Prob. 5 (c), Sec. 3, in Munkres' TOPOLOGY, 2nd ed: How to find this equivalence relation? Let $S$ be the following subset of the plane:
$$S \colon= \{ \ x \times y \ | \ y = x+1, \ 0 < x < 2 \ \}.$$
Let $T$ be an equivalence relation on the real line such that $T$ is the intersection of all the equivalence relations on the real line that contain $S$.
Then how to describe the ordered pairs in $T$ completely?
My effort:

The following subset of the plane, which is an equivalence relation on the real line, contains $S$ as a proper subset:
$$S^\prime \colon= \{ \ x \times y \ | \ y-x \in \mathbb{Z} \ \}.$$
So $$T \subset S^\prime.$$


For any $x \in (0,2)$, we have $x \times (x+1) \in T$, and so $(x+1) \times x \in T$ also, by virtue of symmetry. So, by transitivity, we can conclude that $x \times x \in T$ and $(x+1) \times (x+1) \in T$ for all $x \in (0,2)$.

What next?

Hint:
You are almost ready.
I preassume that in your question $x\times y$ is a notation for an ordered pair.
Also $(0,3)^2\cup\{x\times x\mid x\in\mathbb R\}$ is an equivalence relation that contains $S$.
So have a look at its intersection with $S'$.

edit (theoretical part, practicizing my own notation for ordered pair)
For any $A\subseteq\mathbb{R}^{2}$ define $A^{op}=\left\{ \left(x,y\right)\mid\left(y,x\right)\in A\right\} $
and define $\triangle:=\left\{ \left(x,x\right)\mid x\in\mathbb{R}\right\} $.
A relation is reflexive if and only if it contains $\triangle$. 
A relation is $Q$ symmetric if and only if $Q^{op}\subseteq Q$ (or equivalently $Q^{op}= Q$). 
Note that $\left(A^{op}\right)^{op}=A$.
Then $T:=S\cup S^{op}$ is the smallest symmetric relation
that contains $S$. 
Now define $T^{n}$ by stating $T^{0}=\triangle$
and $T^{n+1}=T^{n}\circ T$. 
Then the relation $E:=\cup_{n=0}^{\infty}T^{n}$ is transitive and
is the smallest equivalence relation containing $S$.
