**A = {0, 2, 4, 6, 8, 10} B = {0, 1, 2, 3, 4, 5, 6} C = {4, 5, 6, 7, 8, 9, 10} a. A\cap B\cap CA∩B∩C, requires us to determine the elements common to set A, set B and set C. Therefore, A\cap B\cap C=\{4,6\}A∩B∩C={4,6} b. A\cup B\cup CA∪B∪C requires us to determine all elements in set A, B and C. Therefore, A\cup B\cup C=\{0,1,2,3,4,5,6,7,8,9,10\}A∪B∪C={0,1,2,3,4,5,6,7,8,9,10} C. (A\cup B)\cap C(A∪B)∩C, First, we need to determine (A\cup B)(A∪B), which are the elements in both set A and set B. A\cup B=\{0,1,2,3,4,5,6,8,10\}A∪B={0,1,2,3,4,5,6,8,10}. Set C=\{4,5,6,7,8,9,10\}C={4,5,6,7,8,9,10} Therefore, (A\cup B)\cap C(A∪B)∩C requires us to determine elements common to (A\cup B)(A∪B) and set C.C. Hence, (A\cup B)\cap C=\{4,5,6,8,10\}(A∪B)∩C={4,5,6,8,10} d. (A\cap B)\cup C(A∩B)∪C We first find elements common to set A and set B. (A\cap B)=\{0,2,4,6\}(A∩B)={0,2,4,6} Set C=\{4,5,6,7,8,9,10\}C={4,5,6,7,8,9,10} We now find all elements in the intersection(A\cap B)(A∩B) and set C.C. Thus, (A\cap B)\cup C=\{0,2,4,5,6,7,8,9,10\}(A∩B)∪C={0,2,4,5,6,7,8,9,10}.**

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**Here's the Solution to this Question**

(a) Given:$A \cup B=A$

Using the definition of the union, an element of $A \cup B$ is an element that is in A or in B.

$A \cup B=\{x \mid x \in A \vee x \in B\}$

Let x be an element of B.

$x \in B$

By the definition of the union, $x$ then has to be an element of $A \cup B$ .

$\Rightarrow x \in A \cup B$

By $A \cup B=A$ , we then know that x also has to be an element of A :

$\Rightarrow x \in A$

Every element of B is thus also an element of A. By the definition of a subset, we then know that B is a subset of A

$B \subseteq A$

(b) Given: $A \cap B=A$

Using the definition of the intersection, an element of $A \cup B$ is an element that is in A and in B.

$A \cap B=\{x \mid x \in A \wedge x \in B\}$

Let $x$ be an element of A.

$x \in A$

By $A \cap B=A$ , $x$ then has to be an element of $A \cap B$ .

$\Rightarrow x \in A \cap B$

By the definition of the intersection $x$ then has to be an element of B as well.

$\Rightarrow x \in B$

Every element of A is thus also an element of B. By the definition of a subset, we then know that A is a subset of B.

$A \subseteq B$

(c) Given: $A-B=A$

Using the definition of the difference, an element of $A-B$ is an element that is in A and not in B.

$A-B=\{x \mid x \in A \wedge x \notin B\}$

Let $x$ be an element of B.

$x \in B$

By the definition of the difference, $x$ then is not an element of $A-B$ ,

$\Rightarrow x \notin A-B$

By $A-B=A$ , we then know that $x$ also cannot be an element of A :

$\Rightarrow x \notin A$

Every element of B is thus also an element of the complement of A. By the definition of a subset, we then know that B is a subset of $\bar{A}$ .

$B \subseteq \bar{A}$

Since the intersection of A and $\bar{A}$ is empty and $B \subseteq \bar{A}$ , the intersection of A and B then also has to be empty.

$A \cap B=\emptyset$

(d) Commutative law for set identities:

$A \cap B=B \cap A$

The given statement is the commutative law and is thus true for all sets A and B. This then means that we cannot say anything about the sets (since there are no conditions in using the commutative law).

(e) Given: $A-B=B-A$

Using the definition of the difference, an element of $A-B$ is an element that is in A and not in B.

$A-B=\{x \mid x \in A \wedge x \notin B\}$

Using the definition of the difference, an element of $B-A$ is an element that is in B and not in A.

$B-A=\{x \mid x \in B \wedge x \notin A\}$

Let us assume $x \in A-B$ . Then $x \in A$ is true and $x \notin A$ is false. By the definition of $B-A,$ $x$ can then not be in $B-A$ . Since none of the elements of $A-B$ are in $B-A$ and since $A-B=B-A$ , the differences then have to be the empty set.

$A-B=B-A=\emptyset$

The difference A-B does not contain any elements, if all elements of A are also an element of B.

$A \subseteq B$

The difference B-A does not contain any elements, if all elements of B are also an element of A.

$B \subseteq A$

Since $A \subseteq B$ and $B \subseteq A$ , the two sets then have to be equal.

$A=B$