Problem 3 - A Primer of Real Analysis

Question

Let \(I=\\{x: x \in \mathbb{R}, 0 \leq x<1\\}\) and suppose \(a\) and \(b\) are real numbers with \(a

Step-by-Step Solution

Answer

The intervals \([0, 1)\) and \([a, b)\) are bijective, hence have the same cardinality.

Step 1: Identify Key Sets and Intervals

We have two sets to compare: \(I=\{x: x \in \mathbb{R}, 0 \leq x<1\}\) and \(J=\{x: x \in \mathbb{R}, a \leq x

Step 2: Define a Function to Map Between Sets

Define a function \(f: I \to J\) by linear scaling: \(f(x) = (b-a)x + a\). This function transforms the interval \([0, 1)\) to \([a, b)\).

Step 3: Prove that the Function is Bijective

To show \(|I| = |J|\), demonstrate \(f\) is both injective and surjective. - **Injective proof**: Assume \(f(x_1) = f(x_2)\). Then \((b-a)x_1 + a = (b-a)x_2 + a\), implying \(x_1 = x_2\). Thus, \(f\) is injective.- **Surjective proof**: For any \(y \in J\), we solve \((b-a)x + a = y\) for \(x\), giving \(x = \frac{y-a}{b-a}\). This \(x\) belongs to \(I\), hence for every \(y \in J\), there is an \(x \in I\) such that \(f(x) = y\), proving \(f\) is surjective.

Step 4: Conclude Set Cardinality

Since \(f\) is bijective (both injective and surjective), the cardinality of sets \(I\) and \(J\) is the same, fulfilling \(|I| = |J|\).

Key Concepts

Cardinality

Cardinality refers to the 'size' or 'number of elements' in a set. In simpler terms, it's like thinking about how many members there are in a club. For finite sets, cardinality is easy to determine because we can count the elements. But for infinite sets, like the intervals discussed in this problem, we need a different approach.

In this exercise, we are comparing the cardinality of two interval sets:

• Set \(I\), which consists of all real numbers between 0 and 1, inclusive of 0 but exclusive of 1.
• Set \(J\), which consists of all real numbers between \(a\) and \(b\), inclusive of \(a\) but exclusive of \(b\).

To show that these sets have the same cardinality, we need to demonstrate that there is a one-to-one correspondence, meaning a way to pair each element of one set uniquely with an element of the other. This problem uses the concept of bijective functions (injective and surjective) to establish such a correspondence.

Injective Function

An injective function, or "one-to-one" function, is a function where each element of the input set is mapped to a unique element of the output set. Think of it as each person in a class having their own unique student ID. It ensures that no two different inputs have the same output.

In the solution provided, the function \(f(x) = (b-a)x + a\) is defined to map set \(I\) to set \(J\). To prove it is injective, it was shown that if \(f(x_1) = f(x_2)\), then it must follow that \(x_1 = x_2\). This implies that our function doesn't "glue" any two distinct elements from \(I\) to the same element in \(J\).

A fun example of an injective function in every day life is if each club member receives a different colored badge - no badge color belongs to more than one member.

Surjective Function

A surjective function, or "onto" function, is one where every element of the output set is the image of at least one element of the input set. Imagine a situation where every seat in a theater is filled, with no empty seats.

In the context of this exercise, the function \(f(x) = (b-a)x + a\) is shown to be surjective because for every \(y\) in set \(J\), there exists an \(x\) in set \(I\) such that \(f(x) = y\). Essentially, it illustrates that no element in the interval from \(a\) to \(b\) (set \(J\)) is left unmapped, ensuring that the function covers the whole output set.

This concept is key in proving that the two sets have the same cardinality, alongside injectivity, to form a bijective function.

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Problem 2

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Problem 3

Other exercises in this chapter

Problem 2

Let \(I=\\{x: x \in \mathbb{R}, 0 \leq x

View solution

Problem 2

Let \(A\) be the set of even integers. Show that \(|A|=\aleph_{0}\).

View solution

Problem 3

Verify each of the following: a. If \(A\) is a nonempty subset of \(\mathbb{Z}^{+},\) then \(A\) is either finite or countable. b. If \(A\) is a nonempty subset

View solution

Problem 4

Suppose \(A\) is uncountable and \(B \subset A\) is countable. Show that \(A \backslash B\) is uncountable.

View solution

Key Concepts

Cardinality

Cardinality refers to the 'size' or 'number of elements' in a set. In simpler terms, it's like thinking about how many members there are in a club. For finite sets, cardinality is easy to determine because we can count the elements. But for infinite sets, like the intervals discussed in this problem, we need a different approach.

In this exercise, we are comparing the cardinality of two interval sets:

• Set \(I\), which consists of all real numbers between 0 and 1, inclusive of 0 but exclusive of 1.
• Set \(J\), which consists of all real numbers between \(a\) and \(b\), inclusive of \(a\) but exclusive of \(b\).

To show that these sets have the same cardinality, we need to demonstrate that there is a one-to-one correspondence, meaning a way to pair each element of one set uniquely with an element of the other. This problem uses the concept of bijective functions (injective and surjective) to establish such a correspondence.

Injective Function

An injective function, or "one-to-one" function, is a function where each element of the input set is mapped to a unique element of the output set. Think of it as each person in a class having their own unique student ID. It ensures that no two different inputs have the same output.

In the solution provided, the function \(f(x) = (b-a)x + a\) is defined to map set \(I\) to set \(J\). To prove it is injective, it was shown that if \(f(x_1) = f(x_2)\), then it must follow that \(x_1 = x_2\). This implies that our function doesn't "glue" any two distinct elements from \(I\) to the same element in \(J\).

A fun example of an injective function in every day life is if each club member receives a different colored badge - no badge color belongs to more than one member.

Surjective Function

A surjective function, or "onto" function, is one where every element of the output set is the image of at least one element of the input set. Imagine a situation where every seat in a theater is filled, with no empty seats.

In the context of this exercise, the function \(f(x) = (b-a)x + a\) is shown to be surjective because for every \(y\) in set \(J\), there exists an \(x\) in set \(I\) such that \(f(x) = y\). Essentially, it illustrates that no element in the interval from \(a\) to \(b\) (set \(J\)) is left unmapped, ensuring that the function covers the whole output set.

This concept is key in proving that the two sets have the same cardinality, alongside injectivity, to form a bijective function.