# The cardinality of the set of countably infinite subsets of an infinite set

Let $A$ be a set with card($A$)=$a$. What is the cardinal number of the set of countably infinite subsets of $A$?

I see that this problem is equivalent to finding the cardinal number of the set of injective functions from $\mathbb{N}\rightarrow{A}$. I also know that the cardinal number of the set of bijections from $A\rightarrow{A}$ is $a^{a}$.

Hints and general heurisitcs would be greatly appreciated.

$A$ has $a^\omega$ countably infinite subsets, and there’s not much more that you can say unless you know something about the cardinal $a$. For example, if $2\le a\le 2^\omega=\mathfrak c$, then $a^\omega=2^\omega$. If $\operatorname{cf}a=\omega$, i.e., if $a$ has cofinality $\omega$, then $a^\omega>a$.

This can have several different answers. Note that assuming the axiom of choice the set of countable subsets, $[A]^{\leq\omega}$ has the same cardinality as $A^\omega$.

• If $a=\omega$ then $\omega^\omega$ is of size continuum, and the cardinality of the continuum is not decidable in ZFC, so it can get quite large, or not.
• If $a=2^\omega$ then $a^\omega=2^\omega$ again.
• If $a=\omega_1$, $2^\omega=\omega_2$, and $2^{\omega_1}=\omega_3$, then $2^\omega\leq\omega_1^\omega\leq(2^\omega)^\omega=\omega_2<2^{\omega_1}=\omega_3$.

We see that there are many possible options, and of course if $a$ is singular with cofinality $\omega$ then $a<a^\omega$, and we need to check whether or not SCH holds for $a$ or not.

One word on the situation without the axiom of choice, in models where all sets of real numbers are Lebesgue measurable the set $[\mathbb R]^\omega$ has cardinality strictly larger than that of the continuum; although there is a surjection from $\mathbb R^\omega\sim\mathbb R$ onto this set. It is peculiar, but this is how things are when you negate the axiom of choice.

Here is my solution to the problem based on Mr. Scott's answer.

Let $C$ denote the set of injective maps from $\omega\rightarrow{A}$ and $C'$ the set of injective maps from $\omega\times{\omega}\rightarrow{A\times{A}}$. Then $card(C)=card(C')=c$. Since $C\subset{A^\omega}$ we find $c\le{a^\omega}$. We now fix $\alpha\in\omega$. On the other hand to each $f\in{A^\omega}$ we associate the map in $C'$ which swaps $(x,\alpha)$ with $(x,f(x))$ and fixes eveything else. This shows $a^\omega\le{c}$, whence $a^\omega=c$.