Cardinality: is it true that $|X^\mathcal{inj}| \leq 2^{|X|}$?

Question

Definition. Whenever $X$ is a set, write $X^\mathcal{inj}$ for the collection of all injections $f$ such that:

1. $f$ has codomain $X$
2. There exists an ordinal $\alpha$ such that $f$ has domain $\alpha$.

Proposition. Assume $\mathrm{ZFC}.$ Then $|X^\mathcal{inj}| \geq 2^{|X|}.$

Proof. Define a function $e : X^\mathcal{inj} \rightarrow 2^{X}$ by writing $e(f) = \mathrm{im}(f)$. Since every subset of $X$ is well-orderable, hence $e$ is a surjection. We conclude that $|X^\mathcal{inj}| \geq 2^{|X|}$.

Question. Does the converse hold for infinite sets?

In particular, assuming $\mathrm{ZFC}$, is it true that for all infinite sets $X$, we have $|X^\mathcal{inj}| \leq 2^{|X|}$?

Answer 1

Let's count stuff.

There are exactly $|X|^+$ many ordinals which can be injectively mapped into $X$. Omitting all the finite ordinals since their cardinality is negligible here.

Let $\alpha<|X|^+$, then there are $[X]^{|\alpha|}$ possible sets which can be the range of an injection from $\alpha\to X$, and there are $2^{|\alpha|}$ injections into each such set (fix one, then apply permutations on $\alpha$). Therefore there are $2^\alpha\cdot|[X]^{|\alpha|}|=|[X]^{|\alpha|}|=|X|^{|\alpha|}$ such injections for a fixed, infinite $\alpha$.

Therefore the result is $|X|^+\cdot|X|^{<|X|^+}=|X|^{<|X|^+}=|X|^{|X|}=2^{|X|}$.