$\omega$ is the least limit ordinal

Let $\mathbb{N}=\omega$ be the natural numbers (The natural numbers $\mathbb{N}$ (von Neumann)) with their usual order (Order on the natural numbers). Then:

(i) for all $m,n\in \mathbb{N}$, $m\in n$ if and only if $m<n$;

(ii) every natural number is an ordinal (Ordinal (von Neumann)), and $\omega$ is an ordinal;

(iii) $\omega$ is a limit ordinal (Successor and limit ordinals);

(iv) every ordinal $\alpha \in \omega$ is $0$ or a successor ordinal, and consequently $\omega$ is the least limit ordinal: $\omega \subseteq \gamma$, that is $\omega \le \gamma$, for every limit ordinal $\gamma$.

So the natural numbers are exactly the ordinals below $\omega$, and $\omega$ is the first ordinal at which induction acquires a limit clause.

Everything here is a theorem of ZF, and no choice principle is used. The only axiom beyond the basic ones that any of it needs is Infinity, which is what makes $\omega$ a set at all (The natural numbers exist: a smallest inductive set).

Why claim (i) has to be proved. The published development builds the order on $\mathbb{N}$ from addition (Order on the natural numbers) and never identifies it with membership; the identification is recorded there as a remark, not a theorem. Ordinals need it as a theorem, because the whole definition of an ordinal is phrased in terms of $\in$. Once claim (i) is available, the two pictures of $n$, as "the number of predecessors" and as "the set of its predecessors", coincide.

The finite ordinals. Claims (ii) and (iv) say the natural numbers are exactly the ordinals $\alpha$ with $\alpha \in \omega$, and that each is $0$ or a successor. This is the precise sense in which $\mathbb{N}$ is an initial segment of the ordinals, and it is why ordinary induction (The principle of mathematical induction) is the special case of Transfinite induction at $W=\omega$.

Limits exist only because $\omega$ does. The Axiom of Infinity is what makes $\omega$ a set (The natural numbers exist: a smallest inductive set), and it is also what gives this lemma its content: ZF without Infinity cannot prove that any limit ordinal exists, assuming that theory consistent. The reason is that every limit ordinal $\gamma$ is itself an inductive set, and so would witness Infinity outright: $0\in \gamma$, because the $\in$-least element of a nonempty ordinal is $\mathrm{\varnothing }$; and $\beta \in \gamma$ gives ${\beta }^{+}\subseteq \gamma$, hence ${\beta }^{+}\in \gamma$ by claim (f) of Basic closure properties of ordinals unless ${\beta }^{+}=\gamma$, which a limit ordinal excludes. Dropping an axiom is not the same as assuming its negation, and nothing here says that without Infinity every ordinal is $0$ or a successor: ZF itself extends ZF without Infinity and has limit ordinals. What is lost is any proof that one exists. The successor operation alone never produces a limit; a limit is always reached by taking a union, here $\omega =\bigcup \{n:n\in \omega \}$.

Ordinal arithmetic is not developed here. Sums and products of ordinals, and the ordinals $\omega +1$, $\omega \cdot 2$ and so on, are defined by transfinite recursion and would fit naturally after this item, but nothing on this page needs them, so they are left to a later page rather than introduced unused.