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

73 Views Asked by At

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?

1

There are 1 best solutions below

0
On BEST ANSWER

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.