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Chapter 1: Problem 10

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

Short Answer

Expert verified
Question: Prove that \(A \cup B = (A \cap B) \cup (A \backslash B) \cup (B \backslash A)\). Answer: To prove the given statement, we showed that \((A \cap B) \cup (A \backslash B) \cup (B \backslash A) \subseteq A \cup B\) and \(A \cup B \subseteq (A \cap B) \cup (A \backslash B) \cup (B \backslash A)\). Since both directions of the subset relationship hold, this implies that the two sets are equal. Therefore, we can conclude that \(A \cup B = (A \cap B) \cup (A \backslash B) \cup (B \backslash A)\).

Step by step solution

01

Define union, intersection, and set difference operations

Union (\(A \cup B\)): The union of two sets A and B is the set that contains all elements that are either in A, or in B, or in both. Intersection (\(A \cap B\)): The intersection of two sets A and B is the set that contains all elements that are common to both A and B. Set Difference (\(A \backslash B\)): The set difference of A and B is the set of all elements in A that are not in B. Now that we have defined these operations, we can proceed with the proof.
02

Show that \((A \cap B) \cup (A \backslash B) \cup (B \backslash A) \subseteq A \cup B\)

Let \(x \in (A \cap B) \cup (A \backslash B) \cup (B \backslash A)\). By definition of the union operation, \(x\) should be in one of the sets \(A \cap B\), \(A \backslash B\), or \(B \backslash A\). If \(x \in A \cap B\), then \(x \in A\) and \(x \in B\), so \(x \in A \cup B\) by the definition of union. If \(x \in A \backslash B\), then \(x \in A\) and \(x \notin B\), but \(x\) is still in \(A \cup B\) by the definition of union. If \(x \in B \backslash A\), then \(x \in B\) and \(x \notin A\), but \(x\) is still in \(A \cup B\) by the definition of union. Since \(x\) belongs to \(A \cup B\) in all cases, we have that \((A \cap B) \cup (A \backslash B) \cup (B \backslash A) \subseteq A \cup B\).
03

Show that \(A \cup B \subseteq (A \cap B) \cup (A \backslash B) \cup (B \backslash A)\)

Let \(x \in A \cup B\). If \(x \in A\) and \(x \in B\), then \(x \in A \cap B\), so \(x \in (A \cap B) \cup (A \backslash B) \cup (B \backslash A)\). If \(x \in A\) and \(x \notin B\), then \(x \in A \backslash B\), so \(x \in (A \cap B) \cup (A \backslash B) \cup (B \backslash A)\). If \(x \in B\) and \(x \notin A\), then \(x \in B \backslash A\), so \(x \in (A \cap B) \cup (A \backslash B) \cup (B \backslash A)\). Since \(x\) belongs to \((A \cap B) \cup (A \backslash B) \cup (B \backslash A)\) in all cases, we have that \(A \cup B \subseteq (A \cap B) \cup (A \backslash B) \cup (B \backslash A)\).
04

Conclude the proof

By showing that \((A \cap B) \cup (A \backslash B) \cup (B \backslash A) \subseteq A \cup B\) and \(A \cup B \subseteq (A \cap B) \cup (A \backslash B) \cup (B \backslash A)\), we have proved that \(A \cup B = (A \cap B) \cup (A \backslash B) \cup (B \backslash A)\).

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Key Concepts

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

Union
The union of two sets, denoted by \( A \cup B \), is a fundamental concept in set theory. Imagine you have two groups of elements, one in Set A and one in Set B. The union combines these groups into a single set that contains all distinct elements from both groups.

To visualize this, think of combining two boxes of toys, where \( A \) is Box A and \( B \) is Box B. The union \( A \cup B \) is like pouring both boxes into a third box without checking the origin, so every toy appears in the new box.

In mathematical terms, for any element \( x \), if \( x \) is in Set A or Set B, or in both, \( x \) will appear in \( A \cup B \). This means the union takes all elements available in either set without duplication. It's like making an inclusive friendship group鈥攁ll friends from both lists are welcomed.
Intersection
The intersection of two sets, denoted by \( A \cap B \), refers to the set containing only the elements that are common to both sets. This operation is a way of narrowing down to shared elements. If you think of Set A and Set B as circles in a Venn Diagram, the intersection is the overlapping part.

Let's visualize it with the toy box example: if Box A and Box B both have toys and we want to know which toys they both share, the intersection \( A \cap B \) gives us that smaller set. It's like focusing on what both friends lists have in common, highlighting the mutual ones.

In terms of logic, any element \( x \) is in \( A \cap B \) only if \( x \) is present in both A and B. So, this operation is like finding the intersection at a crossroads鈥攖he place where paths cross and elements meet.
Set Difference
The set difference \( A \backslash B \), also known as the relative complement, involves elements in Set A that are not in Set B. It helps in identifying what's unique to one set. Here, you look at everything present in A, excluding what's also in B.

Consider our box of toys: if we want toys only from Box A but not those that also appear in Box B, we calculate \( A \backslash B \). This operation is akin to decluttering, keeping just the elements distinctive to A.

Similarly, the difference \( B \backslash A \) would yield items found exclusively in Box B, without overlapping with Box A. It's useful for focusing on distinctions between groups, sharpening the understanding of what makes each set unique.
  • Set difference means filtering out commonalities.
  • It reveals exclusive elements to one set.
This concept is crucial for selectively approaching elements based on specific criteria.

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Most popular questions from this chapter

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

Determine whether or not the following relations are equivalence relations on the given set. If the relation is an equivalence relation, describe the partition given by it. If the relation is not an equivalence relation, state why it fails to be one. (a) \(x \sim y\) in \(\mathbb{R}\) if \(x \geq y\) (c) \(x \sim y\) in \(\mathbb{R}\) if \(|x-y| \leq 4\) (b) \(m \sim n\) in \(\mathbb{Z}\) if \(m n>0\) (d) \(m \sim n\) in \(\mathbb{Z}\) if \(m \equiv n(\bmod 6)\)

If \(A=\\{a, b, c\\}, B=\\{1,2,3\\}, C=\\{x\\},\) and \(D=\emptyset,\) list all of the elements in each of the following sets. (a) \(A \times B\) (c) \(A \times B \times C\) (b) \(B \times A\) (d) \(A \times D\)

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.

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 \(f^{-1}\left(Y \backslash B_{1}\right)=X \backslash f^{-1}\left(B_{1}\right)\).
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