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Chapter 1: Problem 14

Prove \(A \backslash(B \cup C)=(A \backslash B) \cap(A \backslash C)\).

Short Answer

Expert verified
Question: Prove that \(A \backslash(B \cup C)=(A \backslash B) \cap(A \backslash C)\) for sets A, B, and C. Answer: We have proved that \((A \backslash B) \cap(A \backslash C) = A \backslash(B \cup C)\), which shows that the set difference of A and the union of B and C is equal to the intersection of the set differences of A and B, and A and C.

Step by step solution

01

Understand the given expression

We are given the expression \(A \backslash(B \cup C)=(A \backslash B) \cap(A \backslash C)\), and we need to prove that both sides are equal. Here, \(A, B, C\) are sets, and \(\backslash\), \(\cup\), and \(\cap\) are set operations representing difference, union, and intersection, respectively.
02

Recall the definition of set difference

Set difference of two sets A and B, denoted as \(A \backslash B\), is the set of elements that belong to A but not to B. Mathematically, \(A \backslash B = \{x | x \in A \text{ and } x \notin B\}\).
03

Prove that \((A \backslash B) \cap(A \backslash C) \subseteq A \backslash(B \cup C)\)

Let \(x\) be an element of \((A \backslash B) \cap(A \backslash C)\). This means that \(x \in (A \backslash B)\) and \(x \in (A \backslash C)\). By the definition of set difference, \(x \in A\) and \(x \notin B\), and \(x \in A\) and \(x \notin C\). Therefore, \(x \notin B \cup C\), since it is not in both B and C. So, \(x \in A \backslash (B \cup C)\). This shows that \((A \backslash B) \cap(A \backslash C) \subseteq A \backslash(B \cup C)\).
04

Prove that \(A \backslash(B \cup C) \subseteq (A \backslash B) \cap(A \backslash C)\)

Now let \(y\) be an element of \(A \backslash(B \cup C)\). This means that \(y \in A\) and \(y \notin (B \cup C)\). Since \(y \notin B \cup C\), it must be the case that \(y \notin B\) and \(y \notin C\). Hence, \(y \in A \backslash B\) and \(y \in A \backslash C\). So, \(y \in (A \backslash B) \cap(A \backslash C)\). This shows that \(A \backslash(B \cup C) \subseteq (A \backslash B) \cap(A \backslash C)\).
05

Conclusion

Since we have proved that \((A \backslash B) \cap(A \backslash C) \subseteq A \backslash(B \cup C)\) and \(A \backslash(B \cup C) \subseteq (A \backslash B) \cap(A \backslash C)\), it follows that \((A \backslash B) \cap(A \backslash C) = A \backslash(B \cup C)\).

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Key Concepts

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

Set Difference
Set difference is a fundamental concept in set theory, which involves comparing two sets to identify elements that belong to one set but not to the other. For two sets, say \( A \) and \( B \), the set difference \( A \backslash B \) is defined as all the elements that are in \( A \) but not in \( B \). This can be expressed mathematically as \( A \backslash B = \{x \mid x \in A \text{ and } x otin B\} \).

When dealing with set difference, always remember:
  • It focuses on excluding elements of the second set from the first.
  • Set difference is directional, meaning \( A \backslash B eq B \backslash A \) in general.
  • This operation helps in "subtracting" one set from another.
This concept is particularly useful in various applications like database queries, where you need to filter out unwanted data.
Union of Sets
The union of sets is another key concept that combines elements from multiple sets into a single set. For sets \( B \) and \( C \), the union \( B \cup C \) includes all elements that are either in \( B \), in \( C \), or in both. Mathematically, it's expressed as \( B \cup C = \{x \mid x \in B \text{ or } x \in C\} \).

Key points to remember about set union:
  • Union operation treats repeated elements as a single entity, ensuring no duplicates in the resulting set.
  • It brings together distinct attributes from the involved sets.
  • Typically used to combine information in problems involving simultaneous conditions.
In our exercise, \( B \cup C \) means considering all elements from both sets \( B \) and \( C \), and then comparing it with the set \( A \).
Intersection of Sets
Intersection of sets focuses on identifying common elements between multiple sets. For sets \( A \) and \( B \), the intersection \( A \cap B \) is defined as the set containing all elements that are present in both \( A \) and \( B \). It can be mathematically expressed as \( A \cap B = \{x \mid x \in A \text{ and } x \in B\} \).

Some points to consider regarding set intersection:
  • It's all about finding similarities across sets.
  • If there are no common elements, the result is an empty set, denoted as \( \emptyset \).
  • This operation helps identify overlapping areas of data or elements.
In the context of the given exercise, the intersection \( (A \backslash B) \cap (A \backslash C) \) is used to demonstrate how excluding \( B \) and \( C \) separately from \( A \) leads to the same result as excluding their union.

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Most popular questions from this chapter

Prove \((A \cup B) \times C=(A \times C) \cup(B \times C)\).

Projective Real Line. \(\quad\) Define a relation on \(\mathbb{R}^{2} \backslash\\{(0,0)\\}\) by letting \(\left(x_{1}, y_{1}\right) \sim\left(x_{2}, y_{2}\right)\) if there exists a nonzero real number \(\lambda\) such that \(\left(x_{1}, y_{1}\right)=\left(\lambda x_{2}, \lambda y_{2}\right) .\) Prove that \(\sim\) defines an equivalence relation on \(\mathbb{R}^{2} \backslash(0,0)\). What are the corresponding equivalence classes? This equivalence relation defines the projective line, denoted by \(\mathbb{P}(\mathbb{R})\), which is very important in geometry.

Prove \(A \subset B\) if and only if \(A \cap B=A\).

Let \(f: X \rightarrow Y\) be a map with \(A_{1}, A_{2} \subset X\) and \(B_{1}, B_{2} \subset Y\). (a) Prove \(f\left(A_{1} \cup A_{2}\right)=f\left(A_{1}\right) \cup f\left(A_{2}\right)\). (b) Prove \(f\left(A_{1} \cap A_{2}\right) \subset f\left(A_{1}\right) \cap f\left(A_{2}\right)\). Give an example in which equality fails. (c) Prove \(f^{-1}\left(B_{1} \cup B_{2}\right)=f^{-1}\left(B_{1}\right) \cup f^{-1}\left(B_{2}\right),\) where $$ f^{-1}(B)=\\{x \in X: f(x) \in B\\} $$ (d) Prove \(f^{-1}\left(B_{1} \cap B_{2}\right)=f^{-1}\left(B_{1}\right) \cap f^{-1}\left(B_{2}\right)\). (e) Prove \(f^{-1}\left(Y \backslash B_{1}\right)=X \backslash f^{-1}\left(B_{1}\right)\).

Prove \(A \cap(B \backslash C)=(A \cap B) \backslash(A \cap C)\).
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