Question
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1. Let \( A=\{1,2,3,4\} \) and \( B=\{a, c\} \). Write the following sets by listing their elements between braces: (a) \( A \times B \) (b) \( B \times B \) (c) \( \emptyset \times B \) (d) \( (A \times B) \times B \)

Ask by Mcfarlane Donnelly. in the United States
Mar 12,2025

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Answer

**(a)** \(A \times B = \{ (1,a),\ (1,c),\ (2,a),\ (2,c),\ (3,a),\ (3,c),\ (4,a),\ (4,c) \}\) **(b)** \(B \times B = \{ (a,a),\ (a,c),\ (c,a),\ (c,c) \}\) **(c)** \(\emptyset \times B = \emptyset\) **(d)** \[ (A \times B) \times B = \{ ((1,a),a),\ ((1,a),c),\ ((1,c),a),\ ((1,c),c),\ ((2,a),a),\ ((2,a),c),\ ((2,c),a),\ ((2,c),c),\ ((3,a),a),\ ((3,a),c),\ ((3,c),a),\ ((3,c),c),\ ((4,a),a),\ ((4,a),c),\ ((4,c),a),\ ((4,c),c) \} \]

Solution

**(a)** \(A \times B = \{ (1,a),\ (1,c),\ (2,a),\ (2,c),\ (3,a),\ (3,c),\ (4,a),\ (4,c) \}\) **(b)** \(B \times B = \{ (a,a),\ (a,c),\ (c,a),\ (c,c) \}\) **(c)** \(\emptyset \times B = \emptyset\) **(d)** First note that \(A \times B = \{ (1,a),\ (1,c),\ (2,a),\ (2,c),\ (3,a),\ (3,c),\ (4,a),\ (4,c) \}\). Then, \[ (A \times B) \times B = \{ ((x,y),z) \mid (x,y) \in A \times B,\ z \in B \} \] Listing all elements: \[ \begin{aligned} (A \times B) \times B = \{ & ((1,a),a),\ ((1,a),c),\ ((1,c),a),\ ((1,c),c), \\ & ((2,a),a),\ ((2,a),c),\ ((2,c),a),\ ((2,c),c), \\ & ((3,a),a),\ ((3,a),c),\ ((3,c),a),\ ((3,c),c), \\ & ((4,a),a),\ ((4,a),c),\ ((4,c),a),\ ((4,c),c) \} \end{aligned} \]

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The Deep Dive

Let's tackle each part one by one. (a) The Cartesian product \( A \times B \) consists of all ordered pairs where the first element is from set \( A \) and the second element is from set \( B \). Thus, we have: \[ A \times B = \{(1, a), (1, c), (2, a), (2, c), (3, a), (3, c), (4, a), (4, c)\} \] (b) For \( B \times B \), we take all ordered pairs where both elements are from set \( B \): \[ B \times B = \{(a, a), (a, c), (c, a), (c, c)\} \] (c) The Cartesian product \( \emptyset \times B \) is the set of all ordered pairs where the first element is from the empty set and the second from set \( B \). Since there are no elements in \( \emptyset \), the result is: \[ \emptyset \times B = \emptyset \] (d) Finally, for \( (A \times B) \times B \), we first consider \( A \times B \) from part (a) and treat each element of it as a first element in a new ordered pair with the second element from set \( B \): \[ (A \times B) \times B = \{((1, a), a), ((1, a), c), ((1, c), a), ((1, c), c), ((2, a), a), ((2, a), c), ((2, c), a), ((2, c), c), ((3, a), a), ((3, a), c), ((3, c), a), ((3, c), c), ((4, a), a), ((4, a), c), ((4, c), a), ((4, c), c)\} \]

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