From 690a53516410e58bd13816fe1b1dcfb54238bfc0 Mon Sep 17 00:00:00 2001 From: tomit4 Date: Sat, 18 Jul 2026 14:36:32 -0700 Subject: [PATCH] :construction: Setup for 6.2 --- chapter_6/exercises.md | 306 +++++++++++++++++++++++++++++++++++++ chapter_6/notes.md | 262 +++++++++++++++++++++++++++++++ chapter_6/test_yourself.md | 25 +++ 3 files changed, 593 insertions(+) diff --git a/chapter_6/exercises.md b/chapter_6/exercises.md index 40decce..f1d0e53 100644 --- a/chapter_6/exercises.md +++ b/chapter_6/exercises.md @@ -1390,3 +1390,309 @@ Omitted. given set that is represented as the array $a[1], a[2], \dots, a[n]$. Omitted. + +--- + +Page 427 + +**Exercise Set 6.2** + +1. + +a. To say that an element is in $A \cap (B \cup C)$ means that it is in __ (1) +__ and in __ (2) __. + +b. To say that an element is in $(A \cap B) \cup C$ means that it is in __ (1) +__ or in __ (2) __. + +c. To say that an element is in $A - (B \cap C)$ means that it is in __ (1) __ +and not in __ (2)__. + +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) __. + +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) __. + +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. + +a. **Proof:** Suppose $A$ and $B$ are any sets. To show that +$A - B \subseteq A$, 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 $A$. + +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. + +3. **Theorem:** For all sets $A$, $B$, and $C$, if $A \subseteq C$, + $B \subseteq C$, then $A \subseteq C$. + +**Proof:** + +| Statement | Explanation | +| ------------------------------------------------------------------------------------ | ------------------------------------- | +| Suppose $A$, $B$, and $C$ are any sets such that $A \subseteq B$ and $B \subseteq C$ | starting point | +| We must show that $A \subseteq C$. | conclusion to be shown | +| Let $x$ be any element in $A$. | start of an element proof | +| Then $x$ is in $B$. | __ (a) __ | +| It follows that $x$ is in $C$. | __ (b) __ | +| 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]._ | __ \(c\) __ | + +4. **Theorem:** For all sets $A$ and $B$, if $A \subseteq B$, then + $A \cup B \subseteq B$. + +**Proof:** + +| Statement | Explanation | +| ----------------------------------------------------------------- | -------------------------------------------- | +| Suppose $A$, $B$, and $C$ are any sets such that $A \subseteq B$. | starting point | +| We must show that $A \cup B \subseteq B$ | conclusion to be shown | +| Let $x$ be any element in $A \cup B$. | start of an element proof | +| Then $x$ is in $A$ or $x$ is in $B$. | __ (a) __ | +| In case $x$ is in $A$, then $x$ is in $B$ | __ (b) __ | +| In case $x$ is in $B$, then $x$ is in $B$. | tautology ($p \to p$) | +| So in either case $x$ is in $B$. | proof by division into cases | +| 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]._ | __ \(c\) __ | + +5. Prove that for all sets $A$ and $B$, $(B - A) = B \cap A^c$. + +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 $C$, $A \cap (B \cup C) = (A \cap C) \cup (A \cap C)$. + +**Proof:** Suppose $A$, $B$, and $C$ are any sets. + +(1) Proof that $A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)$: + +Let $x \in A \cap (B \cup C)$. _[We must show that $x \in$ __ (a) __ ]._ + +By definition of $\cap$, $x \in$ __ (b) __ and $x \in B \cup C$. + +Thus $x \in A$ and, by definition of $\cup$, $x \in B$ or __ \(c\) __. + +_Case 1 $(x \in A \text{ and } x \in B)$:_ In this case, $x \in A \cap B$ by +definition of $\cap$. + +_Case 2 $(x \in A \text{ and } x \in C)$:_ IN this case, $x \in A \cap C$ by +definition of $\cap$. + +By cases 1 and 2, $x \in A \cap B$ or $x \in A \cap C$, and so, by definition of +$\cup$, __ (d) __. + +_[So $A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)$ by definition of +subset.]_ + +(2) Proof that $(A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)$: + +Let $x \in (A \cap B) \cup (A \cap C)$. _[We must show that +$x \in A \cap (B \cup C)$.]_ + +By definition of $\cup$, $x \in A \cap B$ __ (a) __ $x \in A \cap C$. + +_Case 1 $(x \in A \cap B)$:_ In this case, by definition of $\cap$, $x \in A$ +and $x \in B$$. + +Since $x \in B$, then $x \in B \cup C$ by definition of $\cup$. + +_Case 2 $(x \in A \cap C)$:_ In this case, by definition of $\cap$, $x \in A$ __ +(b) __ $x \in C$. + +Since $x \in C$, then $x \in B \cup C$ by definition of $\cup$. + +In both cases $x \in A$ and $$ix \in B \cup C, and so, by definition of $\cap$, +__ \(c\) __. + +_[So $(A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)$ by definition of +__ (d) __ .]_ + +(3) Conclusion: _[Since both subset relations have been proved, it follows, by +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 $U$. + +7. For all sets $A$ and $B$, $(A \cap B)^c = A^c \cup B^c$. + +8. For all sets $A$ and $B$, $(A \cap B) \cup (A \cap B^c) = A$. + +(This property is used in Section 9.9.) + +9. For all sets $A$, $B$, and $C$, + +$$ (A - B) \cup (C - B) = (A \cup C) - B $$ + +10. For all sets $A$, $B$, and $C$, + +$$ (A \cup B) \cap C \subseteq A \cup (B \cap C) $$ + +11. For all sets $A$, $B$, and $C$, + +$$ A \cap (B - C) \subseteq (A \cap B) - (A \cap C) $$ + +12. For all sets $A$, $B$, and $C$, + +$$ (A \cup B) - C \subseteq (A - C) \cup (B - C) $$ + +13. For all sets $A$, $B$, and $C$, + +$$ (A - B) \cap (C - B) = (A \cap C) - B $$ + +14. For all sets $A$ and $B$, $A \cup (A \cap B) = A$. + +15. For every set $A$, $A \cup \emptyset = A$. + +16. For all sets $A$, $B$, and $C$, if $A \subseteq B$ then + $A \cap C \subseteq B \cap C$. + +17. For all sets $A$, $B$, and $C$, if $A \subseteq B$ then + $A \cup C \subseteq B \cup C$. + +18. For all sets $A$ and $B$, if $A \subseteq B$ then $B^c \subseteq A^c$. + +19. For all sets $A$, $B$, and $C$, if $A \subseteq B$ and $A \subseteq C$ then + $A \subseteq B \cap C$. + +20. For all sets $A$, $B$, and $C$, if $A \subseteq C$ and $B \subseteq C$ then + $A \cup B \subseteq C$. + +21. For all sets $A$, $B$, and $C$, + +$$ A \times (B \cup C) = (A \times B) \cup (A \times C) $$ + +22. For all sets $A$, $B$, and $C$, + +$$ A \times (B \cap C) = (A \times B) \cap (A \times C) $$ + +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 $A \subseteq C$. + +**Proof:** Suppose $A$, $B$, and $C$ are any sets such that $A \subseteq B$ and +$B \subseteq C$. Since $A \subseteq B$, there is an element $x$ such that +$x \in A$ 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 $A \subseteq C$. + +24. Find the mistake in the following "proof." + +**Theorem:** For all sets $A$ and $B$, $A^c \cup B^c \subseteq (A \cup B)^c^c$ + +**Proof:** Suppose $A$ and $B$ are any sets, and $x \in A^c \cup B^c$. Then +$x \in A^c$ or $x \in B^c$ by definition of union. It follows that $x \notin A$ +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 $A^c \cup B^c \subseteq (A \cup B)^c$. + +25. Find the mistake in the following "proof" that for all sets $A$ and $B$, + $(A - B) \cup (A \cap B) \subseteq A$. + +**Proof:** Suppose $A$ and $B$ are any sets, and suppose +$x \in (A - B) \cup (A \cap B)$. 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. + +26. Consider the Venn diagram below. + +(See page 429 for image.) + +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. + +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. + +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.) + +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.) + +27. Fill in the blanks in the following proof that for all sets $A$ and $B$, + $(A - B) \cap (B - A) = \emptyset$. + +**Proof:** + +Let $A$ and $B$ be any sets and suppose $(A - B) \cap (B - A) \neq \emptyset$. +That is, suppose there is an element $x$ in __ (a) __. BY definition of __ (b) +__, $x \in A - B$ and $x \in$ __ \(c\) __. Then by definition of set difference, +$x \in A$ and $x \notin B$ and $x \in$ __ (d) __ and $x \notin$ __ (e) __. IN +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) __. + +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.) + +29. For all sets $A$, $B$, and $C$, + +$$ (A - C) \cap (B - C) \cap (A - B) = \emptyset $$ + +30. For every subset $A$ of a universal set $U$, $A \cap A^c = \emptyset$. + +31. If $U$ denotes a universal set, then $U^c = \emptyset$. + +32. For every set $A$, $A \times \emptyset = \emptyset$. + +33. For all sets $A$ and $B$, if $A \subseteq B$ then $A \cap B^c = \emptyset$. + +34. For all sets $A$ and $B$, if $B \subseteq A^c$ then $A \cap B = \emptyset$. + +35. For all sets $A$, $B$, and $C$, if $A \subseteq B$ and + $B \cap C = \emptyset$ then $A \cap C = \emptyset$. + +36. For all sets $A$, $B$, and $C$, if $C \subseteq B - A$, then + $A \cap C = \emptyset$. + +37. For all sets $A$, $B$, and $C$, if $B \cap C \subseteq A$, then + $(C - A) \cap (B - A) = \emptyset$. + +38. For all sets $A$, $B$, $C$, and $D$, if $A \cap C = \emptyset$ then + $(A \times B) \cap (C \times D) = \emptyset$. + +Prove each statement in 39-44. + +39. For all sets $A$ and $B$, + +a. $(A - B) \cup (B - A) \cup (A \cap B) = A \cup B$ + +b. The sets $(A - B)$, $(B - A)$, and $(A \cap B)$ are mutually disjoint. + +40. For every positive integer $n$, if $A$ and $B_1, B_2, B_3, \dots$ are any + sets, then + +$$ A \cap \left(\bigcup_{i = 1}^{n}B_i\right) = \bigcup_{i = 1}^{n}(A \cap B_i) $$ + +41. For every positive integer $n$, if $A_1, A_2, A_3, \dots$ and $B$ are any + sets, then + +$$ \bigcap_{i = 1}^{n}(A_i - B) = \left(\bigcup_{i = 1}^{n}A_i\right) - B $$ + +42. For every positive integer $n$, if $A_1, A_2, A_3, \dots$ and $B$ are any + sets, then + +$$ \bigcap_{i = 1}^{n}(A_i - B) = \left(\bigcap_{i = 1}^{n}A_i\right) - B $$ + +43. For every positive integer $n$, if $A$ and $B_1, B_2, B_3, \dots$ are any + sets, then + +$$ \bigcup_{i = 1}^{n}(A \times B_i) = A \times \left(\bigcup_{i = 1}^{n}B_i\right) $$ + +44. For every positive integer $n$, if $A$ and $B_1, B_2, B_3, \dots$ are any + sets, then + +$$ \bigcap_{i = 1}^{n}(A \times B_i) = A \times \left(\bigcap_{i = 1}^{n}B_i\right) $$ diff --git a/chapter_6/notes.md b/chapter_6/notes.md index 51ebba6..c70365d 100644 --- a/chapter_6/notes.md +++ b/chapter_6/notes.md @@ -170,3 +170,265 @@ $b[1], b[2], \dots, b[n]$ [a one-dimensional array representing the set $B$]_ $i := 1, \text{answer} := A \subseteq B\\ \text{\textbf{while}} (i \leq m \text{ and answer } = A \subseteq B )\\ \ \ j := 1, \text{found} := \text{"no"}\\ \ \ \text{\textbf{while }} (j \neq n \text{ and } \text{found}= \text{"no"})\\ \ \ \ \ \text{\textbf{if }} a[i] = b[j] \text{\textbf{ then }} \text{found} := \text{"yes"}\\ \ \ \ \ j := j + 1\\ \ \ \text{\textbf{end while}}\\ \ \ \text{[If found has not been given the value "yes" when execution reaches this point, then } a[i] \neq B\text{ .]}\\ \ \ \text{\textbf{if }} \text{found} = \text{"no"} \text{\textbf{ then }} \text{answer} := A \nsubseteq B\\ \ \ i := i + 1\\ \text{\textbf{end while}}$ **Output:** _answer [a string]_ + +--- + +Page 414 + +**Theorem 6.2.1 Some Subset Relations** + +1. _Inclusion of Intersection:_ For all sets $A$ and $B$, + +$$ \text{(a) } A \cap B \subseteq A \quad \text{ and } \quad \text{ (b) } A \cap B \subseteq B $$ + +2. _Inclusion in Union:_ For all sets $A$ and $B$, + +$$ \text{(a) } A \subseteq A \cup B \quad \text{ and } \quad \text{ (b) } B \subseteq A \cup B $$ + +3. _Transitive Property of Subsets:_ For all sets $A$, $B$, $C$, + +$$ \text{if } A \subseteq B \text{ and } B \subseteq C \text{, then } A \subseteq C $$ + +--- + +Page 415 + +**Procedural Versions of Set Definitions** + +Let $X$ and $Y$ be subsets of a universal set $U$ and suppose $x$ and $y$ are +elements of $U$. + +1. $x \in X \cup Y \Leftrightarrow x \in X \text{ or } x \in Y$ + +2. $x \in X \cap Y \Leftrightarrow x \in X \text{ and } x \in Y$ + +3. $x \in X - Y \Leftrightarrow x \in X \text{ and } x \notin Y$ + +4. $x \in X^c \Leftrightarrow x \notin X$ + +5. $(x, y) \in X \times Y \Leftrightarrow x \in X \text{ and } y \in Y$ + +--- + +Page 417 + +**Theorem 6.2.2 Set Identities** + +Let all sets referred to below be subsets of a universal set $U$. + +1. _Commutative Laws:_ For all sets $A$ and $B$, + +$$ \text{(a) } A \cup B = B \cup A \quad \text{ and } \quad \text{ (b) } A \cap B = B \cap A $$ + +2. _Associative Laws:_ For all sets $A$, $B$, and $C$, + +$$ \text{(a) } (A \cup B) \cup C = A \cup (B \cup C) \quad \text{ and } \quad \text{ (b) } (A \cap B) \cap C = A \cap (B \cap C) $$ + +3. _Distributive Laws:_ For all sets $A$, $B$, and $C$, + +$$ \text{(a) } A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \quad \text{ and } \quad \text{ (b) } A \cap (B \cup C) = (A \cap B) \cup (A \cap C) $$ + +4. _Identity Laws:_ For every set $A$, + +$$ \text{(a) } A \cup \emptyset = A \quad \text{ and } \quad \text{ (b) } A \cap U = A $$ + +5. _Complement Laws:_ For every set $A$, + +$$ \text{(a) } A \cup A^c = U \quad \text{ and } \quad A \cap A^c = \emptyset $$ + +6. _Double Complement Law:_ For every set $A$, + +$$ (A^c)^c = A $$ + +7. _Idempotent Laws:_ For every set $A$, + +$$ \text{(a) } A \cup A = A \quad \text{ and } \quad \text{ (b) } A \cap A = A $$ + +8. _Universal Bound Laws:_ For every set $A$, + +$$ \text{(a) } A \cup U = U \quad \text{ and } \quad \text{ (b) } A \cap \emptyset = \emptyset $$ + +9. _De Morgan's Laws:_ For all sets $A$ and $B$, + +$$ \text{(a) } (A \cup B)^c = A^c \cap B^c \quad \text{ and } \quad \text{ (b) } (A \cap B)^c = A^c \cup B^c $$ + +10. _Absorption Laws:_ For all sets $A$ and $B$, + +$$ \text{(a) } A \cup (A \cap B) = A \quad \text{ and } \quad \text{ (b) } A \cap (A \cup B) = A $$ + +11. _Complements of $U$ and $\emptyset$:_ + +$$ \text{(a) } U^c = \emptyset \quad \text{ and } \quad \text{ (b) } \emptyset^c = U $$ + +12. _Set Difference Law:_ For all sets $A$ and $B$, + +$$ A - B = A \cap B^c $$ + +--- + +Page 418 + +**Basic Method for Proving That Sets Are Equal** + +Let sets $X$ and $Y$ be given. To prove that $X = Y$: + +1. Prove that $X \subseteq Y$. + +2. Prove that $Y \subseteq X$. + +--- + +Page 420 + +**Theorem 6.2.2(3)(a) A Distributive Law for Sets** + +(Too lengthy, see page 420) + +--- + +Page 422 + +**Theorem 6.2.2(9)(a) A De Morgan's Law for Sets** + +For all sets $A$ and $B$, $(A \cup B)^c = A^c \cap B^c$. + +**Proof:** Suppose $A$ and $B$ are sets. + +_Proof that $(A \cup B)^c \subseteq A^c \cap B^c$:_ + +_[We must show that +$\forall x, \text{ if } x \in (A \cup B)^c \text{ then } x \in A^c \cap B^c$.]_ + +Suppose $x \in (A \cup B)^c$. _[We must show that $x \in A^c \cap B^c$.]_ By +definition of complement, + +$$ x \notin A \cup B $$ + +Now to say that $x \notin A \cup B$ means that + +it is false that ($x$ is in $A$ or $x$ is in $B$). + +By De Morgan's laws of logic, this implies that + +$x$ is not in $A$ and $x$ is not in $B$, + +which can be written + +$$ x \notin A \quad \text{ and } \quad x \notin B $$ + +Hence $x \in A^c$ and $x \in B^c$ by definition of complement. It follows, by +definition of intersection, that $x \in A^c \cap B^c$ _[as was to be shown]._ So +$(A \cup B)^c \subseteq A^c \cap B^c$ by definition of subset. + +_Proof that $A^c \cap B^c \subseteq (A \cup B)^c$:_ + +_[We must show that +$\forall x, \text{ if } x \in A^c \cap B^c \text{ then } x \in (A \cup B)^c$.]_ + +Suppose $x \in A^c \cap B^c$. _[We must show that $x \in (A \cup B)^c$.]_ By +definition of intersection, $x \in A^c$ and $x \in B^c$, and by definition of +complement, + +$$ x \notin A \quad \text{ and } \quad x \notin B $$ + +In other words, + +$x$ is not in $A$ and $x$ is not in $B$. + +By De Morgan's laws of logic this implies that + +it is false that ($x$ is in $A$ or $x$ is in $B$), + +which can be written + +$$ x \notin A \cup B $$ + +by definition of union. Hence, by definition of complement, $x \in (A \cup B)^c$ +_[as was to be shown]._ It follows that $A^c \cap B^c \subseteq (A \cup B)^c$ by +definition of subset. + +_Conclusion:_ Since both set containments have been proved, +$(A \cup B)^c = A^c \cap B^c$ by definition of set equality. + +--- + +Page 423 + +**Theorem 6.2.3 Intersection and Union with a Subset** + +For any sets $A$ and $B$, if $A \subseteq B$, then + +$$ \text{(a) } A \cap B = A \quad \text{ and } \quad \text{ (b) } A \cup B = B $$ + +**Proof:** + +_Part (a):_ Suppose $A$ and $B$ are sets with $A \subseteq B$. To show part (a) +we must show both that $A \cap B \subseteq A$ and that $A \subseteq A \cap B$. +We already know that $A \cap B \subseteq A$ by the inclusion of intersection +property. To show that $A \subseteq A \cap B$, let $x$ be any element in $A$. +_[We must show that $x$ is in $A \cap B$.]_ But, because of the hypothesis that +$A \subseteq B$, we can conclude that $x$ is also in $B$ by definition of +subset. Hence + +$$ x \in A \quad \text{ and } x \in B $$ + +and thus + +$$ x \in A \cap B $$ + +by definition of intersection _[as was to be shown]._ + +**Proof:** + +_Part (b):_ The proof of part (b) is left as an exercise. + +--- + +Page 424 + +**Theorem 6.2.4 A Set with No Elements Is a Subset of Every Set** + +If $E$ is a set with no elements and $A$ is any set, then $E \subseteq A$. + +**Proof (by contradiction):** + +Suppose not. _[We take the negation of the theorem and suppose it to be true.]_ +Suppose there exists a set $E$ with no elements and a set $A$ such that +$E \nsubseteq A$. _[We must deduce a contradiction.]_ Then there would be an +element of $E$ that is not an element of $A$ _[by definition of subset]_. But +there can be no such element since $E$ has no elements. This is a contradiction. +_[Hence the supposition that there are sets $E$ and $A$, where $E$ has no +elements and $E \nsubseteq A$, is false, and so the theorem is true.]_ + +--- + +Page 424 + +**Corollary 6.2.5 Uniqueness of the Empty Set** + +There is only one set with no elements. + +**Proof:** Suppose $E_1$ and $E_2$ are both sets with no elements. By Theorem +6.2.4, $E_1 \subseteq E_2$ since $E_1$ has no elements. Also $E_2 \subseteq E_1$ +since $E_2$ has no elements. Thus $E_1 = E_2$ by definition of set equality. + +--- + +Page 425 + +**Proposition 6.2.6** + +For all sets $A$, $B$, and $C$, if $A \subseteq B$ and $B \subseteq C^c$, then +$A \cap C = \emptyset$. + +**Proof:** + +Suppose $A$, $B$, and $C$ are sets such that $A \subseteq B$ and +$B \subseteq C^c$. We must show that $A \cap C = \emptyset$. Suppose not. That +is, suppose there is an element $x$ in $A \cap C$. By definition of +intersection, $x \in A$ and $x \in C$. Then, since $A \subseteq B$, $x \in B$ by +definition of subset. Also, since $B \subseteq C^c$, then $x \in C^c$ by +definition of subset again. It follows by definition of complement that +$x \notin C$. Thus $x \in C$ and $x \notin C$, which is a contradiction. So the +supposition that there is an element $x$ in $A \cap C$ is false, and thus +$A \cap C = \emptyset$ _[as was to be shown]_. diff --git a/chapter_6/test_yourself.md b/chapter_6/test_yourself.md index 0c32179..583dde5 100644 --- a/chapter_6/test_yourself.md +++ b/chapter_6/test_yourself.md @@ -52,3 +52,28 @@ all $A_i$ are a subset of $A$, but are also disjoint. $A$ is the union of all the sets $A_1, A_2, A_3, \dots$ and $A_i \cap A_j = \emptyset$ whenever $i \neq j$. + +--- + +Page 426 + +**Test Yourself** + +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 $x \in A$ _____ $x \in B$. + +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 $x \in A$ _____ $x \in B$. + +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 _____. + +4. To prove that a set $A \cap B$ is a subset of $X$, you suppose that _____ and + you show that _____. + +5. To prove that a set $X$ equals a set $Y$, you prove that _____ and that + _____. + +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 _____.