Problem 3
Question
True or False The intersection of two sets is always a subset of their union.
Step-by-Step Solution
Verified Answer
True
1Step 1: Understand the key concepts
Review the definitions of 'intersection' and 'union' of sets. The intersection of two sets contains all elements that are common to both sets, while the union contains all elements from both sets.
2Step 2: Define two sets
Consider two sets, A and B. For example, let A = {1, 2, 3} and B = {3, 4, 5}.
3Step 3: Find the intersection
Determine the intersection of A and B, denoted as A ∩ B. In this case, A ∩ B = {3}, because 3 is the only element common to both sets.
4Step 4: Find the union
Determine the union of A and B, denoted as A ∪ B. In this case, A ∪ B = {1, 2, 3, 4, 5}, which includes all elements from sets A and B.
5Step 5: Check if intersection is a subset of union
By definition, set C is a subset of set D if every element of C is also an element in D. In this case, confirm that every element in A ∩ B is also in A ∪ B.
6Step 6: Conclusion
Since the element '3' in A ∩ B is also in A ∪ B, we confirm that the intersection of two sets is always a subset of their union.
Key Concepts
Intersection of SetsUnion of SetsSubset
Intersection of Sets
Understanding the concept of intersection is key in set theory. The intersection of two sets, denoted by \( A \cap B \), includes only the elements that are common to both sets.
For example, if we have sets A = {1, 2, 3} and B = {3, 4, 5}, their intersection \( A \cap B \) is {3}.
It’s important to note that if there are no common elements, the intersection is an empty set, denoted as \( \varnothing \).
For example, if we have sets A = {1, 2, 3} and B = {3, 4, 5}, their intersection \( A \cap B \) is {3}.
It’s important to note that if there are no common elements, the intersection is an empty set, denoted as \( \varnothing \).
- The intersection helps in understanding shared characteristics or elements.
- This concept is pivotal in mathematical proofs and applications.
Union of Sets
The union of sets, written as \( A \cup B \), combines all elements from both sets, eliminating any duplicates.
For instance, with sets A = {1, 2, 3} and B = {3, 4, 5}, their union \( A \cup B \) is {1, 2, 3, 4, 5}.
Every element from both sets is included in the union.
Key points to remember:
For instance, with sets A = {1, 2, 3} and B = {3, 4, 5}, their union \( A \cup B \) is {1, 2, 3, 4, 5}.
Every element from both sets is included in the union.
Key points to remember:
- The union encompasses all unique elements from the sets involved.
- Union operations are often useful in areas such as probability, logical operations, and database queries.
Subset
A set C is considered a subset of set D, denoted \( C \subseteq D \) if every element of C is also an element of D.
This means that C can be fully contained within D.
For example, let’s say C = {3} and D = {1, 2, 3, 4, 5}; clearly, every element of C is also in D.
This concept helps in defining hierarchical relationships among sets.
Important notes on subsets:
This means that C can be fully contained within D.
For example, let’s say C = {3} and D = {1, 2, 3, 4, 5}; clearly, every element of C is also in D.
This concept helps in defining hierarchical relationships among sets.
Important notes on subsets:
- If \( C \subseteq D \), there’s no element in C that isn’t in D.
- This is crucial in proofs, as subsets demonstrate containment and ordering within sets.
Other exercises in this chapter
Problem 2
The binomial coefficient \(\left(\begin{array}{l}6 \\ 4\end{array}\right)\) equals (a) \(\frac{6 !}{4 !}\) (b) \(\frac{6 !}{4 ! \cdot 2 !}\) (c) \(\frac{6 !}{2
View solution Problem 4
True or False. In a probability model, the sum of all probabilities is \(1 .\)
View solution Problem 4
True or False If \(A\) is a set, the complement of \(A\) is the set of all the elements in the universal set that are not in \(A\).
View solution Problem 5
In a probability model, which of the following numbers could be the probability of an outcome? $$\begin{array}{llllll}0 & 0.01 & 0.35 & -0.4 & 1 & 1.4\end{array
View solution