Understanding Relations and Their Properties

This question involves understanding the properties of binary relations, specifically focusing on the concepts of subset inclusion (represented by '≤'), inverse relations (denoted by a superscript '-1'), and complement relations (denoted by an overline).

Analyzing Statement (i): Inverse Relation Property

Statement (i) claims: If \( R \subseteq S \), then \( S^{-1} \subseteq R^{-1} \).

Let's break this down:

• Relation Inclusion: The notation \( R \subseteq S \) means that every pair in relation R is also in relation S.
• Inverse Relation: The inverse of a relation R, denoted \( R^{-1} \), is defined as \( R^{-1} = \{ (y, x) \mid (x, y) \in R \} \).
• The Claim: The statement asserts that if R is a subset of S, then the inverse of S (\( S^{-1} \)) must be a subset of the inverse of R (\( R^{-1} \)).

To check if this is true, let's consider the definition of subset inclusion for inverse relations. If \( R \subseteq S \), does it necessarily follow that \( S^{-1} \subseteq R^{-1} \)?

Consider an element \( (y, x) \in S^{-1} \). By the definition of the inverse relation, this implies that \( (x, y) \in S \).

Now, we know \( R \subseteq S \). If \( (x, y) \in S \), does it automatically mean \( (x, y) \in R \)? No, it does not. S could contain pairs that are not in R.

Let's test with an example: Let A = {a, b} and B = {a, b}. Let R = { (a, b) } Let S = { (a, b), (b, a) } Here, \( R \subseteq S \) is true. Now let's find the inverses: \( R^{-1} = \{ (b, a) \} \) \( S^{-1} = \{ (b, a), (a, b) \} \) Is \( S^{-1} \subseteq R^{-1} \)? This means is \( \{ (b, a), (a, b) \} \subseteq \{ (b, a) \} \)? No, this is false because \( (a, b) \) is in \( S^{-1} \) but not in \( R^{-1} \).

However, the reverse property, that if \( R \subseteq S \), then \( R^{-1} \subseteq S^{-1} \), is true. Let's prove this briefly: If \( (y, x) \in R^{-1} \), then \( (x, y) \in R \). Since \( R \subseteq S \), we have \( (x, y) \in S \). By the definition of inverse relation, this implies \( (y, x) \in S^{-1} \). Therefore, \( R^{-1} \subseteq S^{-1} \).

Since the statement requires \( S^{-1} \subseteq R^{-1} \), which is generally false, Statement (i) is false.

Analyzing Statement (ii): Complement Relation Property

Statement (ii) claims: If \( R \subseteq S \), then \( \overline{S} \subseteq \overline{R} \).

Let's break this down:

• Complement Relation: The complement of a relation R, denoted \( \overline{R} \), consists of all possible pairs in the universal relation (e.g., \( A \times B \)) that are *not* in R. So, \( \overline{R} = (A \times B) \setminus R \).
• The Claim: The statement asserts that if R is a subset of S, then the complement of S (\( \overline{S} \)) must be a subset of the complement of R (\( \overline{R} \)).

To check if this is true, let's assume \( R \subseteq S \). We need to determine if \( \overline{S} \subseteq \overline{R} \).

Consider an element \( (x, y) \in \overline{S} \). By the definition of the complement relation, this means \( (x, y) \notin S \).

Since \( R \subseteq S \), if a pair \( (x, y) \) is not in S, it cannot be in R either. Therefore, \( (x, y) \notin R \).

If \( (x, y) \notin R \), then by the definition of the complement relation, \( (x, y) \in \overline{R} \).

So, we have shown that if \( (x, y) \in \overline{S} \), then \( (x, y) \in \overline{R} \). This implies that \( \overline{S} \subseteq \overline{R} \).

Therefore, Statement (ii) is true.

Conclusion

Based on the analysis:

• Statement (i) is false.
• Statement (ii) is true.

The question asks which statements are false. Only statement (i) is false.