From 2601f42c1fd05f54e9e10e5a8f75c4fb137fbe0b Mon Sep 17 00:00:00 2001 From: tomit4 Date: Mon, 20 Jul 2026 18:56:05 -0700 Subject: [PATCH] :construction: Mid of 6.2 --- chapter_6/exercises.md | 997 ++++++++++++++++++++++++++++++++++++- chapter_6/test_yourself.md | 12 + 2 files changed, 1008 insertions(+), 1 deletion(-) diff --git a/chapter_6/exercises.md b/chapter_6/exercises.md index f1d0e53..45f20f2 100644 --- a/chapter_6/exercises.md +++ b/chapter_6/exercises.md @@ -1402,19 +1402,39 @@ Page 427 a. To say that an element is in $A \cap (B \cup C)$ means that it is in __ (1) __ and in __ (2) __. +(1) $A$ + +(2) $B \cup C$ + b. To say that an element is in $(A \cap B) \cup C$ means that it is in __ (1) __ or in __ (2) __. +(1) $A \cap B$ + +(2) $C$ + c. To say that an element is in $A - (B \cap C)$ means that it is in __ (1) __ and not in __ (2)__. +(1) $A$ + +(2) $B \cap C$ + d. To prove that $(A \cup B) \cap C \subseteq A \cup (B \cap C)$, we suppose that $x$ is any element in __ (1) __. Then we must show that __ (2) __. +(1) $(A \cup B) \cap C$ + +(2) $x \in A \cup (B \cap C)$ + e. If $A$, $B$, and $C$ are any sets such that $B \subseteq C$, to prove that $A \cap B \subseteq A \cap C$, we suppose that $x$ is any element in __ (1) __. Then we must show that __ (2) __. +(1) $A \cap B$ + +(2) $A \cap C$ + 2. The following are two proofs that for all sets $A$ and $B$, $A - B \subseteq A$. The first is less formal, and the second is more formal. Fill in the blanks. @@ -1424,12 +1444,28 @@ $A - B \subseteq A$, we must show that every element in __ (1) __ is in __ (2) __. But any element in $A - B$ is in __ (3) __ and not in __ (4) __ (by definition of $A - B$). In particular, such an element is in $A$. +(1) $A - B$ + +(2) $A$ + +(3) $A$ + +(4) $B$ + b. **Proof:** Suppose $A$ and $B$ are any sets and $x \in A - B$. _[We must show that __ (1) __.]_ By definition of set difference, $x \in$ __ ( 2 ) __ and $x \notin$ __ (3) __. In particular, $x \in$ __ (4) __ _[which is what was to be shown]._ -In 3 and 4, supply explanations of the stesp in the given proofs. +(1) $x \in A$ + +(2) $A$ + +(3) $B$ + +(4) $A$ + +In 3 and 4, supply explanations of the steps in the given proofs. 3. **Theorem:** For all sets $A$, $B$, and $C$, if $A \subseteq C$, $B \subseteq C$, then $A \subseteq C$. @@ -1446,6 +1482,12 @@ In 3 and 4, supply explanations of the stesp in the given proofs. | Thus every element in $A$ is in $C$ | since $x$ could be any element of $A$ | | Therefore, $A \subseteq C$ _[as was to be shown]._ | __ \(c\) __ | +a. by definition of a subset (because $A$ is a subset of $B$) + +b. by definition of a subset (because $B$ is a subset of $C$) + +c. by definition of a subset + 4. **Theorem:** For all sets $A$ and $B$, if $A \subseteq B$, then $A \cup B \subseteq B$. @@ -1463,8 +1505,54 @@ In 3 and 4, supply explanations of the stesp in the given proofs. | Thus every element in $A \cup B$ is in $B$ | since $x$ could be any element of $A \cup B$ | | Therefore, $A \cup B \subseteq B$ _[as was to be shown]._ | __ \(c\) __ | +a. by the definition of a union (because $A \cup B$) + +b. by definition of a subset (because $A \subseteq B$) + +c. by definition of a subset + 5. Prove that for all sets $A$ and $B$, $(B - A) = B \cap A^c$. +**Proof:** + +Let $A$ and $B$ be any sets. + +To prove that $(B - A) = B \cap A^c$, we must first prove +$(B - A) \subseteq B \cap A^c$ and then prove $B \cap A^c \subseteq (B - A)$. + +_Proof ($(B - A) \subseteq B \cap A^c$):_ + +Suppose $x$ is some element such that $x \in (B - A)$. + +By the definition of the difference of sets, this means that $x \in B$ and +$x \notin A$. It then follows by the definition of the complement of sets that +$x \in B$ and $x \in A^c$. + +By definition of an intersection, it then follows further that +$x \in B \cap A^c$. + +Therefore every element that is in $(B - A)$ is also in $B \cap A^c$. This is +what was to be shown. + +_Proof ($B \cap A^c \subseteq (B - A)$):_ + +Suppose $x$ is some element such that $x \in B \cap A^c$. + +By definition of the intersection of sets, this means that $x \in B$ and +$x \in A^c$. By definition of the complement of sets, this means that $x \in B$ +and $x \notin A$. + +It follows that if $x \in B$ and $x \notin A$, then by the definition of the +difference of sets $x \in (B - A)$. + +Therefore every element that is in $B \cap A^c$ is in $(B - A)$. This is what +was to be shown. + +Since both relations have been proved, it is concluded that +$(B - A) = B \cap A^c$, by definition of set equality. + +Q.E.D. + 6. Let $\cap$ and $\cup$ stand for the words "intersection" and "union", respectively. Fill in the blanks in the following proof that for all sets $A$, $B$, and $C$, $A \cap (B \cup C) = (A \cap C) \cup (A \cap C)$. @@ -1520,58 +1608,724 @@ definition of set equality, that __ (a) __.]_ Use an element argument to prove each statement in 7-22. Assume that all sets are subsets of a universal set $U$. +(1a) $x \in (A \cap B) \cup (A \cap C)$ + +(1b) $A$ + +(1c) $x \in C$ + +(1d) $x \in (A \cap B) \cup (A \cap C)$ + +(2a) or + +(2b) and + +(2c) $x \in A \cap (B \cup C)$ + +(2d) subset + +(3a) for all sets $A$, $B$, and $C$, +$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ + 7. For all sets $A$ and $B$, $(A \cap B)^c = A^c \cup B^c$. +**Proof:** + +Let $A$ and $B$ be any sets. + +To prove that $(A \cap B)^c = A^c \cup B^c$, we must show that +$(A \cap B)^c \subseteq A^c \cup B^c$ and also show that +$A^c \cup B^c \subseteq (A \cap B)^c$. + +_Proof ($(A \cap B)^c \subseteq A^c \cup B^c$):_ + +Suppose $x$ is some element such that $x \in (A \cap B)^c$. + +By the definition of complement, this means that $x \notin (A \cap B)$. + +By the definition of intersection of sets (and by De Morgan's Laws of negation +of sets), this means that $x \notin A$ or $x \notin B$. It follows by the +definition of complement of sets that $x \in A^c$ or $x \in B^c$. + +Thus, by definition of the union of sets, $x \in A^c \cup B^c$. + +Therefore all elements in $(A \cap B)^c$ are in $A^c \cup B^c$. Specifically, +$(A \cap B)^c \subseteq A^c \cup B^c$. + +This is what was to be shown. + +_Proof ($A^c \cup B^c \subseteq (A \cap B)^c$):_ + +Suppose $x$ is some element such that $x \in A^c \cup B^c$. By the definition of +the union of sets, $x \in A^c$ or $x \in B^c$. By the definition of complement +of sets, this means that $x \notin A$ or $x \notin B$. + +By definition of the union of sets, it follows that $x \in A^c \cup B^c$. Then +by De Morgan's Laws of Sets, $x \in (A \cap B)^c$. + +Therefore all elements in $A^c \cup B^c$ are in $(A \cap B)^c$. Specifically, +$A^c \cup B^c \subseteq (A \cap B)^c$. + +This is what was to be shown. + +_Conclusion:_ + +Since both sets have been shown to be subsets of the other, it is concluded that +$A^c \cup B^c = (A \cap B)^c$, by definition of set equality. + +Q.E.D. + 8. For all sets $A$ and $B$, $(A \cap B) \cup (A \cap B^c) = A$. (This property is used in Section 9.9.) +**Proof:** + +Let $A$ and $B$ be any sets. + +To prove that $(A \cap B) \cup (A \cap B^c) = A$, it must be shown that +$(A \cap B) \cup (A \cap B^c) \subseteq A$ and also that +$A \subseteq (A \cap B) \cup (A \cap B^c)$. + +_Proof ($(A \cap B) \cup (A \cap B^c) \subseteq A$):_ + +Suppose $x$ is any element such that $x \in (A \cap B) \cup (A \cap B^c)$. + +By definition of $\cup$, this means that $x \in A \cap B$ or $x \in A \cap B^c$ + +_Case $x \in A \cap B$:_ + +By definition of $\cap$, this means that $x \in A$ and $x \in B$. By definition +of $\subseteq$, if $x \in A$ and $x \in B$, then $A \cap B \subseteq A$. + +_Case $x \in A \cap B^c$:_ + +By definition of $\cap$, this means that $x \in A$ and $x \in B^c$. By +definition of complement, this means that $x \in A$ and $x \notin B$. By +tautology and by definition of $\subseteq$, if $x \in A$ and $x \notin B$, then +$x \in A \cap B^c \subseteq A$. + +Thus in both cases, it has been shown that any element in $A \cap B$ or +$A \cap B^c$ is in $A$. Specifically $(A \cap B) \cup (A \cap B^c) \subseteq A$. + +This is what was to be shown. + +_Proof ($A \subseteq (A \cap B) \cup (A \cap B^c)$):_ + +Suppose $x$ is any element such that $x \in A$. + +To prove $x \in A \subseteq (A \cap B) \cup (A \cap B^c)$, we must prove that +either $x \in B$ or $x \in B^c$ (by definition of $\cup$ and the complement of +sets.) + +_Case $x \in B$:_ + +By the supposition, it is known that $x \in A$ and $x \in B$. By the definition +of $\cap$, this means that $x \in A \cap B$. + +_Case $x \notin B$:_ + +By the supposition, it is known that $x \in A$ and $x \notin B$. By the +definition of $\cap$ and the complement of sets, this means that +$x \in A \cap B^c$. + +In the case that $x \in B$, it has been shown that then $x \in A \cap B$. In the +case that $x \notin B$, it has been shown that $x \in A \cap B^c$. Thus it can +be stated that $x \in A \cap B$ or $x \in A \cap B^c$. By the definition of +$\cup$, it follows that $x \in (A \cap B) \cup (A \cap B^c)$. + +Therefore it can be said that every element in $A$ is in +$(A \cap B) \cup (A \cap B^c)$. Specifically, +$A \subseteq (A \cap B) \cup (A \cap B^c)$. + +This is what was to be shown. + +_Conclusion:_ + +It has been shown that $(A \cap B) \cup (A \cap B^c) \subseteq A$ and +$A \subseteq (A \cap B) \cup (A \cap B^c)$. By the definition of the equality of +sets, this means that $(A \cap B) \cup (A \cap B^c) = A$. + +Q.E.D. + 9. For all sets $A$, $B$, and $C$, $$ (A - B) \cup (C - B) = (A \cup C) - B $$ +**Proof:** + +Let $A$, $B$, and $C$ be any sets. + +To prove $(A - B) \cup (C - B) = (A \cup C) - B$, it must be shown that +$(A - B) \cup (C - B) \subseteq (A \cup C) - B$ and that +$(A \cup C) - B \subseteq (A - B) \cup (C - B)$. + +_Proof ($(A - B) \cup (C - B) \subseteq (A \cup C) - B$):_ + +Suppose $x$ is some element such that $x \in (A - B) \cup (C - B)$. + +By the definition of $\cup$, this means that $x \in (A - B)$ or $x \in (C - B)$. + +In the case that $x \in (A - B)$, $x \in A$ and $x \notin B$. In the case that +$x \in (C - B)$, $x \in C$ and $x \notin B$. In both cases $x \notin B$. + +It follows that $x \in A$ or $x \in C$. Specifically $x \in (A \cup C)$. In +either case, $x \notin B$. Hence, by the definition of difference of sets, +$x \in (A \cup C) - B$. + +Therefore every element in $(A - B) \cup (C - B)$ is in $(A \cup C) - B$. +Specifically, $(A - B) \cup (C - B) \subseteq (A \cup C) - B$. + +This is what was to be shown. + +_Proof ($(A \cup C) - B \subseteq (A - B) \cup (C - B)$):_ + +Suppose $x$ is some element such that $x \in (A \cup C) - B$. + +By the definition of difference of sets, this means that $x \in (A \cup C)$ and +$x \notin B$. By the definition of $\cup$, this means that $x \in A$ or +$x \in C$. + +In the case that $x \in A$, then $x \in A$ and $x \notin B$. By the definition +of complements, this means that $x \in A \cap B^c$. It follows by the set +difference law, that $x \in A - B$. + +In the case that $x \in C$, then $x \in A$ and $x \notin B$. By the definition +of complements, this means that $x \in C \cap B^c$. It follows by the set +difference law, that $x \in C - B$. + +Thus it can be said that $x \in A - B$ or $x \in C - B$. By the definition of +$\cup$, it follows that $x \in (A - B) \cup (C - B)$. + +Therefore every element in $(A \cup C) - B$ is in $(A - B) \cup (C - B)$. +Specifically, $A \cup C - B \subseteq (A - B) \cup (C - B)$. + +This is what was to be shown. + +_Conclusion:_ + +Since both subset relations have been proved, it is concluded that +$(A - B) \cup (C - B) = (A \cup C) - B$ by definition of set equality. + +Q.E.D. + 10. For all sets $A$, $B$, and $C$, $$ (A \cup B) \cap C \subseteq A \cup (B \cap C) $$ +**Proof:** + +Let $A$, $B$, and $C$ be any sets. + +Suppose $x$ is some element such that $x \in (A \cup B) \cap C$. + +By the definition of $\cap$, this means that $x \in (A \cup B)$ and $x \in C$. +By the definition of $\cup$, this means that $x \in A$ or $x \in B$. + +_Case $x \in A$:_ + +Since $x \in A$ and $x \in C$, it follows that $x \in A \cup (B \cap C)$, since +$x \in A$. + +_Case $x \in B$:_ + +Since $x \in A$ and $x \in C$, this means that $x \in B \cap C$. It follows that +$x \in A \cup (B \cap C)$, since $x \in B \cap C$. + +Thus in both cases $x \in A \cup (B \cap C)$. + +Therefore every element in $(A \cup B) \cap C$ is in $A \cup (B \cap C)$. By the +definition of a subset, this means that +$(A \cup B) \cap C \subseteq A \cup (B \cap C)$. + +This is what was to be shown. + +Q.E.D. + 11. For all sets $A$, $B$, and $C$, $$ A \cap (B - C) \subseteq (A \cap B) - (A \cap C) $$ +**Proof:** + +Let $A$, $B$, and $C$ be any sets. + +Suppose $x$ is some 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 the +definition of difference of sets, this means that $x \in A$ and $x \in B$ and +$x \notin C$. + +Since $x \in A$ and $x \in B$, it follows that $x \in A \cap B$. + +Since $x \in A$ and $x \notin C$, by the definition of complement, it can be +said that $x \in A \cap C^c$, or $x \notin A \cap C$. + +Thus $x \in A \cap B$ and $x \notin A \cap C$. Hence, by the difference of sets, +$x \in (A \cap B) - (A \cap C)$. + +Therefore it can be said that every element in $A \cap (B - C)$ is in +$(A \cap B) - (A \cap C)$. Specifically +$A \cap (B - C) \subseteq (A \cap B) - (A \cap C)$. + +This is what was to be shown. + +Q.E.D. + 12. For all sets $A$, $B$, and $C$, $$ (A \cup B) - C \subseteq (A - C) \cup (B - C) $$ +**Proof:** + +Let $A$, $B$, and $C$ be any sets. + +Suppose $x$ is some element such that $x \in (A \cup B) - C$. + +By the definition of difference, this means that $x \in A \cup B$ and +$x \notin C$. + +By the definition of $\cup$, it follows that $x \in A$ or $x \in B$. + +_Case $x \in A$:_ + +Since $x \in A$ and $x \notin C$, by the definition of difference, it can be +said that $x \in A - C$. + +_Case $x \in B$:_ + +Since $x \in B$ and $x \notin C$, by the definition of difference, it can be +said that $x \in B - C$. + +Hence it can be said that $x \in A - C$ or $x \in B - C$. By the definition of +$\cup$, it follows that $x \in (A - C) \cup (B - C)$. + +Therefore it can said that any element in $(A \cup B) - C$ is also in +$(A - C) \cup (B - C)$. Specifically, by definition of a subset, +$(A \cup B) - C \subseteq (A - C) \cup (B - C)$. + +This is what was to be shown. + +Q.E.D. + 13. For all sets $A$, $B$, and $C$, $$ (A - B) \cap (C - B) = (A \cap C) - B $$ +Let $A$, $B$, and $C$ be any sets. + +To prove $(A - B) \cap (C - B) = (A \cap C) - B$, it must be shown that +$(A - B) \cap (C - B) \subseteq (A \cap C) - B$ and that +$(A \cap C) - B \subseteq (A - B) \cap (C - B)$. + +_Proof ($(A - B) \cap (C - B) \subseteq (A \cap C) - B$):_ + +Suppose $x$ is some element such that $x \in (A - B) \cap (C - B)$. + +By the definition of $\cap$, this means that $x \in A - B$ and $x \in C - B$. + +By the definition of difference, this means that $x \in A$ and $x \notin B$ and +$x \in C$ and $x \notin B$. + +Thus $x$ is in $A$ and $C$, or (by definition of $\cap$), $x \in A \cap C$. +Since $x \notin B$, it follows then that $x \in (A \cap C) \cap B^c$, by the +definition of complement. + +By the set difference law, it follows that $x \in (A \cap C) - B$. + +Thus every element in $(A - B) \cap (C - B)$ is in $(A \cap C) - B$. By the +definition of subset, it follows that +$(A - B) \cap (C - B) \subseteq (A \cap C) - B$. + +This is what was to be shown. + +_Proof ($(A \cap C) - B \subseteq (A - B) \cap (C - B)$):_ + +Suppose $x$ is some element such that $x \in (A \cap C) - B$. + +By the definition of difference this means that $x \in A \cap C$ and +$x \notin B$. By the definition of $\cap$, this means that $x \in A$ or +$x \in C$ and $x \notin B$. + +_Case $x \in A$:_ + +Since $x \in A$ and $x \notin B$, this means that $x \in A \cap B^c$. By the set +difference law, this means that $x \in A - B$. + +_Case $x \in C$:_ + +Since $x \in C$ and $x \notin B$, this means that $x \in C \cap B^c$. By the set +difference law, this means that $x \in C - B$. + +It follows that $x \in A - B$ or $x \in C - B$. By the definition of $\cap$, +this means that $x \in (A - B) \cap (C - B)$. + +Thus every element in $(A \cap C) - B$ is in $(A - B) \cap (C - B)$. By the +definition of subset, this means that +$(A \cap C) - B \subseteq (A - B) \cap (C - B)$. + +This is what was to be shown. + +Q.E.D. + 14. For all sets $A$ and $B$, $A \cup (A \cap B) = A$. +**Proof:** + +Let $A$ and $B$ be any sets. + +To prove $A \cup (A \cap B) = A$, it must be shown that +$A \cup (A \cap B) \subseteq A$ and $A \subseteq A \cup (A \cap B)$. + +_Proof ($A \cup (A \cap B) \subseteq A$):_ + +Suppose $x$ is some element such that $x \in A \cup (A \cap B)$. + +By the definition of $\cup$, this means that $x \in A$ or $x \in A \cap B$. + +_Case $x \in A$:_ + +Since $x \in A$, by tautology, $x \in A$. + +_Case $x \in A \cap B$:_ + +By the definition of $\cap$, $x \in A$ and $x \in B$. + +In either case $x \in A$. By the definition of subset, this means that +$x \subseteq A$. + +Thus every element in $A \cup (A \cap B)$ is in $A$. By the definition of +subset, this means that $A \cup (A \cap B) \subseteq A$. + +This is what was to be shown. + +_Proof ($A \subseteq A \cup (A \cap B)$):_ + +Suppose $x$ is some element such that $x \in A$. + +By tautology, $x \in A \to x \in A$. + +Since $x \in A$, $x \in A \cap B$, by virtue of $x \in A$. + +It follows that $x \in A$ or $x \in A \cap B$. + +Thus it can be said that every element in $A$ is in $A \cup (A \cap B)$. By +definition of subset, this means that $A \subseteq A \cup (A \cap B)$. + +This is what was to be shown. + +_Conclusion:_ + +Since both subset relations have been proved, it has been shown that +$A \cup (A \cap B) = A$. + +Q.E.D. + 15. For every set $A$, $A \cup \emptyset = A$. +**Proof:** + +Let $A$ be any set. + +To prove that $A \cup \emptyset = A$, it must be shown that +$A \cup \emptyset \subseteq A$, and that $A \subseteq A \cup \emptyset$. + +_Proof ($A \cup \emptyset \subseteq A$):_ + +Suppose $x$ is some element such that $x \in A \cup \emptyset$. + +By the definition of $\cup$, this means that $x \in A$ or $x \in emptyset$. But +$x \notin \emptyset$, as $\emptyset$ can have no elements. + +Hence $x \in A$, and therefore $A \cup \emptyset \subseteq A$. + +_Proof ($A \subseteq A \cup \emptyset$):_ + +Suppose $x$ is some element such that $x \in A$. It follows that $x \in A$ or +$x \in \emptyset$. By definition of $\cup$, this means that +$x \in A \cup \emptyset$. + +Therefore $A \subseteq A \cup \emptyset$. + +Since both subset relations have been proved, it can be said that +$A \cup \emptyset = A$ by the definition of set equality. + +This is what was to be proved. + +Q.E.D. + 16. For all sets $A$, $B$, and $C$, if $A \subseteq B$ then $A \cap C \subseteq B \cap C$. +**Proof:** + +Let $A$, $B$, and $C$ be any sets such that $A \subseteq B$. + +Suppose $x$ is some element such that $x \in A \cap C$. + +By the definition of $\cap$, this means that $x \in A$ and $x \in C$. + +Since $x \in A$ and $A \subseteq B$, then $x \in B$ by definition of subset. + +Since $x \in B$ and $x \in C$, by the definition of $\cap$, it can be said that +$x \in B \cap C$. + +Thus it has been shown that any element in $A \cap C$ is in $B \cap C$. +Specifically $A \cap C \subseteq B \cap C$ by the definition of subset. + +This is what was to be shown. + +Q.E.D. + 17. For all sets $A$, $B$, and $C$, if $A \subseteq B$ then $A \cup C \subseteq B \cup C$. +**Proof:** + +Let $A$, $B$, and $C$ be any sets such that $A \subseteq B$. + +Suppose $x$ is some element such that $x \in A \cup C$. + +By definition of $\cup$, this means that $x \in A$ or $x \in C$. + +_Case $x \in A$:_ + +Since $x \in A$ and since $A \subseteq B$, this means that $x \in B$. + +_Case $x \in C$:_ + +By tautology, $x \in C$. + +It follows that $x \in B$ or $x \in C$. By the definition of $\cup$, this is +$x \in B \cup C$. + +Thus it can be said that any element in $A \cup C$ is in $B \cup C$, or +$A \cup C \subseteq B \cup C$ by the definition of subset. + +This is what was to be shown. + +Q.E.D. + 18. For all sets $A$ and $B$, if $A \subseteq B$ then $B^c \subseteq A^c$. +**Proof:** + +Let $A$ and $B$ be any sets such that $A \subseteq B$. + +Suppose $x$ is some element such that $x \in B^c$. + +By the definition of complement, this means that $x \notin B$. + +Since $A \subseteq B$, it follows that any element not in $B$ is not in $A$, +thus $x \notin A$. By the definition of complement, this means that $x \in A^c$ + +Hence it can said that any element in $B^c$ is in $A^c$, or $B^c \subseteq A^c$ +by the definition of subset. + +This is what was to be shown. + +Q.E.D. + 19. For all sets $A$, $B$, and $C$, if $A \subseteq B$ and $A \subseteq C$ then $A \subseteq B \cap C$. +**Proof:** + +Let $A$, $B$, and $C$ be any sets such that $A \subseteq B$ and $A \subseteq C$. + +Suppose $x$ is some element such that $x \in A$. + +Since $x \in A$ and $A \subseteq B$, it follows by the definition of subset that +$x \in B$. + +Since $x \in A$ and $A \subseteq C$, it follows by the definition of subset that +$x \in C$. + +Thus it can be said that $x \in B$ and $x \in C. By the definition of $\cap$, +this is $x \in B \cap C$. + +Therefore it has been shown that any element in $A$ is in $B \cap C$, or +$A \subseteq B \cap C$, by the definition of subset. + +This is what was to be shown. + +Q.E.D. + 20. For all sets $A$, $B$, and $C$, if $A \subseteq C$ and $B \subseteq C$ then $A \cup B \subseteq C$. +**Proof:** + +Let $A$, $B$, and $C$ be any sets such that $A \subseteq C$ and $B \subseteq C$. + +Suppose $x$ is some element such that $x \in A \cup B$. + +By the definition of $\cup$, this means that $x \in A$ or $x \in B$. + +_Case $x \in A$:_ + +Since $x \in A$ and $A \subseteq C$, this means, by definition of subset, that +$x \in C$. + +_Case $x \in B$:_ + +Since $x \in B$ and $B \subseteq C$, this means, by definition of subset, that +$x \in C$. + +In either case, $x \in C$. + +Therefore it can be said that any element in $A \cup B$ is in $C$, or +$A \cup B \subseteq C$, by definition of subset. + +This is what was to be shown. + +Q.E.D. + 21. For all sets $A$, $B$, and $C$, $$ A \times (B \cup C) = (A \times B) \cup (A \times C) $$ +**Proof:** + +Let $A$, $B$, and $C$ be any set. + +To prove $A \times (B \cup C) = (A \times B) \cup (A \times C)$, it must be +shown that $A \times (B \cup C) \subseteq (A \times B) \cup (A \times C)$ and +$(A \times B) \cup (A \times C) \subseteq A \times (B \cup C)$. + +_Proof ($A \times (B \cup C) \subseteq (A \times B) \cup (A \times C)$):_ + +Suppose $(x, y)$ are any Cartesian pair such that +$(x, y) \in A \times (B \cup C)$. + +By the definition of a Cartesian pair, this means that $x \in A$ and +$y \in B \cup C$. + +By definition of $\cup$, this means that $y \in B$ or $y \in C$. + +_Case $y \in B$:_ + +Since $x \in A$ and $y \in B$, by definition of Cartesian product, +$(x, y) \in A \times B$. + +_Case $y \in C$:_ + +Since $x \in A$ and $y \in C$, by definition of Cartesian product, +$(x, y) \in A \times C$. + +Thus it can be said that $(x, y) \in A \times B$ or $(x, y) \in A \times C$. By +definition of $\cup$, this is $(x, y) \in (A \times B) \cup (A \times C)$. + +Thus it can be said that any Cartesian pair of elements in $A \times (B \cup C)$ +are in $(A \times B) \cup (A \times C)$, or +$A \times (B \cup C) \subseteq (A \times B) \cup (A \times C)$. + +This is what was to be shown. + +_Proof ($(A \times B) \cup (A \times C) \subseteq A \times (B \cup C)$):_ + +Suppose $(x, y)$ are some Cartesian pair such that +$(x, y) \in (A \times B) \cup (A \times C)$. + +By the definition of $\cup$, this means that $(x, y) \in (A \times B)$ or +$(x, y) \in (A \times C)$. + +_Case $(x, y) \in (A \times B)$:_ + +This means that $x \in A$ and $y \in B$. Since $y \in B$, it follows that +$y \in B \cup C$, by virtue of $y \in B$. + +Thus it can be said, by the definition of Cartesian product, that +$(x, y) \in A \times (B \cup C)$. + +_Case $(x, y) \in (A \times C)$:_ + +This means that $x \in A$ and $y \in C$. Since $y \in C$, it follows that +$y \in B \cup C$, by virtue of $y \in C$. + +Thus it can be said, by the definition of Cartesian product, that +$(x, y) \in A \times (B \cup C)$. + +Hence in both cases $(x, y) \in A \times (B \cup C)$. + +Thus it has been shown that every Cartesian pair in +$(A \times B) \cup (A \times C)$ is in $A \times (B \cup C)$, or +$(A \times B) \cup (A \times C) \subseteq A \times (B \cup C)$. + +This is what was to be shown. + +_Conclusion:_ + +Since both subset relations have been proven, it can be concluded that +$A \times (B \cup C) = (A \times B) \cup (A \times C)$ by the definition of set +equality. + +This is what was to be shown. + +Q.E.D. + 22. For all sets $A$, $B$, and $C$, $$ A \times (B \cap C) = (A \times B) \cap (A \times C) $$ +**Proof:** + +Let $A$, $B$, and $C$ be any sets. + +To prove $A \times (B \cap C) = (A \times B) \cap (A \times C)$, it must be +shown that $A \times (B \cap C) \subseteq (A \times B) \cap (A \times C)$ and +that $(A \times B) \cap (A \times C) \subseteq A \times (B \cap C)$. + +_Proof ($A \times (B \cap C) \subseteq (A \times B) \cap (A \times C)$):_ + +Suppose $(x, y)$ be some elements such that $(x, y) \in A \times (B \cap C)$. + +This means that $x \in A$ and $y \in B \cap C$. + +By the definition of $\cap$, this means that $y \in B$ and $y \in C$. + +Since $x \in A$ and $y \in B$, this means that $(x, y) \in A \times B$ (by the +definition of Cartesian product). + +Furthermore, since $x \in A$ and $y \in C$, this means that +$(x, y) \in A \times C$ (by the definition of Cartesian product). + +Thus $(x, y) \in A \times B$ and $(x, y) \in A \times C$ or +$(x, y) \in (A \times B) \cap (A \times C)$ (by the definition of $\cap$). + +Hence it has been shown that +$A \times (B \cap C) \subseteq (A \times B) \cap (A \times C)$. + +_Proof ($(A \times B) \cap (A \times C) \subseteq A \times (B \cap C)$):_ + +Suppose $(x, y)$ be some elements such that +$(x, y) \in (A \times B) \cap (A \times C)$. + +By the definition of $\cap$, this means that $(x, y) \in A \times B$ and +$(x, y) \in A \times C$. + +Since $(x, y) \in A \times B$, $x \in A$ and $y \in B$. + +Since $(x, y) \in A \times C$, this means that $x \in A$ and $y \in C$. + +Since $y \in B$ and $y \in C$, $y \in B \cap C$ (by the definition of $\cap$). + +Since $x \in A$ and $y \in B \cap C$, by the definition of Cartesian product, +$(x, y) \in A \times (B \cap C)$. + +Hence it has been shown that +$(A \times B) \cap (A \times C) \subseteq A \times (B \cap C)$. + +_Conclusion:_ + +Since both subset relations have been proven, it is concluded that +$A \times (B \cap C) = (A \times B) \cap (A \times C)$ by the definition of set +equality. + +This is what was to be shown. + +Q.E.D. + 23. Find the mistake in the following "proof" that for all sets $A$, $B$, and $C$, if $A \subseteq B$ and $B \subseteq C$ then $A \subseteq C$. @@ -1581,6 +2335,19 @@ $x \in A$ and $x \in B$, and since $B \subseteq C$, there is an element $x$ such that $x \in B$ and $x \in C$. Hence there is an element $x$ such that $x \in A$ and $x \in C$ and so $A \subseteq C$. +There is more than one error in this "proof." The most serious is the misuse of +the definition of subset. To say that $A$ is a subset of $B$ means that for +every $x$, **if** $x \in A$ **then** $x \in B$. It does not mean that there +exists an element of $A$ that is also an element of $B$. The second error in the +proof occurs in the last sentence. Even if there is an element in $A$ that is in +$B$ and an element in $B$ that is in $C$, it does not follow that there is an +element in $A$ that is in $C$. + +For instance, suppose $A = \{1, 2\}$, $B = \{2, 3\}$, and $C = \{3, 4\}$. Then +there is an element in $A$ that is in $B$ (namely $2$) and there is an element +in $B$ that is in $C$ (namely, $3$), but there is no element in $A$ that is in +$C$. + 24. Find the mistake in the following "proof." **Theorem:** For all sets $A$ and $B$, $A^c \cup B^c \subseteq (A \cup B)^c^c$ @@ -1591,6 +2358,13 @@ or $x \notin B$ by definition of complement, and so $x \notin A \cup B$ by definition of union. Thus $x \in (A \cup B)^c$ by definition of complement, and hence $A^c \cup B^c \subseteq (A \cup B)^c$. +The mistake in this "proof" occurs when the author misuses the definition of +union in the sentence "and so $x \notin A \cup B$ by definition of union." + +For example, take $x = 1$, $A = \{1\}$ and $B = \{2\}$, then $A^c \cup B^c$ is +true since $B^c$ is true, but $x \notin A \cup B$ ($x \in (A \cup B)^c$) is +false since $1 \in \{1, 2\}$. + 25. Find the mistake in the following "proof" that for all sets $A$ and $B$, $(A - B) \cup (A \cap B) \subseteq A$. @@ -1599,6 +2373,14 @@ $x \in (A - B) \cup (A \cap B)$. If $x \in A$ then $x \in A - B$, and so, by definition of difference, $x \in A$ and $x \notin B$. In particular, $x \in A$, and, therefore, $(A - B) \cup (A \cap B) \subseteq A$ by definition of subset. +The author of this "proof" makes a mistake when they assume the conclusion, "If +$x \in A$." The supposition should be "Suppose $x$ is some element such that +$x \in (A - B) \cup (A \cap B)$" and follow from there. + +Furthermore it does not follow that if $x \in A$, then $x \in A - B$. Suppose +$A = B = \{x\}$, then $x \in A$, but $A - B = \emptyset$, and by definition of +$\emptyset$, $x \notin \emptyset$, so $x \notin A - B$. + 26. Consider the Venn diagram below. (See page 429 for image.) @@ -1607,18 +2389,26 @@ a. Illustrate one of the distributive laws by shading in the region corresponding to $A \cup (B \cap C)$ on one copy of the diagram and $(A \cup B) \cap (A \cup C)$ on another. +Omitted. + b. Illustrate the other distributive law by shading in the region corresponding to $A \cap (B \cup C)$ on one copy of the diagram and $(A \cap B) \cup (A \cap C)$ on another. +Omitted. + c. Illustrate one of De Morgan's laws by shading in the region corresponding to $(A \cup B)^c$ on one copy of the diagram and $A^c \cap B^c$ on the other. (Leave the set $C$ out of your diagrams.) +Omitted. + d. Illustrate the other De Morgan's law by shading in the region corresponding to $(A \cap B)^c$ on one copy of the diagram and $A^c \cup B^c$ on the other. (Leave the set $C$ out of your diagrams.) +Omitted. + 27. Fill in the blanks in the following proof that for all sets $A$ and $B$, $(A - B) \cap (B - A) = \emptyset$. @@ -1632,38 +2422,243 @@ particular $x \in A$ and $x \notin$ __ (f) __, which is a contradiction. Hence _[the supposition that $(A - B) \cap (B - A) \neq \emptyset$ is false, and so]_ __ (g) __. +a. $(A - B) \cap (B - A)$ + +b. intersection + +c. $B - A$ + +d. $B$ + +e. $A$ + +f. $A$ + +g. $(A - B) \cap (B - A) = \emptyset$ + Use the element method for proving a set equals the empty set to prove each statement in 28-38. Assume that all sets are subsets of a universal set $U$. 28. For all sets $A$ and $B$, $(A \cap B) \cap (A \cap B^c) = \emptyset$. (This property is used in Section 9.9.) +**Proof (by contradiction):** + +Let $A$ and $B$ be any sets and suppose +$(A \cap B) \cap (A \cap B^c) \neq \emptyset$. + +Suppose $x$ is some element such that $x \in (A \cap B) \cap (A \cap B^c)$. + +By the definition of $\cap$, this means that $x \in (A \cap B)$ and +$x \in (A \cap B^c)$. + +Since $x \in (A \cap B)$, this means that $x \in A$ and $x \in B$. + +Since $x \in (A \cap B^c)$, this means that $x \in A$ and $x \notin B$. + +So $x \in B$ and $x \notin B$, which is a contradiction. + +Hence the supposition is false, and therefore +$(A \cap B) \cap (A \cap B^c) = \emptyset$. + +Q.E.D. + 29. For all sets $A$, $B$, and $C$, $$ (A - C) \cap (B - C) \cap (A - B) = \emptyset $$ +**Proof (by contradiction):** + +Let $A$, $B$, and $C$ be any sets, and suppose +$(A - C) \cap (B - C) \cap (A - B) \neq \emptyset$. + +Suppose $x$ is some element such that $x \in (A - C) \cap (B - C) \cap (A - B)$. + +By the definition of $\cap$, this means that $x \in (A - C)$ and $x \in (B - C)$ +and $x \in (A - B)$. + +By the definition of difference, this means that $x \in A$ and $x \notin C$ and +$x \in B$ and $x \notin C$ and $x \in A$ and $x \notin B$. + +So $x \in B$ and $x \notin B$, which is a contradiction. + +Hence the supposition is false, and therefore +$(A - C) \cap (B - C) \cap (A - B) = \emptyset$. + +Q.E.D. + 30. For every subset $A$ of a universal set $U$, $A \cap A^c = \emptyset$. +**Proof (by contradiction):** + +Let $A$ be any set and suppose $A \cap A^c \neq \emptyset$. + +Suppose $x$ is some element such that $x \in A \cap A^c$. + +By the definition of $\cap$, this means that $x \in A$ and $x \in A^c$. + +By the definition of complement, this means that $x \in A$ and $x \notin A$, +which is a contradiction. + +Hence the supposition is false, and therefore $A \cap A^c = \emptyset$. + +Q.E.D. + 31. If $U$ denotes a universal set, then $U^c = \emptyset$. +**Proof (by contradiction):** + +Let $U$ be the universal set of all elements, and suppose $U^c \neq \emptyset$. + +Suppose $x$ is some element such that $x \in U^c$. + +By definition of complement, this means that $x \notin U$. + +Since $U$ is the universal set of all elements, it follows that $x \in U$. + +So $x \notin U$ and $x \in U$, which is a contradiction. + +Hence the supposition is false, and therefore $U^c = \emptyset$. + +Q.E.D. + 32. For every set $A$, $A \times \emptyset = \emptyset$. +**Proof (by contradiction):** + +Let $A$ be any set and suppose $A \times \emptyset \neq \emptyset$. + +Suppose $(x, y)$ are any element pair such that $(x, y) \in A \times \emptyset$. + +By the definition of Cartesian product, this means that $x \in A$ and +$y \in emptyset$. By the definition of $\emptyset$, $y \notin \emptyset$. + +So $y \in \emptyset$ and $y \notin \emptyset$, which is a contradiction. + +Hence the supposition is false, and therefore $A \times \emptyset = \emptyset$. + +Q.E.D. + 33. 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 $A \subseteq B$. + +Suppose $A \cap B^c \neq \emptyset$. Then let $x$ be some 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 complement, this means that $x \in A$ and $x \notin B$. + +Since $x \in A$ and $A \subseteq B$, it follows that $x \in B$ by definition of +subset. + +So $x \notin B$ and $x \in B$, which is a contradiction. + +Hence the supposition is false, therefore $A \cap B^c = \emptyset$. + +Q.E.D. + 34. For all sets $A$ and $B$, if $B \subseteq A^c$ then $A \cap B = \emptyset$. +**Proof (by contradiction):** + +Let $A$ and $B$ be any sets such that $B \subseteq A^c$. + +Suppose $A \cap B \neq \emptyset$. Then let $x$ be some element such that +$x \in A \cap B$. + +By the definition of $\cap$, this means that $x \in A$ and $x \in B$. + +Since $x \in B$ and $B \subseteq A^c$, it follows that $x \notin A$. + +So $x \in A$ and $x \notin A$, which is a contradiction. + +Hence the supposition is false, and therefore $A \cap B = \emptyset$. + +Q.E.D. + 35. For all sets $A$, $B$, and $C$, if $A \subseteq B$ and $B \cap C = \emptyset$ then $A \cap C = \emptyset$. +**Proof (by contradiction):** + +Let $A$, $B$, and $C$ be any sets such that $A \subseteq B$ and +$B \cap C = \emptyset$. + +Suppose $A \cap C \neq \emptyset$, then let $x$ be some element such that +$x \in A \cap C$. + +By the definition of $\cap$, this means that $x \in A$ and $x \in C$. + +Since $x \in A$ and $A \subseteq B$, then $x \in B$ by definition of subset. + +Thus $x \in B$ and $x \in C$, which is, by definition of $\cap$, +$x \in B \cap C$. + +$B \cap C = \emptyset$, so $x \in emptyset$. + +But by the definition of $\emptyset$, $x \notin \emptyset$. + +So $x \in \emptyset$ and $x \notin \emptyset$, which is a contradiction. + +Hence the supposition is false, and therefore $A \cap C = \emptyset$. + +Q.E.D. + 36. For all sets $A$, $B$, and $C$, if $C \subseteq B - A$, then $A \cap C = \emptyset$. +**Proof (by contradiction):** + +Let $A$, $B$, and $C$ be any sets such that $C \subseteq B - A$. + +Suppose $A \cap C \neq \emptyset$, then let $x$ be some element such that +$x \in A \cap C$. + +By the definition of $\cap$, this means that $x \in A$ and $x \in C$. + +Since $x \in C$, and $C \subseteq B - A$, this means that $x \in B - A$. +Furthermore, by the definition of difference, this means that $x \in B$ and +$x \notin A$. + +So $x \in A$ and $x \notin A$, which is a contradiction. + +Hence the supposition is false, and therefore $A \cap C = \emptyset$. + +Q.E.D. + 37. For all sets $A$, $B$, and $C$, if $B \cap C \subseteq A$, then $(C - A) \cap (B - A) = \emptyset$. +**Proof (by contradiction):** + +Let $A$, $B$, and $C$ be any sets such that $B \cap C \subseteq A$. + +Suppose $(C - A) \cap (B - A) \neq \emptyset$, then let $x$ be some element such +that $x \in (C - A) \cap (B - A)$. + +By the definition of $\cap$, this means that $x \in (C - A)$ and +$x \in (B - A)$. + +By the definition of difference, this means that $x \in C$ and $x \notin A$ and +$x \in B$ and $x \notin A$. + +Since $x \in B$ and $x \in C$, this means that $x \in B \cap C$. + +$B \cap C \subseteq A$, so $x \in A$, by definition of subset. + +So $x \notin A$ and $x \in A$, which is a contradiction. + +Hence the supposition is false, and therefore +$(C - A) \cap (B - A) = \emptyset$. + 38. For all sets $A$, $B$, $C$, and $D$, if $A \cap C = \emptyset$ then $(A \times B) \cap (C \times D) = \emptyset$. +Omitted. + Prove each statement in 39-44. 39. For all sets $A$ and $B$, diff --git a/chapter_6/test_yourself.md b/chapter_6/test_yourself.md index 583dde5..617e090 100644 --- a/chapter_6/test_yourself.md +++ b/chapter_6/test_yourself.md @@ -62,18 +62,30 @@ Page 426 1. To prove that a set $X$ is a subset of a set $A \cap B$, you suppose that $x$ is any element of $X$ and you show that $x \in A$ _____ $x \in B$. +and + 2. To prove that a set $X$ is a subset of a set $A \cup B$, you suppose that $x$ is any element of $X$ and you show that $x \in A$ _____ $x \in B$. +or + 3. To prove that a set $A \cup B$ is a subset of a set $X$, you start with any element $x$ in $A \cup B$ and consider the two cases _____ and _____. You then show that in either case _____. +$x \in A$; $x \in B$; $x \in X$ + 4. To prove that a set $A \cap B$ is a subset of $X$, you suppose that _____ and you show that _____. +$x \in A \cap B$; $x \in X$ + 5. To prove that a set $X$ equals a set $Y$, you prove that _____ and that _____. +$X \subseteq Y$; $Y \subseteq X$ + 6. To prove that a set $X$ does not equal a set $Y$, you need to find an element that is in _____ and not _____ or that is in _____ and not _____. + +$X$; in $Y$; $Y$; in $X$