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Chapter 1: Problem 1

If \(A\) and \(B\) are sets, show that \(A \subseteq B\) if and only if \(A \cap B=A\).

Short Answer

Expert verified
This exercise demonstrates that if a set A is a subset of set B, then the intersection of A and B is equal to A and vice versa. Using basic set theory and definitions for union, intersection and subset, the exercise is easily proven.

Step by step solution

01

Expand the bidirectional condition

The stated exercise is a bidirectional condition, hence, it needs to be expanded and verified in both directions.
02

Show first condition, \(A \subseteq B\) implies \(A \cap B = A\)

It's given that A is a subset of B. By definition, this means that every element of A is also an element of B. The intersection of sets A and B (denoted \(A \cap B\)) is the set of elements that are common to both A and B. Therefore, as all elements of A are in B, \(A \cap B\) will also contain all elements of A (since the intersection includes all common elements). Hence, \(A \cap B = A\).
03

Show second condition, \(A \cap B = A\) implies \(A \subseteq B\)

For this condition, it's given that the intersection of sets A and B equals A. This means that all elements common to both A and B make up set A. Hence, every element of A is also an element of B. And by definition, this means that A is a subset of B (\(A \subseteq B\)).

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Most popular questions from this chapter

(a) Show that if \(f: A \rightarrow B\) is injective and \(E \subseteq A\), then \(f^{-1}(f(E))=E\). Give an example to show that equality need not hold if \(f\) is not injective. (b) Show that if \(f: A \rightarrow B\) is surjective and \(H \subseteq B\), then \(f\left(f^{-1}(H)\right)=H .\) Give an example to show that equality need not hold if \(f\) is not surjective.

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