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?

For $x\leq 0$ or $x\geq 3$ let $E(x)=\{x\}$ . Let $[x]$ denote the largest integer not exceeding $x$ . For $x\in (0,1)\cup (1,2)\cup (2,3)$ let $E(x)=\{x-[x],x-[x]+1,x-[x]+2\} $ . Let $E(1)=E(2)=\{1,2\}$. Now let $T= \{E(x)\times E(x) : x\in R\}.$ Now let let $H$ be the set of  equivalence relations $G$ on $R$ that satisfy $S\subset G .  $  Then for every $x\in R$ we have  $E(x)\subset \cap \{[x]_G :G \in H\}=[x]_T=E(x)$ because $T \in H$ and $[x]_T=E(x)$.
