Let $A$ be a set. $\kappa$ a cardinal and assume that $\omega \leq \kappa \leq \vert A \vert \leq 2^{\kappa}$. Determine the cardinality of $C \colon=\{B\subseteq A \colon \vert B \vert \leq \kappa \}$
I have to admit I'm stuck on this one. I got a hint that I should consider well orderings. Cardinals are a subclass of ordinals so they are well ordered under the relation $|A|\leq |B| \Leftrightarrow$ exists an injection $i\colon A \to B$. So think I can view $C$ at the set of injections into $\kappa$ but I'm not sure how this helps me.
Since $C$ is a subset of the power set of $A$ I can at least establish a upper bound
$$\vert C \vert \leq \vert P(A)\vert = 2^{\vert A \vert} \leq 2^{2^{\kappa} } \geq 2^{2^{\omega} } $$
And since we can define $\phi \colon A \to C, a\mapsto \{a\}$ and $\vert A \vert \geq \omega \implies \vert C \vert \geq \omega$ and then I get no further. My intuition tells me that the cardinality probably is $2^{\kappa}$ but I can't show it.
Any hints, preferably on how do do it with well orderings would be highly appreciated. No full solutions please.
Here's a hint for you. This [almost] has nothing to do with well-ordering. It's about counting, so it's about cardinal arithmetic. And every subset of size at most $\kappa$ has a function from $\kappa$ onto that subset (well, except the empty set). So we're counting functions (modulo some equivalence relation).
Try to calculate first the case where $A=\kappa$, or $A=2^\kappa$. Then see that anything in between must have the same result. For a specific case, try to take the case where $\kappa=\aleph_0$. This is a classical result, and we use it, for example, when we count how many continuous functions are there from $\Bbb R$ to itself.