Let `S={1,2,3,...,10}` .Suppose "M" is the set of all the subsets of "S" ,then the relation
`[R={(A,B):A nn B!=phi,A,B in M}` is

Options

A. symmetric only
B. reflexive only
C. symmetric and reflexive only
D. symmetric and transitive only

Answer

To determine the properties of the relation \( R = \{(A, B) : A \cap B \neq \emptyset, A, B \in M\} \), where \( M \) is the set of all subsets of \( S = \{1, 2, 3, \ldots, 10\} \), we will check if the relation is reflexive, symmetric, or transitive.

### Step 1: Check for Reflexivity
A relation \( R \) is reflexive if for every element \( A \in M \), the pair \( (A, A) \) is in \( R \). This means we need to check if \( A \cap A \neq \emptyset \).

- For any non-empty subset \( A \), \( A \cap A = A \), which is not empty.
- However, if \( A = \emptyset \), then \( \emptyset \cap \emptyset = \emptyset \), which is equal to \( \emptyset \).

Since \( (A, A) \) is not in \( R \) when \( A = \emptyset \), the relation \( R \) is **not reflexive**.

### Step 2: Check for Symmetry
A relation \( R \) is symmetric if whenever \( (A, B) \in R \), then \( (B, A) \in R \) as well. This means we need to check if \( A \cap B \neq \emptyset \) implies \( B \cap A \neq \emptyset \).

- Since intersection is commutative, \( A \cap B = B \cap A \). Therefore, if \( A \cap B \neq \emptyset \), it follows that \( B \cap A \neq \emptyset \).

Thus, the relation \( R \) is **symmetric**.

### Step 3: Check for Transitivity
A relation \( R \) is transitive if whenever \( (A, B) \in R \) and \( (B, C) \in R \), then \( (A, C) \in R \) must also hold. This means we need to check if \( A \cap B \neq \emptyset \) and \( B \cap C \neq \emptyset \) imply \( A \cap C \neq \emptyset \).

- Consider the sets \( A = \{1, 2\} \), \( B = \{2, 3\} \), and \( C = \{4, 5\} \).
- Here, \( A \cap B = \{2\} \neq \emptyset \) and \( B \cap C = \emptyset \).
- However, \( A \cap C = \emptyset \).

Since we can find sets \( A, B, C \) such that \( A \cap B \neq \emptyset \) and \( B \cap C \neq \emptyset \) but \( A \cap C = \emptyset \), the relation \( R \) is **not transitive**.

### Conclusion
The relation \( R \) is symmetric but not reflexive and not transitive.

### Summary of Properties
- Reflexive: **No**
- Symmetric: **Yes**
- Transitive: **No**