Countably Compact But Noncompact Long Lines

Countable Sets and the Long Line

The long line resembles the real line in a neighborhood of each point, but its global order extends much farther: every countable collection of its points is bounded, while the whole line is not. This contrast produces a useful example of a space that is countably compact but not compact. We will prove the accumulation-point property directly from the order, then build an open cover that has no finite subcover.

Let \(\omega_1\) denote the first uncountable ordinal, so each ordinal \(\alpha<\omega_1\) is countable. The long ray is the set \(\omega_1\times[0,1)\), ordered lexicographically: \((\alpha,s)<(\beta,t)\) if either \(\alpha<\beta\), or \(\alpha=\beta\) and \(s<t\). The long line \(L\) is formed from two copies of this ray by identifying their initial points. We write the shared point as \(0\), and write the other points as \(+(\alpha,t)\) and \(-(\alpha,t)\). The positive copy has its long-ray order; the negative copy has the reverse order, with every negative point less than \(0\). We give \(L\) its order topology, whose basic open sets are open intervals and open rays.

For a finite set, an accumulation point need not exist. Thus the accumulation-point assertion below concerns every countably infinite subset, or equivalently every countable subset with infinitely many distinct points. The empty set and finite subsets cause no exception to the definition of countable compactness.

Bounded Intervals Are Compact

We first record the order fact used in the proof. Every bounded closed interval \([a,b]\) in \(L\) is order-complete: each nonempty subset has a least upper bound in \([a,b]\). To see why, on a positive ray take the supremum of the ordinal coordinates occurring in the subset. Because the subset is bounded by a point with countable ordinal coordinate, this supremum is still a countable ordinal. If the coordinates are cofinal below that ordinal, the supremum point is the start of its block, \((\alpha,0)\). If the largest ordinal coordinate occurs in the subset, take the real supremum of the second coordinates in that block; if that supremum is \(1\), the least upper bound is the start of the next block. The negative ray has the reversed version of the same construction. A subset meeting both sides is handled by taking its supremum on the side containing its greatest points, or \(0\) if it has no positive points. Restricting to \([a,b]\) keeps the resulting supremum in that interval.

Proof. Consider a nonempty interval \([a,b]\) and an open cover \(\mathcal{U}\) of it. The order on \(L\) is dense: between two points in the same block there is another real coordinate, and between \((\alpha,t)\) and \((\alpha+1,0)\) there is a coordinate in the \(\alpha\)-block strictly larger than \(t\). The same holds on the negative side and across \(0\).

Let \(E\) be the set of points \(x\in[a,b]\) for which \([a,x]\) has a finite subcover from \(\mathcal{U}\). An open set containing \(a\) contains \([a,y]\) for some \(y>a\), unless \(a=b\); hence \(E\) is nonempty (and if \(a=b\), the interval is already covered by one member). Let \(c=\sup E\), which exists by order completeness. Choose \(U\in\mathcal{U}\) containing \(c\). Openness gives points \(u<c<v\) such that \((u,v)\cap[a,b]\subseteq U\), with the corresponding one-sided interval if \(c\) is an endpoint. Since \(u<c=\sup E\), some \(x\in E\) satisfies \(u<x\le c\). A finite cover of \([a,x]\), together with \(U\), covers \([a,c]\); thus \(c\in E\). If \(c<b\), openness of \(U\) and density of the order give \(y>c\) with \([c,y]\subseteq U\). The finite cover of \([a,c]\), together with \(U\), then covers \([a,y]\), contradicting that \(c\) is an upper bound for \(E\). Therefore \(c=b\), so \([a,b]\) has a finite subcover. \(\square\)

For the long line, this lemma applies to any interval with finite endpoints. It does not assert that the whole line is compact: the whole line has no rightmost or leftmost point, and is not a bounded closed interval.

Every Countably Infinite Subset Has an Accumulation Point

Proof. Let \(S\subseteq L\) be countably infinite. The set of ordinal coordinates of its points is countable. The supremum of countably many countable ordinals is countable: it is the union of countably many countable sets, hence is countable. Choose a countable ordinal \(\gamma\) strictly larger than all the coordinates occurring in \(S\). Then every point of \(S\) lies strictly between \(-(\gamma,0)\) and \(+(\gamma,0)\). In particular, \(S\) is contained in the compact interval \(K=[-(\gamma,0),+(\gamma,0)]\).

Suppose that \(S\) has no accumulation point in \(L\). For each \(x\in K\), the definition of accumulation point gives an open neighborhood \(U_x\) of \(x\) such that \(U_x\cap(S\setminus\{x\})=\varnothing\). The relative open sets \(U_x\cap K\) cover \(K\). Compactness of \(K\) gives a finite subcover, say \(U_{x_1}\cap K,\ldots,U_{x_m}\cap K\). Each \(U_{x_i}\) contains at most the one point \(x_i\) of \(S\). Since \(S\subseteq K\), these finitely many sets can cover at most \(m\) points of \(S\), contradicting that \(S\) is infinite. Thus \(S\) has an accumulation point. \(\square\)

The key is not that every countable subset is compact; the key is that every countable subset is contained in a compact interval. To bound the set, we use the countability of its ordinal coordinates. The whole long line, in contrast, runs through all countable ordinals and has no such bound.

Countable Compactness and a Cover With No Finite Subcover

The accumulation-point result implies countable compactness here. We include the argument to make explicit the role of the separation property. The order topology on \(L\) is Hausdorff: if \(x<y\), density gives \(z\) with \(x<z<y\), and the open rays \((-\infty,z)\) and \((z,\infty)\) are disjoint neighborhoods of \(x\) and \(y\). In particular, singletons are closed.

Proof. Let \(\{V_n:n\in\mathbb{N}\}\) be a countable open cover. If it has a finite subcover, there is nothing to prove. Otherwise, for each \(n\) choose \(x_n\in L\setminus\bigcup_{j=1}^n V_j\); such a point exists because the first \(n\) sets do not cover \(L\). The points \(x_n\) take infinitely many distinct values: if a point \(x\) occurred for indices \(n\ge m\), then \(x\in V_m\), since the family covers \(L\), but the choice of \(x_n\) excludes \(V_m\) whenever \(n\ge m\). Thus the set \(S=\{x_n:n\in\mathbb{N}\}\) is infinite and countable.

For any \(x\in L\), choose \(m\) with \(x\in V_m\). Every \(x_n\) with \(n\ge m\) lies outside \(V_m\). Remove from \(V_m\) the finitely many points \(x_n\ne x\) with \(n<m\). The resulting set is still open, because singletons are closed; it is a neighborhood of \(x\) and contains no point of \(S\) other than possibly \(x\). Hence \(x\) is not an accumulation point of \(S\). This holds for every \(x\), contradicting the theorem that every countably infinite subset of \(L\) has an accumulation point. Therefore the cover has a finite subcover. \(\square\)

Now index positive points by nonzero countable ordinals, writing \(r_\alpha=+(\alpha,0)\), and consider the open intervals \(U_\alpha=(-r_\alpha,r_\alpha)\). Each interval is bounded: its endpoints are the two specified points \( -r_\alpha\) and \(r_\alpha\). These intervals cover \(L\). Indeed, if a point has coordinate \((\beta,t)\) on either side, then it lies strictly between \(-r_{\beta+1}\) and \(r_{\beta+1}\), since \(\beta<\beta+1<\omega_1\) and \(t<1\). The origin is also in every \(U_\alpha\).

For a concrete finite selection, \(U_2,U_5,U_7\) have union \(U_7\), and \(r_8\notin U_7\). The same maximum-ordinal argument works for every finite selection, not just this one.

A Common Pitfall

Common pitfall. One might treat a countable sequence of long-line points as though it must escape every bounded interval, because its ordinal coordinates increase. For \(S=\{+(n,1/2):n\in\mathbb{N}\}\), that calculation is wrong: the finite ordinals \(n\) are all below the countable ordinal \(\omega\), and every neighborhood of \(+(\omega,0)\) contains points of \(S\). The correction is to distinguish “unbounded in the ordinary real line” from “unbounded in the long line.” Countably many countable ordinal coordinates have a countable supremum, so they cannot run beyond all the blocks of \(L\).

The long line therefore separates two compactness tests that can look similar. Every countable open cover has a finite subcover, yet the specific cover by bounded intervals above has no finite subcover. Countable compactness controls countable covers; compactness controls covers of every cardinality. The distinction is witnessed here by the uncountable family of intervals indexed by \(\omega_1\).