The Long Line And Its Local Euclidean Structure

From Ordinal Intervals to a Long Line

The real line is locally an open interval at every point, but its order has only countably many unit intervals in either direction. The long line is built by concatenating unit intervals along all countable ordinals. This creates an ordered space with uncountably many interval blocks while retaining a real-interval neighborhood around each point. The construction makes the distinction between local structure and the size of the whole space explicit.

Let \(\omega_1\) denote the first uncountable ordinal, so its elements are exactly the countable ordinals. Give \(\omega_1\times[0,1)\) the lexicographic order: \((\alpha,s)<(\beta,t)\) if either \(\alpha<\beta\), or \(\alpha=\beta\) and \(s<t\). Equip this ordered set with its order topology, generated by open order intervals and the corresponding open rays at an endpoint.

In each copy, the points \((\alpha,t)\) for a fixed \(\alpha\) form one interval block. The block has the order of \([0,1)\); its missing right endpoint is supplied as the first point \((\alpha+1,0)\) of the next block. At a limit ordinal \(\lambda<\omega_1\), the point \((\lambda,0)\) follows all the blocks indexed below \(\lambda\). It is not the first point of a block preceded by an immediate last point: there is no largest ordinal below \(\lambda\).

A Criterion for Bounded Intervals

We use a fact about ordered sets to produce local real-interval charts. A linear order is dense if between any two distinct points lies a third. It is order-complete if every nonempty subset that has an upper bound has a least upper bound. A subset is countable-dense if it has a countable subset meeting every nonempty open interval.

Proof. Remove the first and last points and let \(D\) be a countable-dense subset of the remaining order. It is a countable dense linear order with no first or last point. Such an order is isomorphic to \(\mathbb{Q}\cap(0,1)\): enumerate both sets and build finite order-preserving partial bijections in stages. At alternate stages take the first element not yet used in one enumeration and assign it an unused element of the other set between the images of its already assigned lower and upper neighbors. Density and the absence of endpoints provide such an element. The resulting union is an order-preserving bijection \(g:D\to\mathbb{Q}\cap(0,1)\).

Extend \(g\) to the whole order by sending the first and last points to \(0\) and \(1\), and, for every other point \(x\), setting \(h(x)=\sup\{g(d):d\in D,\ d<x\}\). This supremum exists in \([0,1]\). Density of \(D\) and of the rationals shows that \(h\) is strictly order-preserving: if \(x<y\), choose \(d_1,d_2\in D\) with \(x<d_1<d_2<y\); then \(h(x)\le g(d_1)<g(d_2)\le h(y)\). To see that it is onto, fix \(r\in(0,1)\) and take the supremum \(x\) of \(\{d\in D:g(d)<r\}\). The rational values below \(r\) have supremum \(r\), so the definition gives \(h(x)=r\). Thus \(h\) is an order isomorphism onto \([0,1]\). \(\square\)

Here is why bounded closed intervals in \(L\) meet the lemma’s hypotheses. Every bounded interval lies within a portion of the construction involving only countably many ordinal blocks: its two endpoints have countable ordinal coordinates, and every ordinal between those coordinates is countable. Consequently, the rational-coordinate points \((\alpha,q)\), with \(q\in\mathbb{Q}\cap[0,1)\), give a countable-dense subset after restricting to the interval. The same is true of an interval crossing \(o\), by using rational-coordinate points on both sides and \(o\).

These closed intervals are also order-complete. To find the supremum of a nonempty subset, first take the supremum of its ordinal coordinates. This is still a countable ordinal because the coordinates are bounded by a countable ordinal, and a countable union of countable sets is countable. If the coordinate supremum is not attained, the corresponding point with second coordinate \(0\) is the least upper bound. If it is attained, take the supremum of the second coordinates in that block; when this is less than \(1\), it gives the least upper bound in that block, and when it is \(1\), the least upper bound is the first point of the next block. Restricting to a bounded closed interval keeps the resulting supremum within the interval. Reversing the order on the negative side gives the corresponding completeness there. Finally, the order is dense: within a block this follows from density of \([0,1)\), and across blocks one can find another point in the earlier block above any given point there. Thus the lemma applies to every nondegenerate bounded closed interval in \(L\).

Every Point Has a Real-Interval Neighborhood

Proof. Choose points \(a<x<b\) in \(L\). Such points exist even when \(x=o\), since both copies have points other than their first point. The closed order interval \([a,b]\) is nondegenerate and satisfies the lemma, so it is order-isomorphic to \([0,1]\). The order isomorphism is a homeomorphism for the order topologies: it sends open order intervals and open rays to the corresponding intervals and rays. Its restriction sends \((a,b)\) homeomorphically to \((0,1)\). The interval \((a,b)\) is an open neighborhood of \(x\), so it is the required neighborhood. \(\square\)

The theorem includes points at limit ordinals. Although there is no immediately preceding block at such a point, one can still choose a bounded interval around it. The countably many blocks that occur in this bounded interval are enough for the countability and completeness argument. The following examples exhibit charts at three different kinds of points.

Uncountable Extent and a Common Pitfall

There are uncountably many blocks in each long ray, one for each \(\alpha<\omega_1\), even though every point has a neighborhood that looks like an ordinary real interval. A common mistake is to replace the ordinal indexing by a sequence of blocks and then treat that sequence as if it reached all the way along the long ray. It cannot.

Common pitfall. Suppose someone proposes a sequence of countable ordinals \((\alpha_n)\) that is cofinal in \(\omega_1\), meaning that for every \(\beta<\omega_1\) some \(\alpha_n\) is at least \(\beta\). This is impossible. The supremum \(\gamma=\sup_n\alpha_n\) is countable: it is the union of countably many countable ordinals, hence a countable set. Therefore \(\gamma<\omega_1\), and every \(\alpha_n\le\gamma\). The ordinal \(\gamma+1\) is still countable, but no \(\alpha_n\) reaches it, contradicting cofinality. The correction is to retain all countable ordinals in the construction, rather than assume a countable sequence of blocks reaches the end.

This example is useful because a local description does not specify the global extent of a space. The long line has real-interval neighborhoods, but its order is built from uncountably many blocks, not from the usual countable succession of intervals in \(\mathbb{R}\). When examining a space, check both the neighborhood structure and how the whole space is assembled.