diff --git a/chapter_6/exercises.md b/chapter_6/exercises.md index cc254f0..968443f 100644 --- a/chapter_6/exercises.md +++ b/chapter_6/exercises.md @@ -2705,3 +2705,309 @@ Omitted. $$ \bigcap_{i = 1}^{n}(A \times B_i) = A \times \left(\bigcap_{i = 1}^{n}B_i\right) $$ Omitted. + +--- + +Page 435 + +**Exercise Set 6.3** + +For each of 1-4 find a counterexample to show that the statement is false. +Assume all sets are subsets of a universal set $U$. + +1. For all sets $A$, $B$, and $C$, + +$$ (A \cup B) \cap C = A \cup (B \cap C) $$ + +2. For all sets $A$ and $B$, $(A \cup B)^c = A^c \cup B^c$. + +3. For all sets $A$, $B$, and $C$, if $A \nsubseteq B$ and $B \nsubseteq C$ then + $A \nsubseteq C$. + +4. For all sets $A$, $B$, and $C$, if $B \cup C \subseteq A$ then + +$$ (A - B) \cap (A - C) = \emptyset $$ + +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 +$U$. + +5. For all sets $A$, $B$, and $C$, + +$$ A - (B - C) = (A - B) - C $$ + +6. For all sets $A$ and $B$, $A \cap (A \cup B) = A$. + +7. For all sets $A$, $B$, and $C$, + +$$ (A - B) \cap (C - B) = A - (B \cup C) $$ + +8. For all sets $A$ and $B$, if $A^c \subseteq B$ then $A \cup B = U$. + +9. For all sets $A$ ,$B$, and $C$, if $A \subseteq C$ and $B \subseteq C$ then + $A \cup B \subseteq C$. + +10. For all sets $A$ and $B$, if $A \subseteq B$ then $A \cap B^c = \emptyset$. + +11. For all sets $A$, $B$, and $C$, if $A \subseteq B$ then + $A \cap (B \cap C)^c = \emptyset$. + +12. For all sets $A$, $B$, and $C$, + +$$ A \cap (B - C) = (A \cap B) - (A \cap C) $$ + +13. For all sets $A$, $B$, and $C$, + +$$ A \cup (B - C) = (A \cup B) - (A \cup C) $$ + +14. For all sets $A$, $B$, and $C$, if $A \cap C = B \cap C$ and + $A \cup C = B \cup C$, then $A = B$. + +15. For all sets $A$, $B$, and $C$, $(A - B) \cup C \subseteq A \cup (C - B)$. + +16. For all sets $A$ and $B$, if $A \cap B = \emptyset$ then + $A \times B = \emptyset$. + +17. For all sets $A$ and $B$, if $A \subseteq B$ then + $\mathscr{P}(A) \subseteq \mathscr{P}(B)$. + +18. For all sets $A$ and $B$, + $\mathscr{P}(A \cup B) \subseteq \mathscr{P}(A) \cup \mathscr{P}(B)$. + +19. For all sets $A$ and $B$, + $\mathscr{P}(A) \cup \mathscr{P}(B) \subseteq \mathscr{P}(A \cup B)$. + +20. For all sets $A$ and $B$, + $\mathscr{P}(A \cap B) = \mathscr{P}(A) \cap \mathscr{P}(B)$. + +21. For all sets $A$ and $B$, + $\mathscr{P}(A \times B) = \mathscr{P}(A) \times \mathscr{P}(B)$. + +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 $S \cap T = \emptyset$. + +b. $\exists$ a set $S$ such that $\forall$ sets $T$, $S \cup T = \emptyset$. + +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 + $\mathscr{P}(S)$? + +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 $S_1$. + +b. Find $S_2$. + +c. Are $S_1$ and $S_2$ disjoint? + +d. Compare the sizes of $S_1$ and $S_2$. + +e. How many elements are in $S_1 \cup S_2$? + +f. What is the relation between $S_1 \cup S_2$ and $\mathscr{P}(A)$? + +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. + +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 + subsets of $\{2, 3, \dots, n\}$. For each $S_i \in S$, let $P_i$ be the + product of the elements of $S_i$. Prove or disprove that + +$$ \sum_{i = 1}^{]2^{n - 1} - 1}{P_i} = \frac{(n + 1)!}{2} - 1 $$ + +In 27 and 28 supply a reason for each step in the derivationl.l + +27. For all sets $A$, $B$, and $C$, + +$$ (A \cup B) \cap C = (A \cap C) \cup (B \cap C) $ + +_Proof:_ + +Suppose $A$, $B$, and $C$ are any sets. Then + +$$ (A \cup B) \cap C = C \cup (A \cup B) $$ + +by __ (a) __ + +$$ = (C \cap A) \cup (C \cap B) $$ + +by __ (b) __ + +$$ = (A \cap C) \cup (B \cap C) $$ + +by __ \(c\) __ + +28. For all sets $A$, $B$, and $C$, + +$$ (A \cup B) - (C - A) = A \cup (B - C) $$ + +_Proof:_ + +Suppose $A$, $B$, and $C$ are any sets. Then + +$$ (A \cup B) - (C - A) = (A \cup B) \cap (C - A)^c $$ + +by __ (a) __ + +$$ = (A \cup B) \cap (C \cap A^c)^c $$ + +by __ (b) __ + +$$ = (A \cup B) \cap (A^c \cap C)^c $$ + +by __ \(c\) __ + +$$ = (A \cup B) \cap ((A^c)^c \cup C^c) $$ + +by __ (d) __ + +$$ = (A \cup B) \cap (A \cup C^c) $$ + +by __ (e) __ + +$$ = A \cup (B \cap C^c) $$ + +by __ (f) __ + +$$ = A \cup (B - C) $$ + +by __ (g) __ + +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. + +**Proof:** + +Let any sets $A$ and $B$ be given. Then + +$$ (A \cup C^c) - B = (A \cup B^c) \cap B^c $$ + +by the set difference 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 $\cup$ + +$$ (A - B) \cup B^c $$ + +by the set difference law. + +In 30-40, construct an algebraic proof for the given statement. Cite a property +from Theorem 6.2.2 for every step. + +30. For all sets $A$, $B$, and $C$, + +$$ (A \cap B) \cup C = (A \cup C) \cap (B \cup C) $$ + +31. For all sets $A$ and $B$, $A \cup (B - A) = A \cup B$. + +32. For all sets $A$ and $B$, $(A - B) \cup (A \cap B) = A$. + +33. Fora ll sets $A$ and $B$, $(A - B) \cap (A \cap B) = \emptyset$. + +34. For all sets $A$, $B$, and $C$, + +$$ (A - B) - C = A - (B \cup C) $$ + +35. For all sets $A$ and $B$, $A - (A - B) = A \cap B$. + +36. For all sets $A$ and $B$, $((A^c \cup B^c) - A)^c = A$. + +37. For all sets $A$ and $B$, $(B^c \cup (B^c - A))^c = B$. + +38. For all sets $A$ and $B$, $(A \cap B)^c \cap A = A - B$. + +39. For all sets $A$ and $B$, + +$$ (A - B) \cup (B - A) = (A \cup B) - (A \cap B) $$ + +40. For all sets $A$, $B$, and $C$, + +$$ (A - B) - (B - C) = A - B $$ + +In 41-43 simplify the given expression. Cite a property from Theorem 6.2.2 for +every step. + +41. $A \cap ((B \cup A^c) \cap B^c)$ + +42. $(A - (A \cap B)) \cap (B - (A \cap B))$ + +43. $((A \cap (B \cup C)) \cap (A - B)) \cap (B \cup C^c)$ + +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. + +b. Use an algebraic argument to derive the property (by applying properties from +Theorem 6.2.2). + +c. Comment on which method you found easier. + +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. + +b. Use an algebraic argument to derive the property (by applying properties from +Theorem 6.2.2). + +c. Comment on which method you found easier. + +**Definition:** + +Given sets $A$ and $B$, the **symmetric difference of $A$ and $B$**, denoted +$A \Delta B$, is, is + +$$ A \Delta B = (A - B) \cup (B - A) $$ + +46. Let $A = \{1, 2, 3, 4\}$, $B = \{3, 4, 5, 6\}$, and $C = \{5, 6, 7, 8\}$. + Find each of the following sets: + +a. $A \Delta B$ + +b. $B \Delta C$ + +c. $A \Delta C$ + +d. $(A \Delta B) \Delta C$ + +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. $A \Delta B = B \Delta A$ + +48. $A \Delta \emptyset = A$ + +49. $A \Delta A^c = U$ + +50. $A \Delta A = \emptyset$ + +51. If $A \Delta C = B \Dcelta C$, then $A = B$. + +52. $(A \Delta B) \Delta C = A \Delta (B \Delta C)$ + +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. + +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. diff --git a/chapter_6/notes.md b/chapter_6/notes.md index c70365d..88310e1 100644 --- a/chapter_6/notes.md +++ b/chapter_6/notes.md @@ -432,3 +432,68 @@ 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]_. + +--- + +Page 433 + +**Theorem 6.3.1** + +For every integer $n \geq 0$, if a set $X$ has $n$ elements, then +$\mathscr{P}(X)$ has $2^n$ elements. + +**Proof (by mathematical induction):** + +Let the property $P(n)$ be the sentence + +Any set with $n$ elements has $2^n$ subsets. + +_Show that $P(0)$ is true:_ + +To establish $P(0)$, we must show that + +Any set with $0$ elements has $2^0$ subsets. + +Now the only set with zero elements is the empty set, and the only subset of the +empty set is itself. Thus a set with zero elements has one subset. Since +$1 = 2^0$, we have that $P(0)$ is true. + +_Show that for every integer $k \geq 0$, if $P(k)$ is true then $P(k + 1)$ is +also true:_ + +_[Suppose that $P(k)$ is true for a particular but arbitrarily chosen integer +$k \geq 0$. That is:]_ + +Suppose that $k$ is any integer with $k \geq 0$ such that + +Any set with $k$ elements has $2^k$ subsets. + +_[We must show that $P(k + 1)$ is true. That is:]_ + +We must show that + +Any set with $k + 1$ elements has $2^{k + 1}$ subsets. + +Let $X$ be a set with $k + 1$ elements. Since $k + 1 \geq 1$, we may pick an +element $z$ in $X$. Observe that any subset of $X$ either contains $z$ or does +not. Furthermore, any subset of $X$ that does not contain $z$ is a subset of +$X - \{z\}$. And any subset $A$ of $X - \{z\}$ can be matched up with a subset +$B$, equal to $A \cup \{z\}$, of $X$ that contains $z$. Consequently, there are +as many subsets of $X$ that contain $z$ as do not, and thus there are twice as +many subsets of $X$ as there are subsets of $X - \{z\}$. It follows that since +$X - \{z\}$ has $k$ elements, then, by inductive hypothesis, + +the number of subsets of $X - \{z\} = 2^k$ + +Therefore, + +the number of subsets $X = 2 \cdot (\text{the number of subsets of } X - \{z\})$ + +$$ = 2 \cdot (2^k) $$ + +$$ = 2^{k + 1} $$ + +_[This is what was to be shown.]_ + +_[Since we have proved both the basis step and the inductive step, we conclude +that the theorem is true.]_ diff --git a/chapter_6/test_yourself.md b/chapter_6/test_yourself.md index 617e090..fba5496 100644 --- a/chapter_6/test_yourself.md +++ b/chapter_6/test_yourself.md @@ -89,3 +89,20 @@ $X \subseteq Y$; $Y \subseteq X$ that is in _____ and not _____ or that is in _____ and not _____. $X$; in $Y$; $Y$; in $X$ + +--- + +Page 435 + +**Test Yourself** + +1. Given a proposed set identity involving set variables $A$, $B$, and $C$, the + most common way to show that the equation does not hold in general is to find + concrete sets $A$, $B$, and $C$ that, when substituted for the set variables + in the equation, _____. + +2. When using the algebraic method for proving a set identity, it is important + to _____ for every step. + +3. When applying a property from Theorem 6.2.2, it must be used _____ as it is + stated.