Let $$R$$ be the real line. Consider the following subsets of the plane $$R \times R$$: $$S = \{(x, y) : y = x + 1\ \text{and}\ 0 < x < 2\},\ T = \{(x, y) : x - y\ \text{is an integer}\}$$. Which one of the following is true?

An equivalence relation on $$R$$ must be reflexive, symmetric and transitive. We test these three properties for both $$S$$ and $$T$$.

Case 1: $$S = \{(x,y): y = x+1,\;0 \lt x \lt 2\}$$

• Reflexive test: a relation is reflexive if $$\forall x \in R,\; (x,x) \in S$$.
For an ordered pair $$(x,x)$$ to belong to $$S$$ we need $$x = x+1$$, which is impossible.
Therefore $$(x,x) \notin S$$ for every real $$x$$, so $$S$$ is not reflexive.

Since reflexivity already fails, $$S$$ cannot be an equivalence relation (no need to test symmetry and transitivity further).

Case 2: $$T = \{(x,y): x-y \text{ is an integer}\}$$

• Reflexive: For every $$x \in R$$, $$x-x = 0,$$ and $$0$$ is an integer. Hence $$(x,x) \in T$$ for all $$x$$; so $$T$$ is reflexive.

• Symmetric: Suppose $$(x,y) \in T$$. Then $$x-y$$ is an integer. Its negative $$y-x = -(x-y)$$ is also an integer, so $$(y,x) \in T$$. Hence $$T$$ is symmetric.

• Transitive: Suppose $$(x,y) \in T$$ and $$(y,z) \in T$$. Then $$x-y = m$$ and $$y-z = n$$ for some integers $$m,n$$. Adding, $$x-z = m+n$$, an integer. Thus $$(x,z) \in T$$, so $$T$$ is transitive.

Since $$T$$ satisfies all three properties, it is an equivalence relation on $$R$$.

Combining the two cases: $$S$$ is not an equivalence relation, while $$T$$ is. Hence the correct statement is:

Option D which is: $$T$$ is an equivalence relation on $$R$$ but $$S$$ is not.