# Equivalence Classes for $R = \{ (x, y) \ | \ f(x) = f(y) \}$

Let $f: \mathbb{R}^+ \to \mathbb{N}_0$ be given by $f(x) = \lfloor{x}\rfloor$.

Define a relation $R \subseteq \mathbb{R}^+ \times \mathbb{R}^+$ by

$R = \{ (x, y) \ | \ f(x) = f(y) \}$

Show that $R$ is an equivalence relation and find its equivalence classes.

I'm trying to find its equivalence classes.

I know that, for an equivalence relation on a set $A$, an equivalence class is defined as

$[a]_R = \{b \in A : (a, b) \in R \}$ for all $a \in A$.

My solution is that the equivalence classes are

$[r]_R = \{ n \in \mathbb{R}^+ : n = \lfloor r \rfloor \}$ for all $r \in \mathbb{R}^+$.

We have that

$f(0.5) = 0$, $f(0.3) = 0$, $f(0.8) = 0$, ...

$f(1) = 1$, $f(1.5) = 1$, ...

...

It seems to me like this accounts for every possible element of the relation.

I would greatly appreciate it if people could please check whether my solution is correct and provide feedback.

In general if we have a function $f:A \rightarrow B$ the relation on $A$ $$x \sim y \iff f(x)=f(y)$$

is an equivalence relation named equivalence kernel whose classes are $\{f^{-1}(f(x)) , x\in A\}$

In this particular case:

Let $x \in \mathbb{R^+}$, $[x]=f^{-1}(f(x))= [ \lfloor{x} \rfloor, \lfloor{x} \rfloor+1)$

You seem to have found the equivalence classes defined by means of natural elements in $\;\Bbb R^+\;$ , which of course cannot be as there are much more elements than merely the natural ones.

Observe that $\;(x,y)\in R\iff \exists\,n\in\Bbb N\;\;s.t.\;\;x,\,y\in[n,n+1)\;$ , where the last notation is a half open (or half closed) real interval...and here you have your equivalence classes.

Your definition of $[r]_R$ contains only one element, namely $\lfloor r\rfloor$. You need all elements $s\in\mathbb R^+$ such that $\lfloor s \rfloor = \lfloor r\rfloor$. So, if you have $\lfloor r \rfloor = n$, then $\lfloor s \rfloor = n$ if and only if it is contained in interval...?