \cap B) = \mathscr{P}(A) \cap \mathscr{P}(B)$. + +Q.E.D. + 21. For all sets $A$ and $B$, $\mathscr{P}(A \times B) = \mathscr{P}(A) \times \mathscr{P}(B)$. +**Disproof (by counterexample):** + +Suppose $A$ and $B$ are any sets such that they are disjoint. For example: + +Let $A = \{1\}$ and $B = \{2\}$. + +Then, the left-hand side of the given equality is: + +$$ \mathscr{P}(A \times B) = \mathscr{P}(\{1\} \times \{2\}) $$ + +$$ = \mathscr{P}(\{(1, 2)\}) $$ + +$$ = \{\emptyset, \{(1, 2)\}\} $$ + +Then, the right-hand side of the given equality is: + +$$ \mathscr{P}(A) \times \mathscr{P}(B) = \mathscr{P}(\{1\}) \times \mathscr{P}(\{2\}) $$ + +$$ = \{\emptyset, \{1\}\} \times \{\emptyset, \{2\}\} $$ + +$$ = \{(\emptyset, \emptyset), (\emptyset, \{2\}), (\{1\}, \emptyset), (\{1\}, \{2\})\} $$ + +Note that: + +$$ \mathscr{P}(A \times B) = \{\emptyset, \{(1, 2)\}\} \neq \{(\emptyset, \emptyset), (\emptyset, \{2\}), (\{1\}, \emptyset), (\{1\}, \{2\})\} = \mathscr{P}(A) \times \mathscr{P}(B) $$ + +Hence, the supposition is false since +$\{\emptyset, \{(1, 2)\}\} \neq \{(\emptyset, \emptyset), (\emptyset, \{2\}), (\{1\}, \emptyset), (\{1\}, \{2\})\}$. + +Q.E.D. + 22. Write a negation for each of the following statements. Indicate which is true, the statement or its negation. Justify your answers. a. $\forall$ sets $S$, $\exists$ a set $T$ such that $S \cap T = \emptyset$. +Negation: + +$\exists$ a set $S$ such that $\forall$ sets $T$, $S \cap T \neq \emptyset$. + +The original statement is true, consider if $T = S^c$, then +$S \cap T = S \cap S^c = \emptyset$. + b. $\exists$ a set $S$ such that $\forall$ sets $T$, $S \cup T = \emptyset$. +Negation: + +$\forall$ sets $S$, $\exists$ a set $T$ such that $S \cup T \neq \emptyset$. + +The negation is true. Consider $T \neq \emptyset$, then no matter whether +$S = \emptyset$ or $S \neq \emptyset$, $S \cup T \neq \emptyset$ will always be +true since $T \neq \emptyset$. + 23. Let $S =\{a, b, c\}$, and for each integer $i = 0, 1, 2, 3$, let $S_i$ be the set of all subsets of $S$ that have $i$ elements. List the elements in $S_0, S_1, S_2$, and $S_3$. Is $\{S_0, S_1, S_2, S_3\}$ a partition of $\mathscr{P}(S)$? +$$ S_0 = \{\emptyset\} $$ + +$$ S_1 = \{\{a\}, \{b\}, \{c\}\} $$ + +$$ S_2 = \{\{a, b\}, \{a, c\}, \{b, c\}\} $$ + +$$ S_3 = \{\{a, b, c\}\} $$ + +Since all $S_i$ are mutually disjoint, nonempty, and their union is all of +$\mathscr{P}(S)$, the elements in all $S_i$ are a partition of $\mathscr{P}(S)$. + 24. Let $A = \{t, u, v, w\}$, and let $S_1$ be the set of all subsets of $A$ that do not contain $w$ and $S_2$ the set of all subsets of $A$ that contain $w$. a. Find $S_1$. +$$ S_1 = \{\emptyset, \{t\}, \{u\}, \{v\}, \{t, u\}, \{t, v\}, \{u, v\}, \{t, u, v\}\} $$ + b. Find $S_2$. +$$ S_2 = \{\{w\}, \{t, w\}, \{u, w\}, \{v, w\}, \{t, u, w\}, \{t, v, w\}, \{u, v, w\}, \{t, u, v, w\}\} $$ + c. Are $S_1$ and $S_2$ disjoint? +Yes. + d. Compare the sizes of $S_1$ and $S_2$. +Their sizes are equal (8). + e. How many elements are in $S_1 \cup S_2$? +16. + f. What is the relation between $S_1 \cup S_2$ and $\mathscr{P}(A)$? +$$ S_1 \cup S_2 = \mathscr{P}(A) $$ + 25. Use mathematical induction to prove that for every integer $n \geq 2$, if a set $S$ has $n$ elements, then the number of subsets of $S$ with an even number of elements equals the number of subsets of $S$ with an odd number of elements. +**Proof (by mathematical induction):** + +Let $P(n)$ be the sentence: + +If a set $S$ has $n$ elements, then the number of subsets of $S$ with an even +number of elements equals the number of subsets of $S$ with an odd number of +elements. + +_Basis Step:_ + +Prove $P(2)$, that is: + +If a set $S$ has $2$ elements, then the number of subsets of $S$ with an even +number of elements equals the number of subsets of $S$ with an odd number of +elements. + +Consider $S = \{1, 2\}$. Then the subsets of $S$, or the power set of $S$, would +be: $\mathscr{P}(S) = \{\emptyset, \{1\}, \{2\}, \{1, 2\}\}$. + +The number of subsets with an even number of elements is $2$ +($\emptyset, \{1, 2\}$). + +The number of subsets with an odd number of elements is $2$ ($\{1\}, \{2\}$). + +The number of subsets with an even number of elements is equal to the number of +subsets with an odd number of elements. Therefore $P(2)$ is true. + +_Inductive Step:_ + +Let $k$ be any integer such that $k \geq 2$. + +Suppose $P(k)$, that is: + +If a set $S$ has $k$ elements, then the number of subsets of $S$ with an even +number of elements equals the number of subsets of $S$ with an odd number of +elements. + +This is the inductive hypothesis. + +Prove $P(k + 1)$, that is: + +If a set $S$ has $k + 1$ elements, then the number of subsets of $S$ with an +even number of elements equals the number of subsets of $S$ with an odd number +of elements. + +Let $X$ be some set such that $X$ has $k + 1$ elements, and let $x$ be some +element such that $x \in X$. + +Then, let $Y$ be some set such that $Y = X - \{x\}$. This means that $Y$ has $k$ +elements. + +Every subset of $X$ either contains $x$ or doesn't. The subsets of $X$ that do +not contain $x$ are the subsets of $Y$, and, by the inductive hypothesis, have +an equal number of subsets containing even and odd amounts of elements. + +The subsets containing $x$ are each of the form $Z \cup \{x\}$, where +$Z \subseteq Y$. Adding $x$ flips the parity of each subset (_i.e._ even number +of subsets now becomes odd and odd number of subsets become even.) Note, though, +that the number of subsets with even number of elements and the number of +subsets with odd number of elements remain equal. + +Therefore, $P(k + 1)$ is true. + +Q.E.D. + 26. The following problem, devised by Ginger Bolton, appeared in the January 1989 issue of the _College Mathematics Journal_ (Vol. 20, No. 1, p. 68): Given a positive integer $n \geq 2$, let $S$ be the set of all nonempty @@ -2824,6 +3607,8 @@ f. What is the relation between $S_1 \cup S_2$ and $\mathscr{P}(A)$? $$ \sum_{i = 1}^{]2^{n - 1} - 1}{P_i} = \frac{(n + 1)!}{2} - 1 $$ +Omitted. + In 27 and 28 supply a reason for each step in the derivationl.l 27. For all sets $A$, $B$, and $C$, @@ -2846,6 +3631,12 @@ $$ = (A \cap C) \cup (B \cap C) $$ by __ \(c\) __ +a. by commutative law for $\cap$ + +b. by distributive law + +c. by commutative law for $\cap$ + 28. For all sets $A$, $B$, and $C$, $$ (A \cup B) - (C - A) = A \cup (B - C) $$ @@ -2882,6 +3673,20 @@ $$ = A \cup (B - C) $$ by __ (g) __ +a. by the set difference law + +b. by the set difference law + +c. by the commutative law for $\cap$ + +d. by De Morgan's Law + +e. by the double complement law + +f. by the distributive law + +g. by the set difference law + 29. Some steps are missing from the following proof that for all sets $A$ and $B$, $(A \cup B^c) - B = (A - B) \cup B^c$. Indicate what they are, and then write the proof correctly. @@ -2906,6 +3711,34 @@ $$ (A - B) \cup B^c $$ by the set difference law. +**Proof:** + +Let any sets $A$ and $B$ be given. Then + +$$ (A \cup B^c) - B = (A \cup B^c) \cap B^c $$ + +by the set difference law + +$$ = B^c \cap (A \cup B^c) $$ + +by the commutative law + +$$ = (B^c \cap A) \cup (B^c \cap B^c) $$ + +by the distributive law + +$$ = (B^c \cap A) \cup B^c $$ + +by the idempotent law for $\cap$.r + +$$ = (A \cap B^c) \cup B^c $$ + +by the commutative law + +$$ = (A - B) \cup B^c $$ + +Q.E.D. + In 30-40, construct an algebraic proof for the given statement. Cite a property from Theorem 6.2.2 for every step. @@ -2913,62 +3746,378 @@ from Theorem 6.2.2 for every step. $$ (A \cap B) \cup C = (A \cup C) \cap (B \cup C) $$ +$$ (A \cap B) \cup C = C \cup (A \cap B) $$ + +by commutative law for $\cup$ + +$$ = (C \cup A) \cap (C \cup B) $$ + +by distributive laws + +$$ = (A \cup C) \cap (B \cup C) $$ + +by commutative laws for $\cup$ + 31. For all sets $A$ and $B$, $A \cup (B - A) = A \cup B$. +$$ A \cup (B - A) = A \cup (B \cap A^c) $$ + +by set difference law + +$$ = (A \cup B) \cap (A \cup A^c) $$ + +by distributive laws + +$$ = (A \cup B) \cap U $$ + +by complement laws for $\cup$ + +$$ = A \cup B $$ + +by identity laws for $\cap$ + 32. For all sets $A$ and $B$, $(A - B) \cup (A \cap B) = A$. -33. Fora ll sets $A$ and $B$, $(A - B) \cap (A \cap B) = \emptyset$. +$$ (A - B) \cup (A \cap B) = (A \cap B^c) \cup (A \cap B) $$ + +by set difference law + +$$ = A \cap (B^c \cup B) $$ + +by distributive laws + +$$ = A \cap U $$ + +by complement laws + +$$ = A $$ + +by identity laws + +33. For all sets $A$ and $B$, $(A - B) \cap (A \cap B) = \emptyset$. + +$$ (A - B) \cap (A \cap B) = (A \cap B^c) \cap (A \cap B) $$ + +by set difference law + +$$ = A \cap A \cap B^c \cap B $$ + +by associative laws for $\cap$ + +$$ = A \cap B^c \cap B $$ + +by idempotent laws for $\cap$ + +$$ = A \cap \emptyset $$ + +by complement laws for $\cap$. + +$$ = \emptyset $$ + +by universal bound laws for $\cap$ 34. For all sets $A$, $B$, and $C$, $$ (A - B) - C = A - (B \cup C) $$ +$$ (A - B) - C = (A - B) \cap C^c $$ + +by set difference law + +$$ = (A \cap B^c) \cap C^c $$ + +by set difference law + +$$ = A \cap (B^c \cap C^c) $$ + +by associative laws for $\cap$ + +$$ = A \cap (B \cup C)^c $$ + +by De Morgan's laws + +$$ = A - (B \cup C) $$ + +by set difference law + 35. For all sets $A$ and $B$, $A - (A - B) = A \cap B$. +$$ A - (A - B) = A - (A \cap B^c) $$ + +by set difference law + +$$ = A \cap (A \cap B^c)^c $$ + +by set difference law + +$$ = A \cap (A^c \cup (B^c)^c) $$ + +by De Morgan's laws + +$$ = A \cap (A^c \cup B) $$ + +by double complement law + +$$ = (A \cap A^c) \cup (A \cap B) $$ + +by the distributive laws for $\cap$ + +$$ = \emptyset \cup (A \cap B) $$ + +by complement laws + +$$ = (A \cap B) \cup \emptyset $$ + +by commutative laws for $\cup$ + +$$ = A \cap B $$ + +by identity laws + 36. For all sets $A$ and $B$, $((A^c \cup B^c) - A)^c = A$. +$$ ((A^c \cup B^c) - A)^c $$ + +$$ = ((A^c \cup B^c) \cap A^c)^c $$ + +by set difference law + +$$ = (A^c \cup B^c)^c \cup (A^c)^c $$ + +by De Morgan's laws + +$$ = (A^c \cup B^c)^c \cup A $$ + +by double complement law + +$$ = ((A^c)^c \cap (B^c)^c) \cup A $$ + +by De Morgan's laws + +$$ = (A \cap B) \cup A $$ + +by double complement law + +$$ = A \cup (A \cap B) $$ + +by commutative laws for $\cup$ + +$$ = (A \cup A) \cap (A \cup B) $$ + +by distributive laws + +$$ = A \cap (A \cup B) $$ + +by idempotent laws + +$$ = A $$ + +by absorption laws for $\cap$ + 37. For all sets $A$ and $B$, $(B^c \cup (B^c - A))^c = B$. +$$ (B^c \cup (B^c - A))^c $ + +$$ = (B^c \cup (B^c \cap A^c))^c $ + +by set difference law + +$$ = ((B^c)^c \cap (B^c \cap A^c)^c) $ + +by De Morgan's laws + +$$ = (B \cap (B^c \cap A^c)^c) $ + +by double complement law + +$$ = B \cap ((B^c)^c \cup (A^c)^c) $ + +by De Morgan's laws + +$$ = B \cap (B \cup A) $ + +by double complement law + +$$ = (B \cap B) \cup (B \cap A) $$ + +by distributive laws + +$$ = B \cup (B \cap A) $$ + +by idempotent laws + +$$ = B $$ + +by absorption laws for $\cup$ + 38. For all sets $A$ and $B$, $(A \cap B)^c \cap A = A - B$. +$$ (A \cap B)^c \cap A $$ + +$$ = (A^c \cup B^c) \cap A $$ + +by De Morgan's laws + +$$ = A \cap (A^c \cup B^c) $$ + +by commutative laws for $\cap$ + +$$ = (A \cap A^c) \cup (A \cap B^c) $$ + +by distributive laws + +$$ = \emptyset \cup (A \cap B^c) $$ + +by complement laws + +$$ = (A \cap B^c) \cup \emptyset $$ + +by commutative laws + +$$ = A \cap B^c $$ + +by identity laws + +$$ = A - B $$ + +by set difference law + 39. For all sets $A$ and $B$, $$ (A - B) \cup (B - A) = (A \cup B) - (A \cap B) $$ +$$ (A - B) \cup (B - A) $$ + +$$ = (A \cap B^c) \cup (B \cap A^c) $$ + +by set difference law + +$$ = [(A \cap B^c) \cup B] \cap [(A \cap B^c) \cup A^c] $$ + +by distributive laws + +$$ = [B \cup (A \cap B^c)] \cap [A^c \cup (A \cap B^c)] $$ + +by commutative laws + +$$ = [(B \cup A) \cap (B \cup B^c)] \cap [(A^c \cup A) \cap (A^c \cup B^c)] $$ + +by distributive laws + +$$ = [(A \cup B) \cap (B \cup B^c)] \cap [(A \cup A^c) \cap (A^c \cup B^c)] $$ + +by commutative laws + +$$ = [(A \cup B) \cap U] \cap [U \cap (A^c \cup B^c)] $$ + +by complement laws + +$$ = [(A \cup B) \cap U] \cap [(A^c \cup B^c) \cap U] $$ + +by commutative laws + +$$ = (A \cup B) \cap (A^c \cup B^c) $$ + +by identity laws + +$$ = (A \cup B) \cap (A \cap B)^c $$ + +by De Morgan's laws + +$$ = (A \cup B) - (A \cap B) $$ + +by set difference law + 40. For all sets $A$, $B$, and $C$, $$ (A - B) - (B - C) = A - B $$ +$$ (A - B) - (B - C) $$ + +$$ = (A - B) \cap (B - C)^c $$ + +by set difference law + +$$ = (A \cap B^c) \cap (B \cap C^c)^c $$ + +by set difference law + +$$ = (A \cap B^c) \cap (B^c \cup (C^c)^c) $$ + +by De Morgan's laws + +$$ = (A \cap B^c) \cap (B^c \cup C) $$ + +by double complement law + +$$ = A \cap (B^c \cap (B^c \cup C)) $$ + +by associative laws + +$$ = A \cap ((B^c \cap B^c) \cup (B^c \cap C)) $$ + +by distributive laws + +$$ = A \cap (B^c \cup (B^c \cap C)) $$ + +By idempotent laws + +$$ = A \cap B^c $$ + +by absorption laws + +$$ = A - B $$ + +by set difference law + In 41-43 simplify the given expression. Cite a property from Theorem 6.2.2 for every step. 41. $A \cap ((B \cup A^c) \cap B^c)$ +Omitted. + 42. $(A - (A \cap B)) \cap (B - (A \cap B))$ +Omitted. + 43. $((A \cap (B \cup C)) \cap (A - B)) \cap (B \cup C^c)$ +Omitted. + 44. Consider the following set property: For all sets $A$ and $B$, $A - B$ and $B$ are disjoint. a. Use an element argument to derive the property. +Omitted. + b. Use an algebraic argument to derive the property (by applying properties from Theorem 6.2.2). +Omitted. + c. Comment on which method you found easier. +Omitted. + 35. Consider the following set property: For all sets $A$, $B$, and $C$, $$ (A - B) \cup (B - C) = (A \cup B) - (B \cap C) $$ a. Use an element argument to derive the property. +Omitted. + b. Use an algebraic argument to derive the property (by applying properties from Theorem 6.2.2). +Omitted. + c. Comment on which method you found easier. +Omitted. + **Definition:** Given sets $A$ and $B$, the **symmetric difference of $A$ and $B$**, denoted @@ -2981,33 +4130,57 @@ $$ A \Delta B = (A - B) \cup (B - A) $$ a. $A \Delta B$ +Omitted. + b. $B \Delta C$ +Omitted. + c. $A \Delta C$ +Omitted. + d. $(A \Delta B) \Delta C$ +Omitted. + Refer to the definition of symmetric difference given above. Prove each of 47-52, assuming that $A$, $B$, and $C$ are all subsets of a universal set $U$. 47. $A \Delta B = B \Delta A$ +Omitted. + 48. $A \Delta \emptyset = A$ +Omitted. + 49. $A \Delta A^c = U$ +Omitted. + 50. $A \Delta A = \emptyset$ +Omitted. + 51. If $A \Delta C = B \Dcelta C$, then $A = B$. +Omitted. + 52. $(A \Delta B) \Delta C = A \Delta (B \Delta C)$ +Omitted. + 53. Derive the set identity $A \cup (A \cap B) = A$ from the properties listed 8n Theorem 6.2.2(1)-(9). Start by showing that for every subset $B$ of a universal set $U$, $U \cup B = U$. Then intersect both sides with $A$ and deduce the identity. +Omitted. + 54. Derive the set identity $A \cap (A \cup B) = A$ from the properties listed in Theorem 6.2.2(1)-(9). Start by showing that for every subset $B$ of a universal set $U$, $\emptyset = \emptyset \cap B$. Then take the union of both sides with $A$ and deduce the identity. + +Omitted. diff --git a/chapter_6/test_yourself.md b/chapter_6/test_yourself.md index fba5496..36ecd69 100644 --- a/chapter_6/test_yourself.md +++ b/chapter_6/test_yourself.md @@ -101,8 +101,16 @@ Page 435 concrete sets $A$, $B$, and $C$ that, when substituted for the set variables in the equation, _____. +the equation does not hold. + +make the left-hand side unequal to the right-hand side + 2. When using the algebraic method for proving a set identity, it is important to _____ for every step. +cite the property from 6.2.2 used + 3. When applying a property from Theorem 6.2.2, it must be used _____ as it is stated. + +exactly