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Chapter 2: Problem 27

Prove that \(\mathcal{P}(A) \subseteq \mathcal{P}(B)\) if and only if \(A \subseteq B\)

Short Answer

Expert verified
Prove \(A \backslashsubseteq B\) if \(\backslashmathcal{P}(A)\ \backslashsubseteq \backslashmathcal{P}(B)\) and vice versa.

Step by step solution

01

- Understand the problem statement

We need to prove that the power set of set A, \(\backslashmathcal{P}(A)\), is a subset of the power set of set B, \(\backslashmathcal{P}(B)\), if and only if set A is a subset of set B.
02

- Prove the forward direction

Assume \(\backslashmathcal{P}(A)\ \backslashsubseteq \backslashmathcal{P}(B)\). We need to show that \(A \backslashsubseteq B\). \Let \(x\ \backslashein \backslashmathcal{P}(A)\). By definition, \(x\) is a subset of \(A\). Since \(\backslashmathcal{P}(A)\ \backslashsubseteq \backslashmathcal{P}(B)\), \(x\) must also be an element of \( \backslashmathcal{P}(B) \). This means \(x\) is a subset of \(B\). Consequently, every element of \(A\) must also be an element of \(B\). Therefore, \(A \backslashsubseteq B\).
03

- Prove the backward direction

Assume \(A \backslashsubseteq B\). We need to show that \(\backslashmathcal{P}(A)\ \backslashsubseteq \backslashmathcal{P}(B)\). \Let \(x \backslashein \backslashmathcal{P}(A)\). By definition, \(x \backslasubseteq A\). Since \(A \backslashsubseteq B\), \(x \backslasubseteq B\). Therefore, \(x\) is an element of \( \backslashmathcal{P}(B) \). So, \(\backslashmathcal{P}(A)\ \backslashsubseteq \backslashmathcal{P}(B)\).

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Key Concepts

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

Power Set
The power set of a given set A, denoted as \(\backslashmathcal{P}(A)\), is the set of all subsets of A. This includes every possible combination of elements in A, including the empty set and A itself. For example, if A = \{1, 2\}, the power set of A, \(\backslashmathcal{P}(A)\), would be \{\{}, \{1\}, \{2\}, \{1, 2\}\}. Each element of the power set is itself a set. Power sets are essential in set theory since they help us understand the structure and hierarchy of sets.
Subset
A subset is a set where all of its elements are also contained in another set. Formally, a set A is a subset of set B, denoted as \(A \subseteq B\), if every element of A is also an element of B. For example, if B = \{1, 2, 3\} and A = \{1, 2\}, then \(A \subseteq B\). Note that every set is a subset of itself. Understanding subsets is crucial because it forms the basis for more complex relations between sets, like power sets and their interactions.
Proof by Contradiction
Proof by contradiction is a technique used to prove a statement by assuming the opposite is true and then showing this leads to an inconsistency. The general approach involves:
  • Assuming the negation of the statement you wish to prove.
  • Using logical reasoning to arrive at a contradiction or an absurd result.
  • Concluding that the original assumption must therefore be true.
This method can be particularly effective when dealing with complex statements about sets. It demonstrates the validity of many propositions in discrete mathematics, including subset relationships.
Set Theory
Set theory is a fundamental branch of mathematics that deals with the study of sets, which are collections of objects. Some basic concepts in set theory include:
  • Elements: Objects contained in a set.
  • Subsets: Sets that contain only elements from another set.
  • Power sets: Sets of all subsets of a given set.
  • Union, intersection, and difference: Operations that combine or compare sets.
Set theory provides the foundational language for various other mathematical disciplines and is crucial for understanding logic, probability, and more advanced topics in discrete mathematics.

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Most popular questions from this chapter

Compute each of these double sums. $$ \begin{array}{ll}{\text { a) } \sum_{i=1}^{2} \sum_{j=1}^{3}(i+j)} & {\text { b) } \sum_{i=0}^{2} \sum_{j=0}^{3}(2 i+3 j)} \\ {\text { c) } \sum_{i=1}^{3} \sum_{j=0}^{2} i} & {\text { d) } \sum_{i=0}^{2} \sum_{j=1}^{3} i j}\end{array} $$

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