Analyzing Set Operations for Sets A and B

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.

Statement 1: (A - B) ∪ B = A

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 \(x\).

• If \(x \in A\) and \(x \in B\), then \(x \notin A - B\), but \(x \in B\). So \(x \in (A - B) \cup B\). Here, \(x \in A\).
• If \(x \in A\) and \(x \notin B\), then \(x \in A - B\), and \(x \notin B\). So \(x \in (A - B) \cup B\). Here, \(x \in A\).
• If \(x \notin A\) and \(x \in B\), then \(x \notin A - B\), but \(x \in B\). So \(x \in (A - B) \cup B\). Here, \(x \notin A\).
• If \(x \notin A\) and \(x \notin B\), then \(x \notin A - B\), and \(x \notin B\). So \(x \notin (A - B) \cup B\). Here, \(x \notin A\).

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.

Statement 2: (A - B) ∪ A = A

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 \(x\):

• If \(x \in (A - B) \cup A\), then \(x \in (A - B)\) or \(x \in A\).
  • If \(x \in (A - B)\), then \(x \in A\) and \(x \notin B\). Thus \(x \in A\).
  • If \(x \in A\), then \(x \in A\).
  Thus, if \(x \in (A - B) \cup A\), then \(x \in A\).
• If \(x \in A\), then \(x \in (A - B) \cup A\) is true since A is part of the union.

Both inclusions hold, so \((A - B) \cup A = A\).

Therefore, statement 2 is correct.

Statement 3: (A - B) ∩ B = ϕ

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 must be in \(A - B\), meaning it is in A and not in B.
• It must be in B.

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.

Statement 4: A ⊆ B ⇒ A ∪ B = B

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 \(y \in B\).

• If \(y \in A\), since \(A \subseteq B\), it follows that \(y \in B\).
• If \(y \in B\), then \(y \in B\).

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.

Summary of Correct Statements

Based on our analysis, the correct statements are:

• Statement 2: \((A - B) \cup A = A\)
• Statement 3: \((A - B) \cap B = \varphi\)
• Statement 4: \(A \subseteq B \implies A \cup B = B\)

Statement 1: \((A - B) \cup B = A\) was found to be incorrect; it equals \(A \cup B\).

Final Answer

The statements that are correct are 2, 3, and 4.

Revision Table: Set Operations Properties

Additional Information: Understanding Set Theory

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.

• Venn Diagrams: These diagrams are often used to visually represent sets and their relationships, including the outcomes of set operations. For example, a Venn diagram can clearly show that the region representing \(A - B\) and the region representing B have no overlap, illustrating that \((A - B) \cap B = \varphi\).
• Algebra of Sets: Set operations follow certain laws, similar to algebra with numbers. These include commutative, associative, distributive, and identity laws. For instance, the statement \((A - B) \cup A = A\) can be seen as a form of the absorption law when considering \(A - B\) as a part of A.
• Set Difference Property: The set difference \(A - B\) can be formally defined as \(A \cap B'\), where \(B'\) is the complement of B with respect to a universal set U. This definition is useful in proving identities involving set difference using other known set properties.

Mastering these basic concepts and properties is key to solving more complex problems in set theory.