If $A$ is a set of size $2^{\aleph_0}$ and $S$ is an uncountable partition of $A$, is there an injection from $A$ into $S$?

Question

Let $A$ be a set such that $|A|=2^{\aleph_0}$ and let $S$ be a partition of $A$ such that $S$ is not countable.

Is there a way to define an injective function $f:A\longrightarrow S$ in order to prove that the the cardinality of $S$ is exactly $2^{\aleph_0}$?

This is a problem I've encounter solving a much more difficult one, which I actually asked here some days ago but has not been answered yet.

This is the last thing I need in order to prove that other problem I mentioned.

Answer 1

No, there is no way to define such injection for two major reasons:

1. It is consistent with $\sf ZF$ with the axiom of choice, the usual axioms of set theory, that there is some $S\subseteq\Bbb R$ such that $\aleph_0<|S|<2^{\aleph_0}$. In that case we can easily define the partition of singletons of the elements of $S$, and the rest is just one part. Now we have a partition which has a strictly smaller cardinality than $\Bbb R$, and therefore there is no injection as wanted.

2. The word "define" often refers, implicitly of course, to an explicit definition, without an appeal to the axiom of choice. However it is consistent with $\sf ZF$ that the real numbers can be partitioned into an uncountable number of parts $S$ and neither $2^{\aleph_0}<|S|$ nor $|S|<2^{\aleph_0}$. In that case, as it shows, one cannot define an injection in any direction.