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I'm trying to understand the demonstration of the following proposition

Any partition of an at most countable set has a set of representatives

Proof:

Let $\mathcal{P}$ be a partition of $A$. Then there exists an equivalence relation $\sim$ on $A$ induced by $\mathcal{P}$. Since $A$ is at most countable, the set of equivalence classes, $A / \sim=\left\{[a]_{\sim}: a \in A\right\}$, is at most countable. Hence, $$ A / \sim=\left\langle\left[a_1\right]_{\sim},\left[a_2\right]_{\sim}, \ldots\right\rangle, $$ and so there is a set of representatives: $\left\{a_1, a_2, \ldots\right\}$.

I do not understand why this result has come about:

$$ A / \sim=\left\langle\left[a_1\right]_{\sim},\left[a_2\right]_{\sim}, \ldots\right\rangle, $$

How to get to it?

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    $\begingroup$ Why is the set of equivalence classes at most countable? (This is true, but you need an argument for it.) Also, just because you called one of the classes $[a_1]/{\sim}$, it does not mean that you know what $a_1$ is. Better to list $A/{\sim}$ as $\{c_1,c_2,\dots\}$ Now, for any $a\in c_1$, $c_1=[a]/{\sim}$, but you cannot just say that ``$a$ is the representative of $c_1$'', since if also $b\in c_1$, then $c_1=[b]/{\sim}$. You need to describe a method by which you choose a member of $c_1$. $\endgroup$
    – Andrés E. Caicedo
    May 16 at 0:50
  • $\begingroup$ Thanks for replying. The method is to use the smaller index, as you mention below, right? $\endgroup$
    – Fred
    May 16 at 1:54
  • $\begingroup$ Yes, that is one way. $\endgroup$
    – Andrés E. Caicedo
    May 16 at 2:48

1 Answer 1

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Your set $A$ is at most countable. Enumerate it, say $A=\{a_0,a_1,\dots\}$. Now, given a partition of $A$, each piece, being a subset of $A$, consists of some of the $a_i$, one of which has smallest index $i$. You can use this $a_i$ as the representative for that piece.

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