Math Problem Statement

Jika A = {a, b, {a, c}, ∅} dan B = {a, {a}, d, e}, tentukan himpunan berikut: (a) A − ∅, (b) A − {∅}, (c) {{a, c}} − A, (d) A ⊕ B, (e) {a} − {A}, (f) P(A − B), (g) ∅ − A, (h) B², (i) A ∪ (B ∩ A), (j) A ∩ P(A)

Solution

Let's analyze and solve the set operations given in the image. The problem defines two sets:

  • A={a,b,{a,c},}A = \{ a, b, \{ a, c \}, \emptyset \}
  • B={a,{a},d,e}B = \{ a, \{ a \}, d, e \}

We need to find the results of several set operations:

(a) AA - \emptyset

Subtracting the empty set from AA does not change AA, so: A={a,b,{a,c},}A - \emptyset = \{ a, b, \{ a, c \}, \emptyset \}

(b) A{}A - \{\emptyset\}

This operation removes the empty set from AA: A{}={a,b,{a,c}}A - \{\emptyset\} = \{ a, b, \{ a, c \} \}

(c) {{a,c}}A\{\{ a, c \}\} - A

Here, the set {{a,c}}\{\{a, c\}\} is a set containing the set {a,c}\{a, c\}. Since {a,c}\{a, c\} is already in AA, the result is: {{a,c}}A=\{\{ a, c \}\} - A = \emptyset

(d) ABA \oplus B (Symmetric Difference)

The symmetric difference includes elements in either AA or BB, but not in both: AB={b,{a,c},,{a},d,e}A \oplus B = \{ b, \{ a, c \}, \emptyset, \{ a \}, d, e \}

(e) {a}{A}\{a\} - \{A\}

The set {a}\{a\} is just {a}, and {A}={{a,b,{a,c},}}\{A\} = \{\{a, b, \{a, c\}, \emptyset\}\}. There is no common element, so: {a}{A}={a}\{a\} - \{A\} = \{ a \}

(f) P(AB)P(A - B) (Power Set of ABA - B)

First, calculate ABA - B: AB={b,{a,c},}A - B = \{ b, \{ a, c \}, \emptyset \} The power set P(AB)P(A - B) includes all subsets: P(AB)={,{b},{{a,c}},{},{b,{a,c}},{b,},{{a,c},},{b,{a,c},}}P(A - B) = \{\emptyset, \{b\}, \{\{a, c\}\}, \{\emptyset\}, \{b, \{a, c\}\}, \{b, \emptyset\}, \{\{a, c\}, \emptyset\}, \{b, \{a, c\}, \emptyset\}\}

(g) A\emptyset - A

The empty set minus any set is still the empty set: A=\emptyset - A = \emptyset

(h) B2B^2 (Cartesian Product of BB with itself)

The Cartesian product B×BB \times B: B2={(a,a),(a,{a}),(a,d),(a,e),({a},a),({a},{a}),({a},d),({a},e),(d,a),(d,{a}),(d,d),(d,e),(e,a),(e,{a}),(e,d),(e,e)}B^2 = \{ (a, a), (a, \{a\}), (a, d), (a, e), (\{a\}, a), (\{a\}, \{a\}), (\{a\}, d), (\{a\}, e), (d, a), (d, \{a\}), (d, d), (d, e), (e, a), (e, \{a\}), (e, d), (e, e) \}

(i) A(BA)A \cup (B \cap A)

First, find the intersection BAB \cap A: BA={a}B \cap A = \{ a \} Then the union with AA: A(BA)={a,b,{a,c},}A \cup (B \cap A) = \{ a, b, \{ a, c \}, \emptyset \}

(j) AP(A)A \cap P(A)

The power set P(A)P(A) includes all subsets of AA. Since AA itself is not a subset of P(A)P(A), the intersection is: AP(A)=A \cap P(A) = \emptyset

Would you like any more details on these solutions or further clarifications?

Related Questions:

  1. How do you compute the power set of a given set?
  2. What are the properties of symmetric difference in set theory?
  3. How is the Cartesian product of sets defined, and what are its applications?
  4. Can you explain more about set operations like union, intersection, and difference?
  5. How do nested sets affect set operations?

Tip:

Always carefully distinguish between elements and subsets when performing set operations, as confusing these can lead to incorrect results!

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Math Problem Analysis

Mathematical Concepts

Set Operations
Symmetric Difference
Power Set
Union and Intersection
Cartesian Product

Formulas

A - B: Set Difference
A ⊕ B: Symmetric Difference
P(A): Power Set
A ∪ B: Union
A ∩ B: Intersection
B²: Cartesian Product

Theorems

Properties of Power Set
Properties of Cartesian Product
Properties of Set Difference and Symmetric Difference

Suitable Grade Level

High School / Undergraduate