Sorry for my bad english in advance. On the set $\Bbb R\times\Bbb R$ the relation $\sim$ is defined as follows: $x\sim y$ if and only if $\lfloor x\rfloor=\lfloor y\rfloor$ (where $\lfloor x\rfloor$ is defined as the biggest whole number that is not greater than $x$). Prove or disprove the claim: the interval $(2,3)$ is the equivalence class of the relation $\sim\,$.
In order for this to be a equivalence class, does every element in the interval $(2,3)$ need to be in a relation with some element $z$ from my set ? And if I wanted to disprove that this interval is a class equivalence, is it enough to find one counter example ?
First of all, to talk sensibly about equivalence classes, we need to know that the relation of interest is an equivalence relation. In this case, the relation $\sim$ is indeed an equivalence relation, because if $f: X \to Y$ is any function then the condition $f(x) = f(y)$ defines an equivalence relation on $X$: the relation so defined is reflexive ($f(x) = f(x)$), symmetric (if $f(x) = f(y)$ then $(y) = f(x)$) and transitive (if $f(x) = f(y)$ and $f(y) = f(z)$ then $f(x) = f(z)$). Your relation $\sim$ has this form wth $f(x)$ defined to be the floor function, which you are writing as $[x]$.
Now we need to know what the equivalence classes of $\sim$ look like. Each equivalence class is obtained by picking an element $x$ in the field of the equivalence relation and forming the set $C_x = \{ y : x \sim y \}$. Because $\sim$ is an equivalence relation, if $y \in C_x$, then $C_x = C_y$ (do you see how to prove that?). So to check whether $(2, 3)$ is an equivalence class of $\sim$, we can just pick some $x \in (2, 3)$ and check whether $C_x = (2, 3)$. So let's pick $x = 5/2$. Then for every $y \in (2, 3)$, we have $[y] = 2 = [5/2]$, so $(2, 3)$ is contained in $C_x$. Also, we can see that if $y \ge 3$ or $y < 2$, then $[y] \neq 2$. But what if $y = 2$? If so, we have $[y] = 2 = [5/2]$. so $C_x$ should include $2$ as well as $(2, 3)$. Putting these pieces together, we have proved that $C_x$ is the half-open interval $[2, 3)$ and not the open interval $(2, 3)$.