Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

(a) Prove that if \(X\) and \(Y\) are countable sets, so are \(X \cup Y, X \cap Y, X-Y\), and \(X \times Y\). (Caution: Countable means either finite or countably infinite, so there may be separate cases to consider.) (b) If \(X\) and \(Y\) are countably infinite, which of the following sets must be countably infinite: \(X \cup Y, X \cap Y, X-Y,\) and \(X \times Y ?\)

Short Answer

Expert verified
(a) All sets are countable. (b) \(X \cup Y\) and \(X \times Y\) are countably infinite, \(X \cap Y\) and \(X - Y\) can be finite or countably infinite.

Step by step solution

01

Definitions and Cases for Countable Sets

A set is countable if it is either finite or countably infinite, meaning there is a bijection with the natural numbers. Consider sets \(X\) and \(Y\), each of which is countable.
02

Proving \(X \cup Y\) is Countable

If \(X\) and \(Y\) are both finite, \(X \cup Y\) is also finite, hence countable. If either or both are countably infinite, their union, \(X \cup Y\), can be shown to be countably infinite by constructing a bijection using the fact that the countable union of countable sets is countable (using interlacing or the pairing function).
03

Proving \(X \cap Y\) is Countable

The intersection \(X \cap Y\) is a subset of both \(X\) and \(Y\). A subset of a countable set is either finite or countably infinite; hence \(X \cap Y\) is countable (finite or countably infinite).
04

Proving \(X - Y\) is Countable

The set difference \(X - Y\) is a subset of \(X\), which is countable. A subset of a countable set is countable, so \(X - Y\) is countable (either finite or countably infinite).
05

Proving \(X \times Y\) is Countable

For \(X \times Y\), if both \(X\) and \(Y\) are finite, \(X \times Y\) is finite. If both are countably infinite, use the Cantor pairing function or diagonal argument to show \(X \times Y\) is countably infinite, as a countable product of countable sets is countable.
06

Analyzing Countably Infinite Cases

When both \(X\) and \(Y\) are countably infinite, \(X \cup Y\) can be shown to be countably infinite by interleaving elements or using bijections. \(X \cap Y\) could be finite or countably infinite, depending on the overlap of elements. \(X - Y\) remains countably infinite. \(X \times Y\) is countably infinite, proven by the pairing function or diagonal argument.

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 Vaia!

Key Concepts

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

Countably Infinite Sets
Understanding what countably infinite sets are is essential in mathematics, particularly in set theory. A set is deemed countable if we can list its elements in a sequence similar to that of natural numbers. In other words, we can establish a one-to-one correspondence (bijection) between the elements of the set and the natural numbers.
Furthermore, a countably infinite set implies that though the set is infinite, its elements can be arranged in a sequence, just like natural numbers (e.g., the set of integers). Countable does not necessarily mean all sets are infinite. A finite set is also countable because you can count its elements, even if the counting eventually stops. The distinction between finite and infinite (when there’s a bijection with natural numbers) sets is at the heart of many set-related proofs.
Bijection
A bijection is a fundamental concept in understanding countable sets. It refers to a function that perfectly pairs every element of one set with a unique element of another set. No two elements in the first set are mapped to the same element in the second. Hence, both to-and-from directions of the function achieve this unique pairing.
Bijections help demonstrate if a set is countably infinite. If we can create a bijection between a set's elements and natural numbers, we confirm the set is countably infinite. For example, considering two sets, if every element in the first set corresponds to a unique element in the second set without omission, they are bijective.
Union and Intersection of Sets
The concepts of union and intersection of sets play a crucial role in determining if combined or overlapping sets maintain countability.
  • Union of Sets (X \cup Y"): The union combines all elements from two sets, eliminating duplicates. If sets \(X\) and \(Y\) are countable, \(X \cup Y\) is countable because even in cases where one or both are countably infinite, constructing a bijection (through interleaving or pairing functions) shows it can be sequenced like natural numbers.
  • Intersection of Sets (X \cap Y"): The intersection only includes elements that appear in both sets. This is always a subset of the original sets. A subset of a countable set is countable too. Consequently, \(X \cap Y\) stays countable, whether finite or countably infinite.
Cantor Pairing Function
The Cantor pairing function is an innovative mathematical tool that helps show how the Cartesian product of countable sets is still countable. The Cartesian product \(X \times Y\) involves forming pairs consisting of one element from \(X\) and another from \(Y\).
When both \(X\) and \(Y\) are countably infinite, the Cartesian product will also be countably infinite. The Cantor pairing function explicitly constructs a bijection between these pairs and natural numbers. It uses a formula to uniquely encode each pair of numbers into a single natural number, preserving countability. This concept underpins many proofs concerning the countability of products of infinite sets.

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

A man has 10 black socks and 11 blue socks scrambled in a drawer. Still half- asleep. the man reaches in the drawer to get a pair of matching socks. How many socks should he select, one at a time, before he will be sure that he has a matching pair. How many selections are needed to be sure he has a blue pair?

Let \(F\) be a function, and let \(C, D \subseteq \operatorname{domain}(F)\). a.Prove that range \((F|C\cap D)\) \(\subseteq\) range \((F \mid c) \cap\) range \((F \mid D)\). (b) Show by example that equality need not hold in part (a).

Suppose someone (say, Aesop) is marking days in some leap year (say, 2948). You do not know which days he marks, only how many. Use this to answer the following questions. (Warning: Some, but not all, of these questions use the Pigeon-Hole Principle.) (a) How many days would Aesop have to mark before you can conclude that he marked two days in January? (b) How many days would Aesop have to mark before you can conclude that he marked two days in February? (c) How many days would Aesop have to mark before you can conclude that he marked two days in the same month? (d) How many days would Aesop have to mark before you can conclude that he marked three days in the same month? (e) How many days would Aesop have to mark before you can conclude that he marked three days with the same date (for example, the third of three different months, or the 3 ist of three different months)? (f) How many days would Aesop have to mark before you can conclude that he marked two consecutive days (for example, January 31 and February 1 )? (g) How many days would Aesop have to mark before you can conclude that he marked three consecutive days?

Area codes are used to distinguish phone numbers for which the last seven digits are the same. If you have 35,000,000 phone numbers in a state and an area code can distinguish approximately 900,000 phone numbers, how many area codes are needed to distinguish the phone numbers of this state?

Find two functions \(F, G: \mathbb{R} \rightarrow \mathbb{R}\) where \(F \neq G\) but \(\left.F\right|_{[0,1)}=\left.G\right|_{[0,1)}\)

See all solutions

Recommended explanations on Computer Science Textbooks

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.

Sign-up for free