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.