1
$\begingroup$

In one of my exercises we are shown that the relation "$\sim$" is defined as the following:

$$\langle n,m \rangle \sim \langle k,l \rangle \iff |(n\setminus m)| = |(k\setminus l)|$$

in which $|X|$ symbolizes the number of elements of the set. We need to:

(a) Show that "$\sim$" is a equivalence relation on $\omega\times\omega$.

(b) Find a "representative system" ("system of representatives") of this equivalence relation.


For part (a) it is straight forward, we show that it is reflexive, symmetric and transitive. But for part (b) I am stuck. So far I think I have an understanding of the definition for a representative, in terms of this exercise.(Please correct me if I am wrong):

I understand that if "$\sim$" is an equivalence relation on $\omega\times\omega$, then for $x \in \omega\times\omega$, the set

$[x]_{\sim} = \{ y \in \omega\times\omega : y\sim x\}$ is called an equivalence class of $x$. So every $y \in [x]_{\sim}$ (i.e $y\sim x$) is then called a "representative" of the equivalence class of $x$.

However I am lost on how to apply this definition to the question. Some guidance would be very much appreciated. Thank you in advance.

$\endgroup$
3
  • $\begingroup$ @HenningMakholm I noticed your edits, is there a difference between |n\m|=|k\l| and |(n\m)|=|(k\l)| ? because in my exercise it is written with parenthesis. $\endgroup$ – math189925 Feb 2 '15 at 13:12
  • $\begingroup$ x @math: Sorry, I've reinserted the parentheses. I'm not aware of any context where $|(n\setminus m)|$ would mean something different from $|n\setminus m|$. $\endgroup$ – hmakholm left over Monica Feb 2 '15 at 13:14
  • $\begingroup$ If you put $m=\emptyset$, is there exactly one $n$ such that $\langle n,\emptyset \rangle$ belongs to a given equivalence class? $\endgroup$ – Martin Sleziak Feb 2 '15 at 14:16
1
$\begingroup$

A representative system is simply some set that contains exactly each element from each equivalence class. There are many different possibilities for such a set; the exercise simply asks you to describe one of those possibilities, of your own choice.

Each equivalence class is described by a natural number, and contains all pairs $\langle n,m\rangle$ such that $n\setminus m$ has that many elements. (Since $n$ is always finite, $n\setminus m$ will always be finite too).

So what you need to do is simply, for each natural number $k$, to decide on one particular pair $\langle n_k,m_k\rangle$ where $n_k\setminus m_k$ has $k$ elements.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.