The First Uncountable Ordinal and Cofinal Neighborhoods

The Endpoint and Its Tail Neighborhoods

In an ordinal interval, the order determines the topology: points near a final endpoint are described by tails extending toward that endpoint. For a countable ordinal endpoint, a countable sequence of thresholds can reach arbitrarily far along the interval. The first uncountable ordinal behaves differently. No countable set of thresholds reaches every point below it, and that obstruction prevents a countable local base at the endpoint.

Let \(\omega_1\) denote the first uncountable ordinal, so every ordinal \(\alpha<\omega_1\) is countable. Consider the ordinal interval \(X=[0,\omega_1]\) with its order topology. For \(\alpha<\omega_1\), write \((\alpha,\omega_1]=\{\xi\in X:\alpha<\xi\leq\omega_1\}\). The order-topology basis consists of open intervals and endpoint rays. In particular, each set \((\alpha,\omega_1]\) is an open neighborhood of \(\omega_1\), and every open neighborhood of \(\omega_1\) contains one of these tails. This follows because a basic open set containing the maximum \(\omega_1\) is a ray cut off below it.

The weak inequality in the definition of cofinality is convenient: a threshold equal to \(\gamma\) already reaches the tail beginning just above \(\gamma\). A family of thresholds is cofinal precisely when its associated tails can be made to lie inside every endpoint neighborhood.

Countable Sets of Thresholds Stay Below \(\omega_1\)

Proof. Let \(\delta=\sup_{n\in\mathbb{N}}\alpha_n\), the least ordinal greater than or equal to every \(\alpha_n\). Each \(\alpha_n\) is a countable set, and their union is countable by the standard countable-union theorem for countable sets (using countable choice, as in the usual set-theoretic foundations). The supremum of ordinals is their union, so \(\delta=\bigcup_{n\in\mathbb{N}}\alpha_n\) is countable. Hence \(\delta<\omega_1\). Its successor \(\beta=\delta+1\) is also countable, so \(\beta<\omega_1\); and \(\alpha_n\leq\delta<\beta\) for every \(n\). Thus no member of the sequence reaches \(\beta\), which proves that its range is not cofinal. Any countable subset can be listed as a sequence, repeating elements if the subset is finite, so no countable subset is cofinal either. \(\square\)

The successor in this argument matters. The supremum \(\delta\) is already a countable ordinal, but taking \(\delta+1\) gives a particular point strictly above every threshold in the sequence. This strict inequality will let us find a point lying in each proposed basic neighborhood but outside one chosen neighborhood.

Cofinal Tails and Local Bases

Proof. Suppose first that \(A\) is cofinal. Let \(V\) be any open neighborhood of \(\omega_1\). The tail-neighborhood description gives some \(\gamma<\omega_1\) with \((\gamma,\omega_1]\subseteq V\). By cofinality, choose \(\alpha\in A\) with \(\gamma\leq\alpha\). Every point greater than \(\alpha\) is then greater than \(\gamma\), so \((\alpha,\omega_1]\subseteq(\gamma,\omega_1]\subseteq V\). The tails indexed by \(A\) therefore form a local base.

Conversely, suppose the tails indexed by \(A\) form a local base. Fix any \(\gamma<\omega_1\) and use \(V=(\gamma,\omega_1]\) as a neighborhood. There must be \(\alpha\in A\) with \((\alpha,\omega_1]\subseteq(\gamma,\omega_1]\). If \(\alpha<\gamma\), then \(\alpha+1\leq\gamma\), and \(\alpha+1\in(\alpha,\omega_1]\) but \(\alpha+1\notin(\gamma,\omega_1]\). This contradicts the inclusion. Therefore \(\gamma\leq\alpha\). Since this holds for every \(\gamma<\omega_1\), \(A\) is cofinal. \(\square\)

This characterization concerns tails, but the same obstruction applies to arbitrary neighborhoods. Each open neighborhood contains a tail, so a countable family of neighborhoods can be replaced, for the purpose of testing whether it is a local base, by a countable list of thresholds.

Proof. Suppose, for a contradiction, that \((U_n)_{n\in\mathbb{N}}\) is a countable local base at \(\omega_1\). For each \(n\), the set \(U_n\) is an open neighborhood of \(\omega_1\), so choose \(\alpha_n<\omega_1\) such that \((\alpha_n,\omega_1]\subseteq U_n\). By the lemma, there is a countable ordinal \(\beta<\omega_1\) with \(\alpha_n<\beta\) for every \(n\). The set \(V=(\beta,\omega_1]\) is an open neighborhood of \(\omega_1\). Since \((U_n)\) is a local base, some \(U_k\) must satisfy \(U_k\subseteq V\). But \(\beta\in(\alpha_k,\omega_1]\subseteq U_k\), while \(\beta\notin V\), since \(V\) contains only points strictly greater than \(\beta\). This contradicts \(U_k\subseteq V\). Therefore no countable local base exists. \(\square\)

A Common Mistake and Why the Result Matters

Common mistake. One might argue that the tails \((n,\omega_1]\), for \(n\in\mathbb{N}\), form a local base because for every \(n\) the tail extends all the way to \(\omega_1\). That confuses the upper endpoint of each tail with its lower threshold. To test the local-base condition, take \(V=(\omega+1,\omega_1]\). Each \(U_n=(n,\omega_1]\) contains \(\omega+1\), since \(n<\omega+1\), but \(V\) excludes \(\omega+1\). Thus no \(U_n\) is a subset of \(V\). The correction is to compare thresholds: finite thresholds have supremum \(\omega\), and the countable ordinal \(\omega+1\) lies strictly above all of them.

The cofinal viewpoint explains the topological significance without requiring a separate test for every possible neighborhood. At a final endpoint in an order topology, local-base questions reduce to whether the available thresholds are cofinal below that endpoint. For \(\omega_1\), the countable-union argument rules out countable cofinality, and the theorem transfers that order-theoretic fact directly into a statement about the topology. In particular, the space is not first countable at \(\omega_1\): first countability at a point means that the point has a countable local base.