Question

Prove that the sets \(\mathcal{X}=\\{2 n+1: n \in \mathbb{Z}\\}, \mathcal{Y}=\\{10 j: j \in \mathbb{Z}\\},\) and \(\mathcal{Z}=\\{3 n: n \in \mathbb{Z} \mid\) have the same cardinality.

Short answer

The sets \(\mathcal{X}\), \(\mathcal{Y}\), and \(\mathcal{Z}\) have the same cardinality as \(\mathbb{Z}\), hence they have the same cardinality.

Step-by-step solution

Define the Cardinality

Two sets have the same cardinality if there exists a bijection between them, meaning there is a one-to-one and onto function.

Define the Sets

Define the given sets:\[ \mathcal{X} = \{2n + 1 \mid n \in \mathbb{Z}\} \] is the set of all odd numbers.\[ \mathcal{Y} = \{10j \mid j \in \mathbb{Z}\} \] is the set of all multiples of 10.\[ \mathcal{Z} = \{3n \mid n \in \mathbb{Z}\} \] is the set of all multiples of 3.

Construct a Bijection between \(\mathcal{X}\) and \(\mathbb{Z}\)

Define a function \(f: \mathbb{Z} \to \mathcal{X}\) by \(f(n) = 2n + 1\). To show this is a bijection, check the following:**Injective:** If \(f(a) = f(b)\), then \(2a + 1 = 2b + 1\). Solving gives \(a = b\). Hence, \(f\) is injective.**Surjective:** For any \(x \in \mathcal{X}\), there exists an \(n \in \mathbb{Z}\) such that \(x = 2n + 1\). Therefore, \(f\) is surjective.

Construct a Bijection between \(\mathcal{Y}\) and \(\mathbb{Z}\)

Define a function \(g: \mathbb{Z} \to \mathcal{Y}\) by \(g(n) = 10n\). Check bijection property:**Injective:** If \(g(a) = g(b)\), then \(10a = 10b\). Solving gives \(a = b\). Hence, \(g\) is injective.**Surjective:** For any \(y \in \mathcal{Y}\), there exists an \(j \in \mathbb{Z}\) such that \(y = 10j\). Therefore, \(g\) is surjective.

Construct a Bijection between \(\mathcal{Z}\) and \(\mathbb{Z}\)

Define a function \(h: \mathbb{Z} \to \mathcal{Z}\) by \(h(n) = 3n\). Check bijection property:**Injective:** If \(h(a) = h(b)\), then \(3a = 3b\). Solving gives \(a = b\). Hence, \(h\) is injective.**Surjective:** For any \(z \in \mathcal{Z}\), there exists an \(n \in \mathbb{Z}\) such that \(z = 3n\). Therefore, \(h\) is surjective.

Conclude the Cardinality is the Same

Since the functions \(f\), \(g\), and \(h\) prove that sets \(\mathcal{X}\), \(\mathcal{Y}\), and \(\mathcal{Z}\) all have bijections with \(\mathbb{Z}\), all sets have the same cardinality as \(\mathbb{Z}\). Thus, \(\mathcal{X}\), \(\mathcal{Y}\), and \(\mathcal{Z}\) all have the same cardinality.

Key concepts

Bijection

A bijection is a special type of function between two sets, where every element in one set is paired with exactly one element in another set, and vice versa. This means it is both one-to-one (injective) and onto (surjective). Let's break it down:

• Injective (One-to-One): Each element in the first set maps to a unique element in the second set. No two different elements from the first set map to the same element in the second set.
• Surjective (Onto): Every element in the second set has at least one element from the first set mapping to it.

Together, these properties ensure that every element matches perfectly between the two sets—no leftovers on either side! In our context, for the sets of odd numbers, multiples of 10, and multiples of 3, constructing bijections with the set of integers \( \mathbb{Z} \) demonstrates their shared cardinality. Each set's bijection proves it is indeed equal in size to \( \mathbb{Z} \), establishing a common cardinality across these seemingly different collections.

Odd Numbers

Odd numbers are integers that cannot be evenly divided by 2. They usually appear as numbers with the form \( 2n+1 \), where \( n \) is any integer. Some properties of odd numbers include:

• They follow a sequence alternating with even numbers, starting from 1 (e.g., 1, 3, 5, 7...).
• When added or subtracted among themselves, they yield another odd number.
• When multiplied by an odd number, the product is also odd.

In our problem, the set \( \mathcal{X} = \{2n + 1 : n \in \mathbb{Z}\} \) represents all odd numbers. By defining the function \( f(n) = 2n + 1 \), we map integers to these odd numbers, which shows that the set of odd numbers has the same size as \( \mathbb{Z} \). The construction of such a bijection confirms that there are just as many odd numbers as integers.

Multiples of 10

Multiples of 10 are integers that result from multiplying 10 by any integer. They take the form \( 10j \), where \( j \) is an integer. Multiples of 10 have some unique characteristics:

• They end in zero when written in decimal form.
• They are evenly divisible by both 2 and 5.
• Addition or subtraction of two multiples of 10 results in another multiple of 10.

In our exercise, the set \( \mathcal{Y} = \{10j : j \in \mathbb{Z}\} \) consists of these multiples. By defining the bijection \( g(n) = 10n \), we effectively match every integer to a unique multiple of 10, demonstrating that this set, \( \mathcal{Y} \), holds the same cardinality as the set of integers. This proves that the number of multiples of 10 is infinite, like the whole numbers.

Multiples of 3

Multiples of 3 are numbers that can be expressed as \( 3n \), with \( n \) representing any integer. These numbers have distinct qualities, such as:

• Divisibility by 3, without any remainder.
• Their sum is also a multiple of 3.
• They form a predictable sequence: 0, 3, 6, 9, 12, and so on.

In the problem at hand, the set \( \mathcal{Z} = \{3n : n \in \mathbb{Z}\} \) encompasses all multiples of 3. The function \( h(n) = 3n \) maps whole numbers to these multiples, establishing a bijection that shows this set shares the same cardinality with \( \mathbb{Z} \). This mapping effectively displays the infinite nature and equal size of the set of numbers multiplied by 3, alongside the integers.