Exam 6: Set Theory

Consider the statement $\text { For all sets } A \text { and } B , ( A - B ) \cap B = \emptyset \text {. }$ Complete the proof begun below in which the given statement is derived algebraically from the properties listed in Theorem 6.2.2. Be sure to give a reason for every step that exactly justiﬁes what was done in the step: Proof: Let $A$ and $B$ be any sets. Then the left-hand side of the equation to be shown is $Consider the statement \text { For all sets } A \text { and } B , ( A - B ) \cap B = \emptyset \text {. } Complete the proof begun below in which the given statement is derived algebraically from the properties listed in Theorem 6.2.2. Be sure to give a reason for every step that exactly justiﬁes what was done in the step: Proof: Let A and B be any sets. Then the left-hand side of the equation to be shown is which is the right-hand side of the equation to be shown. [Hence the given statement is true.] (The number of lines in the outline shown above works for one version of a proof. If you write a proof using more or fewer lines, be sure to follow the given format, supplying a reason for every step that exactly justiﬁes what was done in the step.)$ which is the right-hand side of the equation to be shown. [Hence the given statement is true.] (The number of lines in the outline shown above works for one version of a proof. If you write a proof using more or fewer lines, be sure to follow the given format, supplying a reason for every step that exactly justiﬁes what was done in the step.)

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Question 1

Correct Answer:

Verified

Proof:
Let $A$ and $B$ be any sets. Then the left-hand side of the equation to be shown is
$\begin{aligned}( A - B ) \cap B & = \left( A \cap B ^ { c } \right) \cap B & \text { by the set difference law } \\& = A \cap \left( B ^ { c } \cap B \right) & & \text { by the associative law (for } \cap ) \\& = A \cap \left( B \cap B ^ { c } \right) & & \text { by the commutative law (for } \cap ) \\& = A \cap \emptyset & & \text { by the complement law (for } \cap ) \\& = \emptyset & & \text { by the universal bound law (for } \cap )\end{aligned}$
which is the right-hand side of the equation to be shown. [Hence the given statement is true.]

Use the element method for proving a set equals the empty set to prove that $\text { For all sets } A \text { and } C , ( A - C ) \cap ( C - A ) = \emptyset \text {. }$

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Question 2

Correct Answer:

Verified

Element Proof: Suppose the given statement is false.
[We must show that this supposition leads logically to a contradiction.]
Then there exist sets $A$ and $C$ such that $( A - C ) \cap ( C - A ) \neq \emptyset$ .
Thus there is an element, say $x$ , in $( A - C ) \cap ( C - A )$ .
By definition of intersection, $x \in A - C$ and $x \in C - A$ .
By definition of set difference applied to both expressions, $x \in A$ and $x \notin C$ and $x \in C$ and $x \notin A$ .
Hence, in particular, $x \in A$ and $x \notin A$ , which is a contradiction.
[Arriving at this contradiction shows that the given statement is not false; so it is true that for all sets $A$ and $B , ( A - C ) \cap ( C - A ) = \emptyset$ .]

$Element Proof: Suppose the given statement is false. [We must show that this supposition leads logically to a contradiction.] Then there exist sets A and C such that ( A - C ) \cap ( C - A ) \neq \emptyset . Thus there is an element, say x , in ( A - C ) \cap ( C - A ) . By definition of intersection, x \in A - C and x \in C - A . By definition of set difference applied to both expressions, x \in A and x \notin C and x \in C and x \notin A . Hence, in particular, x \in A and x \notin A , which is a contradiction. [Arriving at this contradiction shows that the given statement is not false; so it is true that for all sets A and B , ( A - C ) \cap ( C - A ) = \emptyset .] Suppose the given statement is false. Then there exist sets \bar { B } and C such that ( B - C ) - B \neq \emptyset . Thus there is an element x in (B − C) − B. Applying the definition of set difference twice gives that x \in B and x \notin C and x \notin B . Hence x \in B and x \notin B , which is a contradiction [and so the given statement is true].$ Suppose the given statement is false. Then there exist sets $\bar { B }$ and $C$ such that $( B - C ) - B \neq \emptyset$ .
Thus there is an element x in (B − C) − B. Applying the definition of set difference twice gives that $x \in B$ and $x \notin C$ and $x \notin B$ . Hence $x \in B$ and $x \notin B$ , which is a contradiction [and so the given statement is true].

Fill in the blanks: (a) Given sets $A$ and $B$ , to prove that $( A - B ) \cup ( A \cap B ) \subseteq A$ , we suppose that $x \in$ ___and we must show that $x \in$ _ (b) By definition of union, to say that $x \in ( A - B ) \cup ( A \cap B )$ means that___

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Question 3

Correct Answer:

Verified

a. $( A - B ) \cup ( A \cap B ) ; A$
b. $x \in ( A - B )$ or $x \in ( A \cap B )$

Let $X = \{ l \in \mathrm { Z } \mid l = 5 a + 2$ for some integer $a \} , Y = \{ m \in \mathrm { Z } \mid m = 4 b + 3$ for some integer $b \}$ , and $Z = \{ n \in \mathrm { Z } \mid n = 4 c - 1$ for some integer $c \}$ . (a) Is $X \subseteq Y$ ? (b) Is $Y \subseteq Z$ ? Justify your answers carefully. (In other words, provide a proof if the statement is true or a disproof if the statement is false.)

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Question 4

$\text { If } X = \{ u , v \} , \text { what is the power set of } X ?$

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Question 5

Write a negation for the following statement: $\text { For all } x \text {, if } x \in A \cap B \text { then } x \in B \text {. }$

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Question 6

Consider the statement $\text { For all sets } A \text { and } B , ( A - B ) \cap B = \emptyset \text {. }$ The proof below is the beginning of a proof using the element method for proving that the set equals the empty set. Complete the proof without using any of the set properties from Theorem 6.2.2. Proof: Suppose the given statement is false. Then there exist sets $A$ and $B$ such that $( A -$ $B ) \cap B \neq \emptyset$ . Thus there is an element $x$ in $( A - B ) \cap B$ . By definition of intersection,...

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Question 7

Disprove the following statement by ﬁnding a counterexample. $\text { For all sets } A , B \text {, and } C , A \cup ( B \cap C ) \subseteq ( A \cup B ) \cap C \text {. }$

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Question 8

Is the following sentence a statement: This sentence is false or $- 2 ^ { 2 } = 4$ Justify your answer.

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Question 9

(a) Is $2 \subseteq \{ 2,4,6 \}$ ? (b) Is $\{ 3 \} \in \{ 1,3,5 \}$ ?

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Question 10

Let A and B be sets. Deﬁne precisely (but concisely) what it means for A to be a subset of B.

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Question 11

Derive the following result. You may do so either "algebraically" using the properties listed in Theorem 6.2.2, being sure to give a reason for every step, or you may use the element method for proving a set equals the empty set. $\text { For all sets } B \text { and } C , ( B - C ) - B = \emptyset \text {. }$

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Question 12

Define sets $A$ and $B$ as follows: $A = \{ n \in \mathbf { Z } \mid n = 8 r - 3$ for some integer $r \}$ and $B = \{ m \in \mathbf { Z } \mid m = 4 s + 1$ for some integer $s \}$ . (a) Is $A \subseteq B$ ? (b) Is $B \subseteq A$ ? Justify your answers carefully. (In other words, provide a proof if the statement is true or a disproof if the statement is false.)

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Question 13

Prove the following statement using an element argument and reasoning directly from the deﬁnitions of union, intersection, set diﬀerence. $\text { For all sets } A , B \text {, and } C , ( A \cup B ) \cap C \subseteq A \cup ( B \cap C ) \text {. }$

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Question 14

Prove that for all sets $A$ and $B _ { 1 } , B _ { 2 } , \ldots B _ { n }$ , $A - \bigcap _ { i = 1 } ^ { n } B _ { i } = \bigcup _ { i = 1 } ^ { n } \left( A - B _ { i } \right)$

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Question 15

(a) Prove the following statement using the element method for proving that a set equals the empty set: For all sets $A$ and $B , A \cap ( B - A ) = \emptyset$ . (b) Use the properties in Theorem $6.2 .2$ to prove the statement in part (a). Be sure to give a reason for every step.

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Question 16

Fill in the blanks in the following sentence: If $A , B$ and $C$ are any sets, then by definition of set difference $x \in A - ( B \cap C )$ if, and only if, $x$ and $x$

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Question 17

The following is an outline of a proof that $( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c }$ . Fill in the blanks. $The following is an outline of a proof that ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } . Fill in the blanks. Given sets A and B , to prove that ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } , we suppose x \in then we show that x \in \ldots . So suppose that Then by definition of complement, So by definition of union, it is not the case that ( x is in A or x is in B ). Consequently, x is not in A x is not in B because of De Morgan's law of logic. In symbols, this says that x \notin A and x \notin B . So by definition of complement, x \in \ldots and x \in Thus, by definition of intersection, x \in [as was to be shown].$ Given sets $A$ and $B$ , to prove that $( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c }$ , we suppose $x \in$ $The following is an outline of a proof that ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } . Fill in the blanks. Given sets A and B , to prove that ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } , we suppose x \in then we show that x \in \ldots . So suppose that Then by definition of complement, So by definition of union, it is not the case that ( x is in A or x is in B ). Consequently, x is not in A x is not in B because of De Morgan's law of logic. In symbols, this says that x \notin A and x \notin B . So by definition of complement, x \in \ldots and x \in Thus, by definition of intersection, x \in [as was to be shown].$ then we show that $x \in \ldots$ $The following is an outline of a proof that ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } . Fill in the blanks. Given sets A and B , to prove that ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } , we suppose x \in then we show that x \in \ldots . So suppose that Then by definition of complement, So by definition of union, it is not the case that ( x is in A or x is in B ). Consequently, x is not in A x is not in B because of De Morgan's law of logic. In symbols, this says that x \notin A and x \notin B . So by definition of complement, x \in \ldots and x \in Thus, by definition of intersection, x \in [as was to be shown].$ . So suppose that $The following is an outline of a proof that ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } . Fill in the blanks. Given sets A and B , to prove that ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } , we suppose x \in then we show that x \in \ldots . So suppose that Then by definition of complement, So by definition of union, it is not the case that ( x is in A or x is in B ). Consequently, x is not in A x is not in B because of De Morgan's law of logic. In symbols, this says that x \notin A and x \notin B . So by definition of complement, x \in \ldots and x \in Thus, by definition of intersection, x \in [as was to be shown].$ Then by definition of complement, $The following is an outline of a proof that ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } . Fill in the blanks. Given sets A and B , to prove that ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } , we suppose x \in then we show that x \in \ldots . So suppose that Then by definition of complement, So by definition of union, it is not the case that ( x is in A or x is in B ). Consequently, x is not in A x is not in B because of De Morgan's law of logic. In symbols, this says that x \notin A and x \notin B . So by definition of complement, x \in \ldots and x \in Thus, by definition of intersection, x \in [as was to be shown].$ So by definition of union, it is not the case that ( $x$ is in $A$ or $x$ is in $B$ ). Consequently, $x$ is not in $A$ $The following is an outline of a proof that ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } . Fill in the blanks. Given sets A and B , to prove that ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } , we suppose x \in then we show that x \in \ldots . So suppose that Then by definition of complement, So by definition of union, it is not the case that ( x is in A or x is in B ). Consequently, x is not in A x is not in B because of De Morgan's law of logic. In symbols, this says that x \notin A and x \notin B . So by definition of complement, x \in \ldots and x \in Thus, by definition of intersection, x \in [as was to be shown].$ $x$ is not in $B$ because of De Morgan's law of logic. In symbols, this says that $x \notin A$ and $x \notin B$ . So by definition of complement, $x \in \ldots$ $The following is an outline of a proof that ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } . Fill in the blanks. Given sets A and B , to prove that ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } , we suppose x \in then we show that x \in \ldots . So suppose that Then by definition of complement, So by definition of union, it is not the case that ( x is in A or x is in B ). Consequently, x is not in A x is not in B because of De Morgan's law of logic. In symbols, this says that x \notin A and x \notin B . So by definition of complement, x \in \ldots and x \in Thus, by definition of intersection, x \in [as was to be shown].$ and $x \in$ $The following is an outline of a proof that ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } . Fill in the blanks. Given sets A and B , to prove that ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } , we suppose x \in then we show that x \in \ldots . So suppose that Then by definition of complement, So by definition of union, it is not the case that ( x is in A or x is in B ). Consequently, x is not in A x is not in B because of De Morgan's law of logic. In symbols, this says that x \notin A and x \notin B . So by definition of complement, x \in \ldots and x \in Thus, by definition of intersection, x \in [as was to be shown].$ Thus, by definition of intersection, $x \in$ $The following is an outline of a proof that ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } . Fill in the blanks. Given sets A and B , to prove that ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } , we suppose x \in then we show that x \in \ldots . So suppose that Then by definition of complement, So by definition of union, it is not the case that ( x is in A or x is in B ). Consequently, x is not in A x is not in B because of De Morgan's law of logic. In symbols, this says that x \notin A and x \notin B . So by definition of complement, x \in \ldots and x \in Thus, by definition of intersection, x \in [as was to be shown].$ [as was to be shown].

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Question 18

Derive the following result "algebraically" using the properties listed in Theorem 6.2.2. Give a reason for every step that exactly justiﬁes what was done in the step. $\text { For all sets } A , B \text {, and } C , ( A \cup C ) - B = ( A - B ) \cup ( C - B ) \text {. }$

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Question 19

Showing 1 - 19 of 19

Use the element method for proving a set equals the empty set to prove that $\text { For all sets } A \text { and } C , ( A - C ) \cap ( C - A ) = \emptyset \text {. }$

Fill in the blanks: (a) Given sets $A$ and $B$ , to prove that $( A - B ) \cup ( A \cap B ) \subseteq A$ , we suppose that $x \in$ ___and we must show that $x \in$ _ (b) By definition of union, to say that $x \in ( A - B ) \cup ( A \cap B )$ means that___

$\text { If } X = \{ u , v \} , \text { what is the power set of } X ?$

Write a negation for the following statement: $\text { For all } x \text {, if } x \in A \cap B \text { then } x \in B \text {. }$

Disprove the following statement by ﬁnding a counterexample. $\text { For all sets } A , B \text {, and } C , A \cup ( B \cap C ) \subseteq ( A \cup B ) \cap C \text {. }$

Is the following sentence a statement: This sentence is false or $- 2 ^ { 2 } = 4$ Justify your answer.

(a) Is $2 \subseteq \{ 2,4,6 \}$ ? (b) Is $\{ 3 \} \in \{ 1,3,5 \}$ ?

Let A and B be sets. Deﬁne precisely (but concisely) what it means for A to be a subset of B.

Prove the following statement using an element argument and reasoning directly from the deﬁnitions of union, intersection, set diﬀerence. $\text { For all sets } A , B \text {, and } C , ( A \cup B ) \cap C \subseteq A \cup ( B \cap C ) \text {. }$

Prove that for all sets $A$ and $B _ { 1 } , B _ { 2 } , \ldots B _ { n }$ , $A - \bigcap _ { i = 1 } ^ { n } B _ { i } = \bigcup _ { i = 1 } ^ { n } \left( A - B _ { i } \right)$

(a) Prove the following statement using the element method for proving that a set equals the empty set: For all sets $A$ and $B , A \cap ( B - A ) = \emptyset$ . (b) Use the properties in Theorem $6.2 .2$ to prove the statement in part (a). Be sure to give a reason for every step.

Fill in the blanks in the following sentence: If $A , B$ and $C$ are any sets, then by definition of set difference $x \in A - ( B \cap C )$ if, and only if, $x$ and $x$

Derive the following result "algebraically" using the properties listed in Theorem 6.2.2. Give a reason for every step that exactly justiﬁes what was done in the step. $\text { For all sets } A , B \text {, and } C , ( A \cup C ) - B = ( A - B ) \cup ( C - B ) \text {. }$

Speaking Mathematically

The Logic of Compound Statements

The Logic of Quantified Statements

Elementary Number Theory and Methods of Proof

Sequences, Mathematical Induction, and Recursion

Functions

Relations

Counting and Probability

Graphs and Trees

Analyzing Algorithm Efficiency

Regular Expressions and Finite State Automata

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Exam 6: Set Theory

Use the element method for proving a set equals the empty set to prove that For all sets A and C,(A−C)∩(C−A)=∅. \text { For all sets } A \text { and } C , ( A - C ) \cap ( C - A ) = \emptyset \text {. } For all sets A and C,(A−C)∩(C−A)=∅.

If X={u,v}, what is the power set of X?\text { If } X = \{ u , v \} , \text { what is the power set of } X ? If X={u,v}, what is the power set of X?

Write a negation for the following statement: For all x, if x∈A∩B then x∈B. \text { For all } x \text {, if } x \in A \cap B \text { then } x \in B \text {. } For all x, if x∈A∩B then x∈B.

Disprove the following statement by ﬁnding a counterexample. For all sets A,B, and C,A∪(B∩C)⊆(A∪B)∩C. \text { For all sets } A , B \text {, and } C , A \cup ( B \cap C ) \subseteq ( A \cup B ) \cap C \text {. } For all sets A,B, and C,A∪(B∩C)⊆(A∪B)∩C.

Is the following sentence a statement: This sentence is false or −22=4- 2 ^ { 2 } = 4−22=4 Justify your answer.

(a) Is 2⊆{2,4,6}2 \subseteq \{ 2,4,6 \}2⊆{2,4,6} ? (b) Is {3}∈{1,3,5}\{ 3 \} \in \{ 1,3,5 \}{3}∈{1,3,5} ?

Let A and B be sets. Deﬁne precisely (but concisely) what it means for A to be a subset of B.

Prove that for all sets AAA and B1,B2,…BnB _ { 1 } , B _ { 2 } , \ldots B _ { n }B1​,B2​,…Bn​ , A−⋂i=1nBi=⋃i=1n(A−Bi)A - \bigcap _ { i = 1 } ^ { n } B _ { i } = \bigcup _ { i = 1 } ^ { n } \left( A - B _ { i } \right)A−⋂i=1n​Bi​=⋃i=1n​(A−Bi​)

Fill in the blanks in the following sentence: If A,BA , BA,B and CCC are any sets, then by definition of set difference x∈A−(B∩C)x \in A - ( B \cap C )x∈A−(B∩C) if, and only if, xxx ____ and xxx ____

Speaking Mathematically

The Logic of Compound Statements

The Logic of Quantified Statements

Elementary Number Theory and Methods of Proof

Sequences, Mathematical Induction, and Recursion

Functions

Relations

Counting and Probability

Graphs and Trees

Analyzing Algorithm Efficiency

Regular Expressions and Finite State Automata

Filters

Use the element method for proving a set equals the empty set to prove that $\text { For all sets } A \text { and } C , ( A - C ) \cap ( C - A ) = \emptyset \text {. }$

$\text { If } X = \{ u , v \} , \text { what is the power set of } X ?$

Write a negation for the following statement: $\text { For all } x \text {, if } x \in A \cap B \text { then } x \in B \text {. }$

Disprove the following statement by ﬁnding a counterexample. $\text { For all sets } A , B \text {, and } C , A \cup ( B \cap C ) \subseteq ( A \cup B ) \cap C \text {. }$

Is the following sentence a statement: This sentence is false or $- 2 ^ { 2 } = 4$ Justify your answer.

(a) Is $2 \subseteq \{ 2,4,6 \}$ ? (b) Is $\{ 3 \} \in \{ 1,3,5 \}$ ?

Prove that for all sets $A$ and $B _ { 1 } , B _ { 2 } , \ldots B _ { n }$ , $A - \bigcap _ { i = 1 } ^ { n } B _ { i } = \bigcup _ { i = 1 } ^ { n } \left( A - B _ { i } \right)$

Fill in the blanks in the following sentence: If $A , B$ and $C$ are any sets, then by definition of set difference $x \in A - ( B \cap C )$ if, and only if, $x$ and $x$