If A, B and C are subsets of a given set, then which one of the following relations is not correct?
A∪ (A∩B) = A∪B
Understanding Set Relations
This question asks us to identify which of the given set relations is incorrect. We are given four relations involving sets A, B, and C, which are subsets of a universal set. To solve this, we will analyze each relation using the definitions of set union (\(\cup\)) and intersection (\(\cap\)).
Set Operations Explained
- Union (\(\cup\)): The union of two sets A and B, denoted by \(A \cup B\), is the set of all elements that are in A, or in B, or in both.
- Intersection (\(\cap\)): The intersection of two sets A and B, denoted by \(A \cap B\), is the set of all elements that are common to both A and B.
Analyzing Each Set Relation Option
Option 1: \(A \cup (A \cap B) = A \cup B\)
Let's analyze the left side of the equation: \(A \cup (A \cap B)\).
- \(A \cap B\) represents the elements that are in both set A and set B.
- When we take the union of set A with the set \(A \cap B\), we are combining all elements in A with all elements in \(A \cap B\).
- Since every element in \(A \cap B\) is also an element of A (by definition of intersection), the set \(A \cap B\) is a subset of A (\(A \cap B \subseteq A\)).
- Therefore, the union of A with a subset of A is just A itself.
So, \(A \cup (A \cap B) = A\).
Now, let's look at the right side of the equation: \(A \cup B\).
The relation becomes \(A = A \cup B\). This equality holds true only if set B is a subset of set A (\(B \subseteq A\)). However, this is not true for all possible sets A and B. For example, if B has elements not in A, then \(A \cup B\) will contain elements that are not in A, making \(A \neq A \cup B\).
Since the relation \(A \cup (A \cap B) = A \cup B\) is equivalent to \(A = A \cup B\), which is not universally true for all subsets A and B, this relation is incorrect.
Option 2: \(A \cap (A \cup B) = A\)
Let's analyze the left side of the equation: \(A \cap (A \cup B)\).
- \(A \cup B\) represents the elements that are in set A or set B (or both).
- When we take the intersection of set A with the set \(A \cup B\), we are finding the elements that are common to both A and \(A \cup B\).
- Since every element in A is also an element of \(A \cup B\) (by definition of union), set A is a subset of \(A \cup B\) (\(A \subseteq A \cup B\)).
- Therefore, the intersection of A with a set that contains A is just A itself.
So, \(A \cap (A \cup B) = A\).
The right side is A. The relation \(A = A\) is always correct. This relation is a correct set identity.
Option 3: \((A \cap B) \cup C = (A \cup C) \cap (B \cup C)\)
This relation represents one of the distributive laws of set theory: the union is distributive over the intersection. This is a standard and correct set identity.
It states that taking the intersection of A and B first, then uniting the result with C, is equivalent to uniting A with C and uniting B with C separately, and then taking the intersection of those two results.
This relation is correct.
Option 4: \((A \cup B) \cap C = (A \cap C) \cup (B \cap C)\)
This relation represents the other distributive law of set theory: the intersection is distributive over the union. This is also a standard and correct set identity.
It states that taking the union of A and B first, then intersecting the result with C, is equivalent to intersecting A with C and intersecting B with C separately, and then taking the union of those two results.
This relation is correct.
Conclusion
Based on our analysis, the relation that is not correct is \(A \cup (A \cap B) = A \cup B\). The correct identity is \(A \cup (A \cap B) = A\).
| Relation | Analysis | Correctness |
|---|---|---|
| \(A \cup (A \cap B) = A \cup B\) | LHS simplifies to \(A\). RHS is \(A \cup B\). \(A = A \cup B\) is not always true. | Not Correct |
| \(A \cap (A \cup B) = A\) | LHS simplifies to \(A\). RHS is \(A\). \(A = A\) is always true. | Correct |
| \((A \cap B) \cup C = (A \cup C) \cap (B \cup C)\) | Distributive Law of Union over Intersection. | Correct |
| \((A \cup B) \cap C = (A \cap C) \cup (B \cap C)\) | Distributive Law of Intersection over Union. | Correct |
Revision Table: Key Set Identities
| Identity Name | Relation |
|---|---|
| Idempotent Laws | \(A \cup A = A\) \(A \cap A = A\) |
| Commutative Laws | \(A \cup B = B \cup A\) \(A \cap B = B \cap A\) |
| Associative Laws | \((A \cup B) \cup C = A \cup (B \cup C)\) \((A \cap B) \cap C = A \cap (B \cap C)\) |
| Distributive Laws | \(A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\) \(A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\) |
| Identity Laws | \(A \cup \emptyset = A\) \(A \cap U = A\) \(A \cup U = U\) \(A \cap \emptyset = \emptyset\) |
| Complement Laws | \(A \cup A' = U\) \(A \cap A' = \emptyset\) \((A')' = A\) \(\emptyset' = U\) \(U' = \emptyset\) |
| De Morgan's Laws | \((A \cup B)' = A' \cap B'\) \((A \cap B)' = A' \cup B'\) |
| Absorption Laws | \(A \cup (A \cap B) = A\) \(A \cap (A \cup B) = A\) |
Additional Information: Absorption Laws
The first two options in the question relate to what are known as the Absorption Laws in set theory. These laws describe how union and intersection operations interact when one set is a subset of the result of the operation involving the other set.
- The first absorption law is \(A \cup (A \cap B) = A\). We used this to show that Option 1 was incorrect because it claimed this expression was equal to \(A \cup B\).
- The second absorption law is \(A \cap (A \cup B) = A\). This was shown to be correct in Option 2.
Understanding these fundamental identities and laws is crucial for working with sets and simplifying set expressions in discrete mathematics and logic.
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