Finite equivalence class same cardinality For an equivalence relation $\sim$, if each partition has a finite number of elements, and $X$ is an infinite set, then is it true that $|X/\sim|=|X|$?
I can prove injectivity one way by defining the map $f:X/\sim\rightarrow X$ by picking an element from the equivalence class, but not sure how I would go about the other way.
 A: The quotient set $X / {\sim}$ is a partition of $X$ into finite sets $[X]_i$ with $i < |X / {\sim}|$ (the axiom of choice here is implicit: $i$ is an ordinal less than the cardinal number $|X / {\sim}|$).
Map each $f_i : [X]_i \to \omega \times \{ i \}$ injectively since they are finite.
Take the union $\displaystyle \bigcup_{i < |X / {\sim}|} f_i : X \to \omega \times |X / {\sim}|$, which is injective, since the images of the component functions are disjoint.
So we have $$|X| \le |\omega \times X / {\sim}| \le |\omega \times X| = |X|$$ since $X$ is an infinite set.
$|\omega \times X / {\sim}| = | X / {\sim}|$ because $X / {\sim}$ is infinite (easily seen because otherwise, $X = \bigcup X / {\sim}$ is the finite union of finite sets).
A: Assuming the axiom of choice (at least for families of finite sets) then the answer is yes.
The reason that you find it somewhat difficult to continue is exactly this, the axiom of choice is needed and we cannot really do this in an explicit manner. Let me give one of the classical examples.
We say that $S$ is a Russell set if $S$ can be written as a disjoint union of pairs $P_n$ where $n\in\Bbb N$, but $S$ does not have a countably infinite subset. This is equivalent to saying that the product of any infinite set of the pairs is empty, or in simple words: there is no choice function from infinite families of pairs into $S$.
Clearly $\mathscr P=\{P_n\mid n\in\Bbb N\}$ is a partition of $S$ into pairs, which are finite sets, but it is not true that $|\mathscr P|=|S|$. In fact it is not even true that $|\mathscr P|\leq|S|$, which is an even more disturbing fact!
So we really do need the axiom of choice. And this example easily generalize to replace the number of pairs by larger cardinals, in which case we can even save some other limited forms of choice (countable choice, for example), and still have these bizarre sets in our universe. 
