(a) RST hold for $S'$ ($x-x=0$ , $y-x=-(x-y)$ and $z-x=(z-y)+(y-x)$ ). The classes of equivalence are those points the pairwise distances between which are integer numbers. Alternatively, we can say that the class of points equivalent to $x$ is $\{x+k|k\in\mathbb{Z}\}$ . $S$ contains some pairs such that the distance is 1, and, therefore, if $xSy$ then $y-x=1$ , implying $xS'y$ .

(b) If RST hold for any relation in the collection then they hold for any relation in the intersection. For example, if $aRa$ for any $R\in\mathcal{R}$ , then $(a,a)\in R$ for any $R\in\mathcal{R}$ , and $(a,a)\in\cap_{R\in\mathcal{R}}R$ , implying $a\cap_{R\in\mathcal{R}}Ra$ . Similarly for the other properties.

(c) Every equivalence relation that contains $S$ must have at least these pairs $(x,y)$ : $y=x$ (reflexivity), $y=x+1,0<x<2$ (contains S), $y=x-1,1<x<3$ (symmetry), and, by transitivity and symmetry, $y=x+2,0<x<1$ and $y=x-2,2<x<3$ . Let the set of these pairs be $T$ . Then $T$ is in the intersection. If we show that $T$ is an equivalence relation itself, then it is the intersection (from (b) it follows that the intersection must be an equivalence relation, and, therefore, the intersection is an equivalence relation that is contained in any other equivalence relation that contains $S$ ).

But before we show that this is an equivalence relation, let us describe T less formally. T contains the following “equivalence classes” (we don’t know yet that these are equivalence classes before we show that T is an equivalence relation, but within these subsets every element is related to every element, while no elements from different subsets are related): $\{x,x+1,x+2\}$ for $0<x<1$ , $\{1,2\}$ , and $\{x\}$ for $x\le0$ and $x\ge3$ .

R and S clearly hold for $T$ . The transitivity needs a proof. Let $xTy$ and $yTz$ . If $x=y$ or $y=z$ then $xTz$ holds. If $x\neq y$ and $y\neq z$ then $x\in(0,3)$ , $y\in(0,3)$ , $y-x$ is an integer number (-2,-1,1 or 2), $z\in(0,3)$ , $z-y\in\{-2,-1,1,2\}$ . Therefore, $z-x$ is an integer and both are in $(0,3)$ , i.e. either $x,z\in\{1,2\}$ or $x,z\in\{a,a+1,a+2\}$ for some $a\in(0,1)$ . In either case, $xTz$ .