2
$\begingroup$

Let $\sim$ be a relation on $\mathbb{R}$ and $x\sim y$ if and only if $x-y\in \mathbb{Z}$.

(a) Show that $\sim$ is an equivalence relation
(b) Give a complete set of equivalence class representatives.


(a) is easy to show, but I really don’t understand (b).

I know that:

$[a]_{\sim}:=\{y\in \mathbb{R} \mid a\sim y\}$

so

$[0]_{\sim}=\{\dots,-2,-1,0,1,2,\dots\}$

and that

$\dots=[-2]_{\sim}=[-1]_{\sim}=[0]_{\sim}=[1]_{\sim}=[2]_{\sim}=\,\dots$

because if $x\sim y$ then $[x]=[y]$.

But how can I find a complete set of equivalence class representatives?

Thanks in advance!

  • 0
    Can you find even one other class than $[0]_\sim$?2012-10-05

2 Answers 2

0

By a complete set of equivalence class representatives, I think the question wants a set $X \subset \mathbb{R}$ such that for each equivalence class, there is exactly one representative in the set - this is the sense in which it is complete (there is a $1-1$ correspondence between $X$ and $\mathbb{R}/\sim$, the set of equivalence classes).

So what you need to do is for each equivalence class, you need to pick one representative and let $X$ be the set of all your choices. As you point out, there are lots of different representatives you can choose for any given equivalence class. To ensure that your set $X$ contains only one representative for each equivalence class and that each equivalence class does have a representative in $X$, it would be good if you were able to make some sort of consistent choice of representative. For example, there is a reason I would choose $\frac{1}{2}$ to be the element in $X$ which represents $[\frac{5}{2}]_{\sim}$, and I would choose all the other representatives of equivalence classes for the same reason.

  • 0
    There is a contradiction in there somewhere, but you haven't said what it is. At some point, you are going to need to use the definition of $x \sim y$ to arrive at a contradiction. Just for the record, you don't have to justify the answer to me, I'm just suggesting what you should do. Having said that, I'm happy to discuss your reasoning.2012-10-03
1

Hint: two numbers are equivalent under this relation iff they have the same fractional part.

  • 0
    You can say that. Is there a shorter way of writing that set? And then, how would you prove that you've got a complete set of equivalence class representatives? (You'll need to prove that no two elements of your set are in the same equivalence class, and also that everything in $\Bbb R$ is related to something in the set).2012-10-03