The Long Line’s Local and Global Topology

One Space, Two Very Different Scales

The long line is built by placing a copy of the unit interval at each countable ordinal, in ordinal order. Its order topology makes each point locally resemble a point of the real line. Globally, however, the long line has uncountably many disjoint open intervals, which prevents any countable set from being dense. We will prove both facts directly from the ordered construction.

Let \(\omega_1\) denote the first uncountable ordinal. The long ray \(P\) is the set \(\omega_1\times[0,1)\), ordered lexicographically: \((\alpha,s)<(\beta,t)\) when either \(\alpha<\beta\), or \(\alpha=\beta\) and \(s<t\). Give \(P\) its order topology. Its least point is \(o=(0,0)\). The long line \(L\) is formed from a reversed copy of \(P\) and a forward copy of \(P\) by identifying their two copies of \(o\). Order the reversed copy first, the shared point next, and the forward copy last, using the reversed order on the first copy and the given order on the second. Give \(L\) the order topology. Thus \(L\) has no endpoints. We use the standard result that an order isomorphism between linearly ordered sets is a homeomorphism for their order topologies.

The Order-Interval Fact Behind the Local Charts

At a point where two ordinal blocks meet, a neighborhood can involve several blocks. The reason such a neighborhood still has the order of a real interval is that every index below \(\omega_1\) is countable. We first record the ordered-interval fact we will use.

Proof. Add \(a\) and \(b\) to a countable dense subset \(D\) of \(K\), retaining a countable dense set that contains both endpoints. Between any two distinct points of \(D\) there is another point of \(D\): density in \(K\) first gives a point strictly between them, and density of \(D\) then gives a point of \(D\) strictly between them. Thus \(D\setminus\{a,b\}\) is a countable dense order without endpoints. The back-and-forth construction gives an order isomorphism from \(D\) to \(\mathbb{Q}\cap[0,1]\): enumerate the two countable sets, alternately match the first still-unmatched element on each side to an element in the required open interval on the other side. Such an element exists because both interior orders are dense and have no endpoints. The matching preserves order at each finite stage, and every element is eventually matched.

Extend this map to \(K\) by sending \(x\in K\) to the supremum in \([0,1]\) of the images of the points \(d\in D\) with \(d\le x\). This extension is strictly increasing: if \(x<y\), density supplies \(d_1,d_2\in D\) with \(x<d_1<d_2<y\), so the cut for \(x\) lies strictly below the cut for \(y\). It is onto: for \(r\in[0,1]\), take the least upper bound in \(K\) of the \(d\in D\) whose images are at most \(r\). The defining cut has image \(r\), since the rationals are dense in \([0,1]\). Therefore the extension is an order isomorphism. The assumption \(a<b\) is necessary: a one-point interval does not have the order type of \([0,1]\). \(\square\)

Consider now a closed interval in the long ray whose block indices lie between \(\beta\) and \(\alpha\), where \(\alpha<\omega_1\). It has only countably many block indices, because every ordinal below \(\omega_1\) is countable. It is densely ordered: within one block, real coordinates lie between distinct coordinates; across blocks, a point with a larger coordinate in the earlier block lies between the given points. It has a countable dense subset, obtained from rational coordinates in each block together with the block-start points that lie in the interval.

Such a closed interval is also order-complete. To see why, consider a nonempty subset bounded above within the interval. If its block indices have no greatest index, their supremum is a limit ordinal \(\gamma\), and the points are cofinal in the blocks below \(\gamma\); the least upper bound is \((\gamma,0)\). If the indices have a greatest index \(\gamma\), take the supremum \(s\) of the coordinates of the subset in that block. When \(s<1\), the least upper bound is \((\gamma,s)\); when \(s=1\), it is \((\gamma+1,0)\). A bound within the interval ensures that the resulting point is in the interval. Consequently, the lemma applies to every nondegenerate closed interval of this form.

Real-Line Neighborhoods at Every Point

Proof. It suffices to work on the forward ray, at its shared initial point, and then use reflection of order for points on the reversed ray. If \(x=(\alpha,t)\) with \(0<t<1\), choose \(s,t'\) with \(0\le s<t<t'<1\). The closed interval from \((\alpha,s)\) to \((\alpha,t')\) is an ordinary real closed interval in one block, so its open interior is homeomorphic to \((0,1)\), and hence to \(\mathbb{R}\).

Now let \(x=(\alpha,0)\) with \(\alpha>0\). Choose \(\beta<\alpha\). If \(\alpha\) is a successor ordinal, we may take its predecessor for \(\beta\); if \(\alpha\) is a limit ordinal, any smaller ordinal will do. Put \(u=(\beta,1/2)\) and \(v=(\alpha,1/2)\). The closed interval \([u,v]\) is nondegenerate, lies within countably many blocks, and has the completeness, dense order, and countable dense subset established above. The lemma makes \([u,v]\) order-isomorphic, and therefore homeomorphic, to \([0,1]\). Its open interval \((u,v)\) is an open neighborhood of \(x\), and it is homeomorphic to \((0,1)\). Finally, \(r\mapsto\log(r/(1-r))\) is a homeomorphism from \((0,1)\) to \(\mathbb{R}\). At the shared point \(o\), the closed interval between the points with coordinate \(1/2\) in the first negative and first positive blocks is order-isomorphic to \([-1/2,1/2]\), by mapping a negative coordinate \(s\) to \(-s\), \(o\) to \(0\), and a positive coordinate \(s\) to \(s\). Its open interior is again a real-line neighborhood. Reflection gives the same conclusion on the reversed ray. Thus every point has the required neighborhood. \(\square\)

Uncountably Many Open Intervals

Local resemblance to \(\mathbb{R}\) does not make the long line globally the same as \(\mathbb{R}\). For each \(\alpha<\omega_1\), define the open interval \(U_\alpha=((\alpha,1/4),(\alpha,3/4))\) in the forward ray. It is a nonempty open set in \(L\), since it is an open order interval. It lies strictly inside the block indexed by \(\alpha\), so if \(\alpha\ne\beta\), then \(U_\alpha\cap U_\beta=\varnothing\).

Proof. Suppose \(D\subseteq L\) were a countable dense set. Density means that every nonempty open set meets \(D\), so \(D\cap U_\alpha\ne\varnothing\) for every \(\alpha<\omega_1\). Since the intervals \(U_\alpha\) are pairwise disjoint, distinct indices require distinct points of \(D\). More formally, enumerate \(D\) if it is infinite; assign to each \(\alpha\) the least enumeration index of a point in \(D\cap U_\alpha\). These indices are distinct, giving an injection from the uncountable set \(\omega_1\) into a countable set, which is impossible. If \(D\) is finite, it cannot meet the uncountably many pairwise disjoint intervals either. Hence no countable dense subset exists. \(\square\)

What the Two Conclusions Do—and Do Not—Say

The local theorem says that around each individual point there is an open neighborhood with the topology of the real line. It does not say that one can choose a countable dense set for the whole space. The nonseparability proof depends on the global arrangement of the blocks, not on a failure of real-line behavior near any particular point.

A common mistake is to infer that a space is separable because every point has a neighborhood homeomorphic to \(\mathbb{R}\). That inference is invalid: local charts concern one neighborhood at a time, while a dense subset must meet every nonempty open set throughout the entire space. Here the uncountable family \(\{U_\alpha:\alpha<\omega_1\}\) makes the obstruction explicit. The argument proves nonseparability directly, without needing any additional claims about how separability behaves under homeomorphisms.