/*! 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 19 Let \(S=\\{a, b, c\\}\) and for ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 19

Let \(S=\\{a, b, c\\}\) and for each integer \(i=0,1,2,3\), let \(S_{i}\) be the set of all subsets of \(S\) that have \(i\) elements. List the elements in \(S_{0}, S_{1}, S_{2}\), and \(S_{3}\). Is \(\left\\{S_{0}, S_{1}, S_{2}, S_{3}\right\\}\) a partition of \(\mathscr{P}(S)\) ?

Short Answer

Expert verified
The sets \(S_{0} = \{ \emptyset \}\), \(S_{1} = \{ \{a\}, \{b\}, \{c\} \}\), \(S_{2} = \{ \{a, b\}, \{a, c\}, \{b, c\} \}\), and \(S_{3} = \{ \{a, b, c\} \}\) have no elements in common and their union is the power set of S: \(\mathscr{P}(S) = \{ \emptyset, \{a\}, \{b\}, \{c\}, \{a, b\}, \{a, c\}, \{b, c\}, \{a, b, c\} \}\). Therefore, \(\{S_{0}, S_{1}, S_{2}, S_{3}\}\) is a partition of \(\mathscr{P}(S)\).

Step by step solution

01

List out subsets of S that have i elements

Here we will list the subsets of S that have i elements for i from 0 to 3. - For i=0, S鈧� = {} (empty set) because there are no elements in it. - For i=1, S鈧�= { {a}, {b}, {c} } as there is only one element in each subset - For i=2, S鈧�= { {a, b}, {a, c}, {b, c} } because there are two elements in each subset - For i=3, S鈧� = { {a, b, c} } as there are three elements in each subset
02

Check if the given sets form a partition of 饾挮(S)

In order to check if {S鈧�, S鈧�, S鈧�, S鈧儅 forms a partition of 饾挮(S), we need to ensure that the union of all sets Si (i=0, 1, 2, 3) contains every possible subset of S and that each subset Si does not have any elements in common. First, let's find the union of the sets Si (i=0,1,2,3): {S鈧�, S鈧�, S鈧�, S鈧儅 = { {}, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} } Next, we must ensure that none of the sets Si have common elements: - S鈧� does not have any elements in common with other sets. - S鈧� has single element subsets and none of them appear in other sets. - S鈧� has two-element subsets and none of them appear in other sets. - S鈧� has three-element subset {a, b, c} which doesn't appear in other sets. Now, we compare the union of Si sets with the power set of S, which should contain all possible subsets of S, 饾挮(S) = { {}, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} } As we can see, {S鈧�, S鈧�, S鈧�, S鈧儅 contains all possible subsets of S and none of the sets Si have common elements, therefore {S鈧�, S鈧�, S鈧�, S鈧儅 is a partition of 饾挮(S).

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.

Power Set
A power set is a fundamental concept in set theory. When we talk about the power set of a given set, denoted as \( \mathscr{P}(S) \), we are referring to the set of all possible subsets of \( S \), including the empty set and the set itself. For example, if we have a set \( S=\{a, b, c\} \), the power set of \( S \), denoted as \( \mathscr{P}(S) \), would include the following: \( \mathscr{P}(S) = \{ \{\}, \{a\}, \{b\}, \{c\}, \{a, b\}, \{a, c\}, \{b, c\}, \{a, b, c\} \} \).

Characteristics of a Power Set

  • A power set includes all the combinations of the elements of the set it is derived from.
  • The number of elements in a power set is always \(|\mathscr{P}(S)|=2^n\), where \(|S|=n\) is the number of elements in the original set. In this case, \(|S|=3\), so the power set should have \(|\mathscr{P}(S)|=2^3=8\) elements.
  • Understanding power sets helps in calculating the number of possible relations, functions, and many other constructs that are based on set theory.
As evident from the exercise, finding the power set is a primary step before we can discuss partitions of that set or its subsets.
Partition of a Set
A partition of a set is an interesting and useful concept that is often represented in the Venn diagram. When we say a set \( A \) has a partition, we mean that \( A \) can be broken down into distinct subsets, \( A_1, A_2, ..., A_n \), such that every element in \( A \) is included in exactly one of these subsets, and none of these subsets are empty.

A partition must satisfy the following conditions:
  • The union of the subsets must give us back the original set \( A \).
  • The intersection of any two distinct subsets must be empty, \( A_i \cap A_j = \varnothing \) for any \( i eq j \).
  • No subset in the partition is empty, excluding the possibility of the empty set being a part of the partition.
From the exercise, \( \{S_0, S_1, S_2, S_3\} \) forms a partition of the power set of \( S \) because it satisfies all the conditions: each subset is distinct and covers the entire power set without any overlap.
Subsets of a Set
The concept of subsets is one of the pillars of set theory. A subset of a set \( A \) is any set \( B \) where every element of \( B \) is also an element of \( A \). In formal terms, \( B \) is a subset of \( A \) if \( \forall b \in B, b \in A \) as well. One special subset that often causes confusion is the empty set \( \{\} \), which is considered a subset of every set.

Types of Subsets

  • Proper Subset: If \( B \) contains some but not all elements of \( A \) or is the empty set, then \( B \) is a proper subset of \( A \), denoted by \( B \subset A \).
  • Improper Subset: If \( B \) contains all elements of \( A \) (including the possibility that \( B = A \) ), it is an improper subset, denoted by \( B \subseteq A \).
To link back to the exercise, \( S_i \) for \( i=0,1,2,3 \) are subsets of the power set \( \mathscr{P}(S) \) with specific numbers of elements, demonstrating the concept of subsets with a level of granularity.

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

For all sets \(A\) and \(B,\left(B^{c} \cup\left(B^{c}-A\right)\right)^{c}=B\).

Assume that \(B\) is a Boolean algebra with operations \(+\) and . Prove each statement without using any parts of Theorem \(5.3 .2\) unless they have already been proved. You may use any part of the definition of a Boolean algebra and the results of previous exercises, however. For all \(x, y\), and \(z\) in \(B\), if \(x+y=x+z\) and \(x \cdot y=x \cdot z\), then \(y=z\).

Write a negation for each of the following statements. Indicate which is true, the statement or its negation. Justify your answers. a. \(\forall \operatorname{sets} S, \exists\) a set \(T\) such that \(S \cap T=\emptyset .\) b. \(\exists\) a set \(S\) such that \(\forall \operatorname{sets} T, S \cup T=\emptyset\).

Prove each statement that is true and find a counterexample for each statement that is false. Assume all sets are subsets of a universal set \(U\). For all sets \(A\) and \(B, A \cap(A \cup B)=A\).

The following problem, devised by Ginger Bolton, appeared in the January 1989 issue of the College Mathematics Journal (Vol. 20, No. 1, p. 68): Given a positive integer \(n \geq 2\), let \(S\) be the set of all nonempty subsets of \(\\{2,3, \ldots, n\\}\). For each \(S_{i} \in S\), let \(P_{i}\) be the product of the elements of \(S_{i}\). Prove or disprove that $$ \sum_{i=1}^{2^{n-1}-1} P_{i}=\frac{(n+1) !}{2}-1 $$ In 23 and 24 supply a reason for each step in the derivation.
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Probability and Statistics

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.