Consider the following in respect of sets A and B: 1. (A - B) ∪ B = A 2. (A - B) ∪ A = A 3. (A - B) ∩ B = ϕ 4. A ⊆ B ⇒ A ∪ B = B
2, 3 and 4
The question asks us to determine which of the given statements regarding set operations on sets A and B are correct. We will examine each statement individually using the definitions of the set operations involved.
This statement involves the set difference and union operations. The set difference \(A - B\) (or \(A \setminus B\)) consists of all elements that are in set A but not in set B. The union \(X \cup Y\) consists of all elements that are in set X or in set Y (or both).
Let's consider an element \(x\).
From this analysis, we see that any element in A is also in \((A - B) \cup B\). However, elements that are in B but not in A are in \((A - B) \cup B\) but not in A. Thus, \((A - B) \cup B\) is actually equal to \(A \cup B\), not necessarily A.
Therefore, statement 1 is incorrect.
This statement also involves set difference and union. \(A - B\) contains elements of A that are not in B. Every element in \(A - B\) is also an element of A by definition.
When we take the union of a set \(A - B\) with a set A, which contains \(A - B\) as a subset, the union is simply the larger set A.
Let's consider an element \(x\):
Both inclusions hold, so \((A - B) \cup A = A\).
Therefore, statement 2 is correct.
This statement involves set difference and intersection. \(A - B\) contains elements that are in A but not in B. The intersection \(X \cap Y\) contains elements that are in both set X and set Y.
For an element to be in \((A - B) \cap B\), it must satisfy two conditions simultaneously:
It is impossible for an element to be both "not in B" and "in B" at the same time. Therefore, there are no elements that can be in both \(A - B\) and B.
The intersection \((A - B) \cap B\) is the empty set, denoted by \(\varphi\).
Therefore, statement 3 is correct.
This statement involves the concept of a subset and the union operation. The symbol \(\subseteq\) means 'is a subset of'. \(A \subseteq B\) means that every element in set A is also an element in set B. The symbol \(\implies\) means 'implies' or 'if... then...'.
The statement says: If A is a subset of B, then the union of A and B is equal to B.
Let's consider the union \(A \cup B\). This set contains all elements that are in A or in B.
If \(A \subseteq B\), then any element \(x\) that is in A is also in B. So, if \(x \in A\), then \(x \in B\).
Consider an element \(y \in A \cup B\). By definition, \(y \in A\) or \(y \in B\).
In either case, if \(y \in A \cup B\) and \(A \subseteq B\), then \(y \in B\). This shows that \(A \cup B \subseteq B\).
Now consider an element \(z \in B\). By definition of union, if \(z \in B\), then \(z \in A \cup B\) is true. This shows that \(B \subseteq A \cup B\).
Since both \(A \cup B \subseteq B\) and \(B \subseteq A \cup B\), we conclude that \(A \cup B = B\) when \(A \subseteq B\).
Therefore, statement 4 is correct.
Based on our analysis, the correct statements are:
Statement 1: \((A - B) \cup B = A\) was found to be incorrect; it equals \(A \cup B\).
The statements that are correct are 2, 3, and 4.
| Operation | Notation | Definition | Example/Property |
|---|---|---|---|
| Union | \(A \cup B\) | Elements in A or B or both | \(A \cup B = B \cup A\) (Commutative) |
| Intersection | \(A \cap B\) | Elements in both A and B | \(A \cap B = B \cap A\) (Commutative) |
| Difference | \(A - B\) or \(A \setminus B\) | Elements in A but not in B | \(A - B = A \cap B'\) (where B' is complement of B) |
| Subset | \(A \subseteq B\) | Every element of A is in B | If \(A \subseteq B\), then \(A \cap B = A\) and \(A \cup B = B\) |
| Empty Set | \(\varphi\) or \(\emptyset\) | Set with no elements | \(A \cap \varphi = \varphi\), \(A \cup \varphi = A\) |
Set theory is a fundamental branch of mathematics that studies sets, which are collections of objects. Understanding basic set operations like union, intersection, and difference is crucial for many areas of mathematics and computer science.
Mastering these basic concepts and properties is key to solving more complex problems in set theory.
Consider the following statements:
Statement 1: The function f : R → R such that f(x) = x 3for all x ∈ R is one-one
Statement 2: f(a) = f(b) ⇒ a = b for all a, b ∈ R if the function f is one-one.
Which one of the following is correct in respect of the above statements?
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How many elements are there in the range of f?
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3. (x, y) ∈ R and (y, z) ∈ R ⇒ (x, z) ∈ R
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