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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 Griffiths Rose. in the United States
Mar 12,2025

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**(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)** \( \emptyset \times B = \emptyset \) *(Because the Cartesian product with the empty set is empty.)* **(d)** \( (A \times B) \times B = \{ ((1,a),a),\; ((1,a),c),\; ((1,c),a),\; ((1,c),c),\; ((2,a),a),\; ((2,a),c),\; ((2,c),a),\; ((2,c),c),\; ((3,a),a),\; ((3,a),c),\; ((3,c),a),\; ((3,c),c),\; ((4,a),a),\; ((4,a),c),\; ((4,c),a),\; ((4,c),c) \} \)

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Let's explore the Cartesian products step by step: (a) The set \( A \times B \) consists of all ordered pairs where the first element comes from set \( A \) and the second element comes from set \( B \). Therefore, we have: \[ A \times B = \{ (1,a), (1,c), (2,a), (2,c), (3,a), (3,c), (4,a), (4,c) \} \] (b) The set \( B \times B \) consists of all ordered pairs where both elements come from set \( B \): \[ B \times B = \{ (a,a), (a,c), (c,a), (c,c) \} \] (c) The set \( \emptyset \times B \) will be the empty set since the first set is empty, meaning there are no elements to create pairs with: \[ \emptyset \times B = \emptyset \] (d) The set \( (A \times B) \times B \) consists of ordered pairs where the first element is an ordered pair from \( A \times B \) and the second element is from \( B \). Thus: \[ (A \times B) \times B = \{ ((1,a),a), ((1,a),c), ((1,c),a), ((1,c),c), ((2,a),a), ((2,a),c), ((2,c),a), ((2,c),c), ((3,a),a), ((3,a),c), ((3,c),a), ((3,c),c), ((4,a),a), ((4,a),c), ((4,c),a), ((4,c),c) \} \]

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