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

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?

• 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 $c_1$. May 16 at 0:50
• Thanks for replying. The method is to use the smaller index, as you mention below, right?
– Fred
May 16 at 1:54
• Yes, that is one way. May 16 at 2:48

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.