1. Interpret the sets as relations on $\mathbb{R}$

A relation on $\mathbb{R}$ is a subset of $\mathbb{R} \times \mathbb{R}$.

We are given:

$S=\{(x,y): y=x+1 \text{ and } 0<x<2\}$

and

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

To be an equivalence relation, a relation must be:

1. Reflexive
2. Symmetric
3. Transitive

2. Check whether $S$ is an equivalence relation

The relation $S$ consists of pairs of the form

$y=x+1, \quad 0<x<2.$

So examples of pairs in $S$ are $(1,2)$, $(0.5,1.5)$, etc.

(i) Reflexive?

For reflexivity, we need $(x,x) \in S$ for every $x \in \mathbb{R}$.

But if $(x,x) \in S$, then it must satisfy

$x=x+1,$

which is impossible.

Hence $S$ is not reflexive.

So $S$ cannot be an equivalence relation.

(ii) Symmetric?

Suppose $(x,y) \in S$. Then

$y=x+1.$

For symmetry, we would need $(y,x) \in S$, i.e.

$x=y+1.$

But since $y=x+1$, this would mean

$x=(x+1)+1=x+2,$

impossible.

So $S$ is not symmetric.

(iii) Transitive?

Even without checking further, since reflexivity already fails, $S$ is not an equivalence relation.

Therefore,

$S \text{ is not an equivalence relation.}$

3. Check whether $T$ is an equivalence relation

Given

$T=\{(x,y): x-y \in \mathbb{Z}\}.$

That is, $x$ is related to $y$ iff their difference is an integer.

We now check the three properties.

(i) Reflexive

For any $x \in \mathbb{R}$,

$x-x=0,$

and $0$ is an integer.

Hence

$(x,x) \in T \quad \forall x \in \mathbb{R}.$

So $T$ is reflexive.

(ii) Symmetric

Suppose $(x,y) \in T$. Then

$x-y \in \mathbb{Z}.$

Now,

$y-x=-(x-y).$

Since the negative of an integer is also an integer,

$y-x \in \mathbb{Z}.$

Thus $(y,x) \in T$.

So $T$ is symmetric.

(iii) Transitive

Suppose $(x,y) \in T$ and $(y,z) \in T$. Then

$x-y \in \mathbb{Z}, \qquad y-z \in \mathbb{Z}.$

Adding,

$(x-y)+(y-z)=x-z \in \mathbb{Z}.$

Therefore $(x,z) \in T$.

So $T$ is transitive.

Hence,

$T \text{ is an equivalence relation.}$

4. Conclusion

• $S$ is not an equivalence relation.
• $T$ is an equivalence relation.

Therefore the correct option is

$\boxed{\text{D: } T \text{ is an equivalence relation on } \mathbb{R} \text{ but } S \text{ is not.}}$

5. Comparison with stored correct answer

Stored correct answer: $D$

Our derived answer is also $D$, so they agree.