Successor versus Limit Ordinals

Definition and intuition

An ordinal is a transitive set well-ordered by $\in$, so the ordinals are built up one step at a time. Past $0$, each ordinal falls into exactly one of two families, and telling them apart is the whole skill.

The successor of an ordinal $\beta$ glues $\beta$'s one new element on top:

$S(\beta) = \beta + 1 = \beta \cup \{\beta\}.$

So $1 = \{0\}$, $2 = \{0,1\}$, $3 = \{0,1,2\}$, and in general $n+1 = \{0,1,\ldots,n\}$. A successor ordinal always has a largest element (an immediate predecessor): the element you just adjoined.

A limit ordinal $\lambda$ is a nonzero ordinal that is not a successor — it has no maximum and no immediate predecessor. Instead it is the supremum (least upper bound) of all the ordinals beneath it:

$\lambda = \sup\{\alpha : \alpha < \lambda\} = \bigcup_{\alpha < \lambda} \alpha, \qquad \lambda \neq 0,\ \ \lambda \neq \beta+1 \text{ for any } \beta.$

The archetype is $\omega = \{0,1,2,3,\ldots\}$, the set of all natural numbers: it is reached only "in the limit" — no single natural number is its predecessor.

The trichotomy

Every ordinal is exactly one of three kinds:

For ordinals the order relations all coincide: $\beta \in \alpha \iff \beta \subsetneq \alpha \iff \beta < \alpha$. Adding a finite number $k$ to a limit ordinal $\lambda$ marches you up the successor chain $\lambda, \lambda+1, \lambda+2, \ldots$, so $\lambda + k$ for $k \ge 1$ is always a successor, never a limit.

How to classify an ordinal

1. Is it $0$? Then it is zero (and, depending on the textbook, sometimes lumped with the limits).
2. Can you write it as $\beta + 1$ for some $\beta$ — equivalently, does it have a single largest element? Then it is a successor.
3. Otherwise, if it is nonzero and equals the sup of everything below it (no top element), it is a limit.

A fast test: find the immediate predecessor. If one exists, successor; if none does, limit.

Mini-example (konkur 1402, Q107)

Which of $\omega+1$ and $\omega+2$ is a limit ordinal? Write each as a successor:

$\omega + 1 = S(\omega) = \omega \cup \{\omega\}, \qquad \omega + 2 = S(\omega+1).$

Each has an immediate predecessor ($\omega$ and $\omega+1$ respectively), so both are successor ordinals and neither is a limit ordinal. (The first limit ordinal at or above $\omega$ is $\omega$ itself; the next is $\omega\cdot 2 = \omega+\omega$.) A companion question (Q105) checks the matching identity directly: $\alpha\cup\{\alpha\}$ is always the successor ordinal $\alpha+1$, hence always an ordinal — even when $\alpha$ is a limit.

Pitfall: $\omega+1$ is not a limit ordinal — it is the successor $S(\omega)$, with immediate predecessor $\omega$. The "$\omega$" buried inside it is a red herring: limitness is decided by the whole ordinal having no maximum, and $\omega+1$ has a maximum, namely $\omega$. Only $\omega$ (and $\omega\cdot2$, $\omega^2$, $\ldots$) are limits; everything of the form $\lambda+k$ with $k\ge1$ is a successor.