[[1]] is the set of all positive integers whose cube is odd.

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Detailed Solution

Concept- 1,3,5,7,9,11,13........is the set of all positive integers whose cube is odd.
Determine the characteristics of the cubes of the odd and even positive integers first, and then establish a set of those numbers solely whose cube is odd, to obtain the set of all positive integers whose cube is odd.
We must first understand what positive odd integers are in order to be able to solve the problem. A natural number of the form (2n-1), where n is a positive whole number, is a positive odd integer. It is therefore not divisible by two. The positive odd numbers are therefore 1,3,5,7,9,11,13, etc.
We shall now locate the natural numbers' cubes and ascertain their nature. So, :

Cube of 1 = {(1)}^{3} = 1 \times 1 \times 1 = 1

Cube of 2 = {(2)}^{3} = 2 \times 2 \times 2 = 8

Cube of 3 = {(3)}^{3} = 3 \times 3 \times 3 = 27

Cube of 4 = {(4)}^{3} = 4 \times 4 \times 4 = 64

Cube of 5 = {(5)}^{3} = 5 \times 5 \times 5 = 125

Cube of 6 = {(6)}^{3} = 6 \times 6 \times 6 = 216

Clearly, the cubes of 1, 3, and 5 are positive odd integers, but the cubes of 2, 4, and 6 are positive even integers. In a similar vein, we can predict that the cubes of 7,9,13, and 15 will also be strange. We must now compile a list of every positive number whose cube is odd. We must first understand what a set is in order to create one. A set is a grouping of clearly defined and unique objects. Suppose the set has the name "S".
Hence, the correct answer is S={1,3,5,7,9,11,13........}.

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