# The set of all countably-infinite subsets of an infinite set

Let $A$ be an infinite set and $D(A)$ denote the set of all countably-infinite subsets of $A$ and let $P(A)$ denote ,as usual, the power set of $A$, then

(i) does there exist a surjection of $D(A)$ onto $P(A)$ ?

(ii) if $A$ is uncountable then does there exist an injection from $A$ into $D(A)$ ?

(iii) if answer to (i) is "no" , then does there exist a surjection of $A$ onto $D(A)$ ?

( I only know that $D(A)$ is always uncountable)

The answers to 1 and 3 are sometimes; whereas the answer to 2 is trivially yes (even if $A$ is countable).

Let me start with 2, as it's a simple positive answer. Pick $X\subseteq A$ which is countably infinite, then consider the function: $$f(a)=\begin{cases} X\cup\{a\} & a\notin X\\ X\setminus\{a\} & a\in X\end{cases}$$ You can show that this is an injection (note that we didn't even care that $A$ is uncountable, just infinite).

The answer to (1) is easily no in some cases. For example if $A=\Bbb R$ then the set of countably infinite subsets of $\Bbb R$ has the same cardinality of $\Bbb R$. Therefore there is no surjection from $D(\Bbb R)$ onto $\mathcal P(\Bbb R)$; and there is certainly a surjection from $\Bbb R$ onto $D(\Bbb R)$.

However it is possible that the answer is yes. Consider the case where $\sf GCH$ holds, in that case if $|A|=\aleph_\omega$ we have that $|D(A)|=|\mathcal P(A)|$, in which case 3 is irrelevant.

But it could be the case that a statement known as the Singular Cardinal Hypothesis, or $\sf SCH$, fails. This statement follows from $\sf GCH$, and it is possible to have, for example $A$ whose cardinality is $\aleph_\omega$ as before, but all the following inequalities are sharp: $$|A|<|D(A)|<|\mathcal P(A)|$$

The consistency of this failure is stronger than that of $\sf ZFC$, and requires large cardinals.

Here is another example for the failure of 1 and 3 without using large cardinals.

Consider a model in which $2^{\aleph_0}=\aleph_2$ and $2^{\aleph_1}=\aleph_3$. Let $A$ be a set of size $\aleph_1$. Then one can show that $|D(A)|=\aleph_2$, and now we have that $$|A|<|D(A)|<|\mathcal P(A)|$$ without resorting to very difficult results and large cardinals.

• Asaf Karagila: I knew that the answer to (ii) is "yes" for countably infinite $A$ , this is why I asked for uncountable $A$ , my method was different and not as general as that of yours , I didn't see that coming , pretty simple solution. – user102232 Oct 31 '13 at 9:41
• Ah, I see. In the countable case it's indeed easier to prove that $|D(A)|=|\mathcal P(A)|$ and then it's easy to show the existence of an injection. – Asaf Karagila Oct 31 '13 at 9:49
• Asaf Karagila : Could it be possible to characterize all infinite sets for which answer to (i) is "yes" ? – user102232 Oct 31 '13 at 9:52
• Not really, unless you want to assume something like $\sf GCH$, or something like that. The opposite question, math.stackexchange.com/questions/534219/… is equally daunting without some restrictions. Generally speaking cardinal exponentiation is very independent from the axioms of $\sf ZFC$ and you have to assume more to actually say something intelligent about it. – Asaf Karagila Oct 31 '13 at 9:57