Successor and Limit Ordinals in the Order Topology

Local Neighborhoods in an Ordinal Interval

In an ordinal interval, the order topology makes successor points behave differently from limit points. A successor has a point immediately before it, so a small order interval can isolate it. A nonzero limit ordinal has no immediate predecessor; its neighborhoods instead contain tails of ordinals below it. The size and form of a local neighborhood basis at a limit are controlled by which subsets are cofinal below that limit.

Fix an ordinal \(\theta\), and let \(X=[0,\theta]=\{\alpha:\alpha\leq\theta\}\), ordered by the ordinal order and equipped with its order topology. For \(a,b\in X\) with \(a<b\), write \((a,b)=\{x\in X:a<x<b\}\). The order-topology basis consists of such open intervals and the endpoint rays \([0,b)=\{x\in X:x<b\}\) and \((a,\theta]=\{x\in X:a<x\}\), when their endpoints are in \(X\). We include \(X\) itself as a basis element. This last provision handles \(\theta=0\), when \(X=\{0\}\) and there are no nonempty intervals or endpoint rays of the displayed forms.

Definition: A neighborhood basis at \(x\) is a family of open sets containing \(x\) such that every open neighborhood of \(x\) contains at least one member of the family. An ordinal \(\alpha\) is a successor ordinal if \(\alpha=\gamma+1\) for some ordinal \(\gamma\). A nonzero ordinal that is not a successor is a limit ordinal.

We treat \(0\) separately: with this convention, it is neither a successor nor a nonzero limit ordinal. All neighborhoods and bases below are relative to \(X=[0,\theta]\), not to the class of all ordinals.

Successor Points Are Isolated

Theorem 1: In \(X=[0,\theta]\), every point that is \(0\) or a successor ordinal has the singleton \(\{\alpha\}\) as a neighborhood basis. In particular, each such point is isolated.

Proof. First suppose \(\alpha=\gamma+1\) and \(\alpha<\theta\). Then \(\alpha+1\leq\theta\), and the order interval \((\gamma,\alpha+1)\) contains exactly \(\alpha\): any ordinal strictly greater than \(\gamma\) is at least \(\gamma+1=\alpha\), and any ordinal strictly less than \(\alpha+1\) is at most \(\alpha\). Thus \((\gamma,\alpha+1)=\{\alpha\}\), which is open. If \(\alpha=\theta\) is a successor, then the endpoint ray \((\gamma,\theta]\) is exactly \(\{\theta\}\), so it too is open. If \(\alpha=0\) and \(\theta>0\), then \([0,1)=\{0\}\) is open. Finally, if \(\theta=0\), then \(X=\{0\}\) is open by the basis convention above. In each case, every neighborhood of \(\alpha\) contains \(\{\alpha\}\), and \(\{\alpha\}\) itself is a neighborhood. Hence the singleton is a neighborhood basis. \(\square\)

This result uses both sides of a successor point when it is not the top endpoint: its immediate predecessor supplies the lower endpoint of a basic interval, and its immediate successor supplies the upper endpoint. At the top endpoint only a left ray is needed.

Worked Example: The Finite Ordinal Interval [0, 4]

Take \(X=[0,4]=\{0,1,2,3,4\}\) with its order topology. We find an open singleton around each point. At \(0\), the initial ray \([0,1)=\{0\}\) is open. At \(1\), the interval \((0,2)=\{1\}\) is open; at \(2\), \((1,3)=\{2\}\) is open; and at \(3\), \((2,4)=\{3\}\) is open. At the top endpoint \(4\), the ray \((3,4]=\{4\}\) is open.

These sets cover \(X\), and each is a basis neighborhood of its only point. Therefore every point is isolated, and every subset of \(X\) is open: it is a union of these open singletons. As a degenerate check, if instead \(\theta=0\), then the only point is \(0\), and the added basis element \(X=\{0\}\) makes the same conclusion valid without any interval or ray.

Neighborhood Bases at Limit Ordinals

Let \(\lambda\) be a nonzero limit ordinal with \(\lambda\leq\theta\). For each \(\beta<\lambda\), define the tail ending at \(\lambda\) by \((\beta,\lambda]=\{x\in X:\beta<x\leq\lambda\}\). These tails are open in \(X\). If \(\lambda=\theta\), the set is the endpoint ray \((\beta,\theta]\). If \(\lambda<\theta\), then \(\lambda+1\leq\theta\), and \((\beta,\lambda+1)\) is an open interval equal to \((\beta,\lambda]\), because there is no ordinal strictly between \(\lambda\) and \(\lambda+1\).

Theorem 2: If \(\lambda\) is a nonzero limit ordinal in \(X=[0,\theta]\), then \(\{(\beta,\lambda]:\beta<\lambda\}\) is a neighborhood basis at \(\lambda\).

Proof. Each set \((\beta,\lambda]\) is open by the preceding description and contains \(\lambda\). Let \(N\) be any open neighborhood of \(\lambda\). Since the order-topology basis generates the topology, there is a basis element \(B\) with \(\lambda\in B\subseteq N\). If \(B\) is an interval \((a,b)\), then \(a<\lambda<b\), and \((a,\lambda]\subseteq(a,b)=B\). If \(B\) is an initial ray \([0,b)\), then \(\lambda<b\), and \((0,\lambda]\subseteq[0,b)\); this uses \(0<\lambda\). If \(B\) is a final ray \((a,\theta]\), then \(a<\lambda\), so \((a,\lambda]\subseteq(a,\theta]=B\). If \(B=X\), then \((0,\lambda]\subseteq X\). In every case, a tail of the stated form is contained in \(N\). Thus the tails form a neighborhood basis at \(\lambda\). \(\square\)

Unlike a successor point, a nonzero limit ordinal is not isolated. Indeed, for every \(\beta<\lambda\), the ordinal \(\beta+1\) is still below \(\lambda\); otherwise \(\lambda=\beta+1\) would be a successor. Thus each basic neighborhood \((\beta,\lambda]\) contains \(\beta+1\) as well as \(\lambda\). Since every neighborhood contains one of these tails, no neighborhood of \(\lambda\) can be the singleton \(\{\lambda\}\).

Worked Example: A Cofinal Sequence Below \(\omega\cdot 2\)

Let \(\lambda=\omega\cdot 2\), the limit ordinal whose elements below it are the finite ordinals and the ordinals \(\omega+k\) for finite \(k\geq0\). Consider \(c_n=\omega+n\) for \(n\in\mathbb{N}\). Each \(c_n\) is below \(\lambda\), and the sequence is strictly increasing.

To check cofinality, take any \(\beta<\lambda\). If \(\beta\) is finite, then \(\beta<\omega+1=c_1\). If \(\beta=\omega+k\) for a finite \(k\geq0\), choose \(n=\max(1,k)\). Then \(\omega+k\leq\omega+n=c_n\). Thus for every \(\beta<\lambda\) some \(c_n\) satisfies \(\beta\leq c_n\). Consequently the tails \((\omega+n,\lambda]\), for \(n\in\mathbb{N}\), form a neighborhood basis at \(\lambda\): for any neighborhood, Theorem 2 gives a tail \((\beta,\lambda]\) inside it, and choosing \(n\) with \(\beta\leq\omega+n\) gives \((\omega+n,\lambda]\subseteq(\beta,\lambda]\).

Check: the inclusion has the correct direction because increasing the left endpoint removes points. In particular, if \(\beta\leq\omega+n\), every \(x>\omega+n\) also satisfies \(x>\beta\), so \((\omega+n,\lambda]\subseteq(\beta,\lambda]\).

Cofinality Determines Which Tails Are Needed

Definition: A subset \(C\subseteq\lambda\) is cofinal in \(\lambda\) if for every \(\beta<\lambda\) there is \(c\in C\) with \(\beta\leq c\). A sequence \((c_n)\) in \(\lambda\) is cofinal if its range is cofinal in this sense.

Theorem 3: For a nonzero limit ordinal \(\lambda\leq\theta\), the family \(\{(c,\lambda]:c\in C\}\) is a neighborhood basis at \(\lambda\) if and only if \(C\) is cofinal in \(\lambda\).

Proof. Suppose first that \(C\) is cofinal. Let \(N\) be any open neighborhood of \(\lambda\). By Theorem 2, there is \(\beta<\lambda\) with \((\beta,\lambda]\subseteq N\). Cofinality gives \(c\in C\) with \(\beta\leq c\). Every point greater than \(c\) is then greater than \(\beta\), so \((c,\lambda]\subseteq(\beta,\lambda]\subseteq N\). Therefore the family indexed by \(C\) is a neighborhood basis. Conversely, suppose \(C\) is not cofinal. Then there is \(\beta<\lambda\) such that no \(c\in C\) satisfies \(\beta\leq c\). Thus every \(c\in C\) satisfies \(c<\beta\). The neighborhood \((\beta,\lambda]\) cannot contain any \((c,\lambda]\), because \(\beta\in(c,\lambda]\) but \(\beta\notin(\beta,\lambda]\). Hence this family is not a neighborhood basis. Both directions hold, including the possibility that \(C\) is empty: it is not cofinal, and its family cannot be a neighborhood basis. \(\square\)

The theorem separates two facts that are easy to conflate. All tails \((\beta,\lambda]\) give a local basis, but a smaller family of tails does so only when its left endpoints reach arbitrarily far up toward \(\lambda\), in the precise cofinal sense. If a countable cofinal sequence exists, its tails give a countable neighborhood basis. If no countable subset is cofinal, no countable family of these tails can be a neighborhood basis.

Worked Example: A Noncofinal Family Does Not Give a Basis

Again take \(\lambda=\omega\cdot2\), but now use \(C=\mathbb{N}\), viewed as the finite ordinals. This set is not cofinal: \(\beta=\omega\) is below \(\lambda\), and no finite ordinal \(n\) satisfies \(\omega\leq n\).

The failure can also be checked directly on neighborhoods. The set \((\omega,\lambda]\) is an open neighborhood of \(\lambda\). For every \(n\in\mathbb{N}\), the point \(\omega+1\) lies in \((n,\lambda]\), since \(n<\omega+1\), but does not lie in \((\omega,\lambda]\), since \(\omega+1>\omega\) does lie there—so this proposed witness would not show failure. Instead use the endpoint \(\omega\) itself: it belongs to \((n,\lambda]\) for every finite \(n\), because \(n<\omega\), and it does not belong to \((\omega,\lambda]\), because the left inequality is strict. Thus \((n,\lambda]\nsubseteq(\omega,\lambda]\) for every \(n\in C\). No member of the proposed family lies inside this neighborhood, so the family is not a neighborhood basis.

Check: membership uses the strict condition \(n<x\leq\lambda\). For \(x=\omega\), \(n<\omega\) holds for every finite \(n\); for the target tail, \(\omega<\omega\) is false. This verifies the failure without relying only on the cofinality criterion.

A Common Pitfall and the Local Picture

Common pitfall. It is incorrect to treat every ordinal point as though it had a greatest ordinal immediately below it. For example, \(\omega\cdot2\) is a limit ordinal, not a successor: there is no \(\gamma<\omega\cdot2\) with \(\gamma+1=\omega\cdot2\). A mistaken calculation might claim that \((\omega+4,\omega\cdot2]\) is the singleton \(\{\omega\cdot2\}\). But \(\omega+5\) is strictly greater than \(\omega+4\) and strictly less than \(\omega\cdot2\), so it belongs to this interval. The correct conclusion is that the interval is a tail neighborhood containing intermediate ordinals as well as its endpoint. By contrast, at the successor \(\omega+5\), the interval \((\omega+4,\omega+6)\) is exactly \(\{\omega+5\}\).

The distinction is useful whenever continuity or convergence is tested at points of an ordinal space: isolated points have a singleton neighborhood, while neighborhoods of a limit point must be checked on tails below it. Cofinality tells us whether a chosen collection of those tails is sufficient. The endpoint case is part of the same description: when the limit ordinal is \(\theta\), the tails are open final rays; when it lies below \(\theta\), they are open intervals ending just above it. The singleton interval \(X=[0,0]\) is handled separately by including \(X\) in the basis, so the description does not depend on nonexistent endpoints.

Check Your Understanding

Use the order topology on the indicated ordinal interval in each question.

For \(X=[0,\omega+2]\), give an open singleton neighborhood of \(\omega+1\), and explain why \(\omega\) does not have one.
For a nonzero limit \(\lambda\leq\theta\), prove that \((\beta,\lambda]\) is open for every \(\beta<\lambda\), treating \(\lambda=\theta\) and \(\lambda<\theta\) separately.
In \(X=[0,\omega\cdot2]\), is \(C=\{\omega+2n:n\in\mathbb{N}\}\) cofinal in \(\omega\cdot2\)? Use the cofinality criterion to decide whether its tails give a neighborhood basis.
Why does the family \(\{(n,\omega\cdot2]:n\in\mathbb{N}\}\) fail to be a neighborhood basis at \(\omega\cdot2\)? Identify a single neighborhood that witnesses the failure.
Explain why the basis convention including \(X\) is needed for the order topology on \([0,0]\).