Does $\mathbb N$ have a predecessor? Let $\mathbb N$ be the Von Neumann constructed natural numbers. Does $\mathbb N$ have a predecessor? **My try** $$\mathbb N^*:=\mathbb N \backslash \\{\varnothing\\}$$ Let $p$ be a map $$p:\mathbb N^* \to \mathbb N \\\n \to \bigcup n$$ namely the predecessor map. **Example** $$p(3) = \bigcup 3 = \bigcup \\{ \varnothing ,\\{\varnothing \\},\\{\varnothing, \\{\varnothing \\}\\}\\} = \\{\varnothing, \\{\varnothing \\}\\} = 2$$ **My try continuous** $$p(\mathbb N)=\bigcup \mathbb N = \mathbb N$$ thus, $$p(\mathbb N) \notin \mathbb N$$ But it does not make sense, because $\mathbb N \notin \mathbb N$.

No. The first infinite ordinal $\Bbb N$ (or better written $\omega$ in this context) is a limit ordinal, i.e. it has _no_ (immediate) predecessor.