Infinite digraph

Question

Given a digraph $\Gamma$. Let $v\in V(\Gamma)$, then $N^-(v)$ denotes the set of $v$'s incoming vertices.

Let $\Gamma$ has the following properties:

1. There is exactly one vertex $a\in V(\Gamma)$ s.t. $N^-(a)=\emptyset$
2. If $u\neq v$, then $N^-(u)\neq N^-(v)$
3. There is a vertex $b\in V(\Gamma)$ s.t.

\begin{align} a\in N^-(b) \end{align} \begin{align} \forall[u\in N^-(b)]\exists[v \in N^-(b)]:N^-(v)=N^-(u)\cup\{u\} \end{align} Can we prove that $|N^-(b)|\geq\aleph_0$?

Answer 1

I think the answer is yes.
(after dealing with the degenerate case in the comment)

This is true because $b$ satisfies the properties of a limit ordinal, and therefore its cardinality is at least $\aleph_0$.