/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 Prove that every subset of a cou... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 48

Prove that every subset of a countable set is countable.

Short Answer

Expert verified
In conclusion, we have three cases to consider: 1. If the countable set is finite, a bijection exists between the subset and the first few natural numbers up to the size of the subset, making the subset countable. 2. If the countable set is infinite and the subset is finite, a bijection exists between the finite subset and the first few natural numbers, making the subset countable. 3. If both the countable set and its subset are infinite, we can enumerate their elements and define a bijection between the subset and the natural numbers, making the subset countable. Thus, in all cases, every subset of a countable set is countable.

Step by step solution

01

Case 1: Countable set is finite

If the countable set is finite (has a finite number of elements), then it has a finite number of subsets. This means that we can create a bijection between the subset and the first few natural numbers up to the size of the subset. Since the natural numbers are countable, the finite subset is also countable.
02

Case 2: Countable set is infinite and the subset is finite

In this case, we can create a bijection between the finite subset and the first few natural numbers, just like in the first case. Since the natural numbers are countable, the finite subset is countable as well.
03

Case 3: Countable set and the subset are both infinite

Given that both the countable set (A) and its subset (B) are infinite, we can enumerate the elements of the countable set A as \(a_1, a_2, a_3, ...\). For each element in the subset B, we will denote it as \(b_i\), where \(b_i\) is an element of A, and i is the index of the element in B. Now we will define a bijection between the natural numbers (1, 2, 3, ...) and the subset B. Since B is an infinite subset of A, we can list its elements according to the order that they appear in the enumeration of A. For example, the first element of B will be the element with the lowest index i such that \(a_i \in B\). We can continue this process for all elements in B, creating a bijection between the natural numbers and B. Since the natural numbers are countable, B is also countable. To conclude, we have shown that every subset of a countable set is countable, whether the countable set and its subset are finite or infinite.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Finite Subsets
A finite subset contains a specific, limited number of elements. Regardless of how large or small, these elements can be represented completely as they are limited, not infinite. Picture a bag of marbles: once you count them all, that's it, as you have reached the end of the collection. In mathematics, when we mention finite subsets, we mean the set can be paired with a portion of the natural numbers, such as 1 to 5, if the subset has five elements. This pairing, or mapping, is called a *bijection* and allows for each element in the subset to be matched with exactly one natural number. Such a mapping demonstrates that finite subsets are indeed countable. The main takeaway is that finite means limited and graspable, making finite subsets always countable, no matter the starting or parent countable set.
Infinite Subsets
An infinite subset, as opposed to a finite one, extends indefinitely. It could contain elements that do not stop, much like counting natural numbers one after another forever. When we say a set is infinite, it still remains countable if we can establish a systematic way to list all its elements. The intriguing part about infinite subsets of countable sets is that they don't run out of elements either; however, they can still be listed in a sequence. We rely on an ordered, eternal recounting, similar to naming each star in an infinite sky one by one. Through this endless yet structured listing, we can prove that infinite subsets are countable, as each element in these subsets can be mapped in a one-to-one manner with natural numbers, ensuring all elements are accounted for.
Bijection
A bijection is like a perfect matchmaker. It ensures that for every element in one set, there is exactly one partner in another set. This concept is quite useful in proving countability because if every element in a subset can be matched with a unique natural number, the subset can indeed be deemed countable. Think of it as arranging a flawless seating chart: each seat has precisely one guest assigned. In mathematics, creating a bijection between two sets establishes a definitive alignment, whereby each member of one set pairs with one and only one member of another set, with none left out or doubled. For subsets of countable sets, demonstrating a bijection confirms they are also countable, whether they are finite or infinite.
Natural Numbers
Natural numbers are the foundation of counting. Beginning from 1 and progressing upwards indefinitely (such as 1, 2, 3, ...), they serve as a basic set of elements in mathematics. Their endless nature allows them to mesh seamlessly with both finite and infinite subsets. Because natural numbers themselves are infinitely countable, they provide a standard measure against which any countable set or subset can be compared or linked through a bijection. This comparison is key: if a subset can be matched completely with a section of natural numbers, that subset is countable. Even for infinite subsets, aligning them in a one-to-one fashion with natural numbers reveals their countability. In simple terms, natural numbers are like an infinite, ever-ready yardstick for measuring any set that seeks to be countable.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(m, n \in \mathbb{N}\) and let \(x \in \mathbb{R}\) be such that \(x \geq 0\) and \(x \neq 1\). Show that \(\frac{x^{m}-1}{m}>\frac{x^{n}-1}{n} \quad\) if \(m>n \quad\) and \(\quad \frac{x^{m}-1}{m}<\frac{x^{n}-1}{n} \quad\) if \(mn\). Write \(n\left(x^{m}-1\right)-m\left(x^{m}-1\right)\) as \((x-1)\left[n\left(x^{n}+x^{n+1}+\cdots+x^{m-1}\right)-(m-n)\left(1+x+\cdots+x^{n-1}\right)\right]\). Compare the \(m-n\) elements \(x^{n}, x^{n+1}, \ldots, x^{m-1}\) as well as the \(n\) elements \(1, x, \ldots, x^{n-1}\) with \(x^{n}\), when \(x<1\) and \(x>1 .\) )

Show that if \(n \in \mathbb{N}\) and \(a_{1}, \ldots, a_{n}\) are nonnegative real numbers, then \(\left(a_{1}+\cdots+a_{n}\right)^{2} \leq n\left(a_{1}^{2}+\cdots+a_{n}^{2}\right) .\) (Hint: Write \(\left(a_{1}+\cdots+a_{n}\right)^{2}\) as \(t_{1}+\cdots+t_{n}\) where \(\left.t_{k}:=a_{1} a_{k}+a_{2} a_{k+1}+\cdots+a_{n-k+1} a_{n}+a_{n-k+2} a_{1}+\cdots+a_{n} a_{k-1} .\right)\) [Note: Exercise 35 gives an alternative approach to this inequality.]

Prove that every rational number \(r\) can be written as \(r=\frac{p}{q}\), where \(p, q \in \mathbb{Z}, q>0\) and \(p, q\) are relatively prime, and moreover the integers \(p\) and \(q\) are uniquely determined by \(r .\)

Let \(n \in \mathbb{N}\) and \(a_{1}, \ldots, a_{n}\) be positive real numbers. Prove that $$ \sqrt[n]{a_{1} \cdots a_{n}} \geq \frac{n}{r} \quad \text { where } \quad r:=\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}} $$ and that equality holds if and only if \(a_{1}=\cdots=a_{n}\). [Note: The above result is sometimes called the G.M.-H.M. inequality and \(n / r\) is called the harmonic mean of \(a_{1}, \ldots, a_{n}\).]

Let \(p(x), q(x) \in \mathbb{R}[x]\). Show that \(p(x)\) and \(q(x)\) are relatively prime if and only if \(u(x) p(x)+v(x) q(x)=1\) for some \(u(x), v(x) \in \mathbb{R}[x] .\) Is it true that if a nonzero polynomial \(d(x) \in \mathbb{R}[x]\) satisfies \(u(x) p(x)+v(x) q(x)=d(x)\) for some \(u(x), v(x) \in \mathbb{R}[x]\), then \(d(x)\) is a GCD of \(p(x)\) and \(q(x)\) ?
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.