Why The Long Line Is Not Metrizable

Two Obstacles to Metrizability

The long line looks locally like the real line, but its global order has a feature that the real line does not: it contains uncountably many pairwise disjoint nonempty open intervals. This prevents it from having a countable basis. That fact alone does not rule out a metric, because metric spaces can have uncountably many disjoint open sets. The second ingredient is countable compactness: a countably compact metric space must be second countable. Together, these facts show that the long line is not metrizable.

We use the following model. Let \(\omega_1\) be the first uncountable ordinal, so its elements are precisely the countable ordinals. Order \(\omega_1\times[0,1)\) lexicographically: \((\alpha,s)<(\beta,t)\) if either \(\alpha<\beta\), or \(\alpha=\beta\) and \(s<t\). Give this set its order topology, whose basic open sets are open intervals in the order, together with the appropriate open rays at any endpoints.

The long line \(\mathbb L\) is obtained by placing a reversed copy of this long ray to the left of a copy of it, with their initial points identified. We give \(\mathbb L\) the resulting order topology. The proof of non-second-countability will use only intervals lying in the positive copy. We also use the standard countable-compactness theorem for the long line, established in earlier material on countably compact long lines: every countable open cover of \(\mathbb L\) has a finite subcover. Countable compactness is weaker than compactness, which requires a finite subcover for every open cover, including uncountable ones.

Disjoint Open Sets and Countable Bases

Proof. Let \(\mathcal B\) be a countable basis and let \(\mathcal U\) be a family of pairwise disjoint, nonempty open subsets of \(X\). For each \(U\in\mathcal U\), choose a point \(x_U\in U\). Since \(\mathcal B\) is a basis and \(U\) is open, there is some \(B_U\in\mathcal B\) with \(x_U\in B_U\subseteq U\). If \(U,V\in\mathcal U\) are different, then \(U\cap V=\varnothing\). Thus \(B_U\ne B_V\), since a nonempty basis element contained in both would meet both \(U\) and \(V\). The assignment \(U\mapsto B_U\) is injective from \(\mathcal U\) into the countable set \(\mathcal B\), so \(\mathcal U\) is at most countable. \(\square\)

Why Countable Compactness Matters for Metric Spaces

The preceding argument does not say that a space with uncountably many disjoint open sets cannot be metrizable. Instead, we need the additional hypothesis of countable compactness. Here is the result that connects that hypothesis to the basis obstruction.

Proof. Let \((X,d)\) be a countably compact metric space. If \(X=\varnothing\), the empty family is a countable basis, so assume \(X\ne\varnothing\). First we prove that every sequence in \(X\) has a convergent subsequence.

Take a sequence \((x_n)\). If some point occurs infinitely often, those terms give a constant, hence convergent, subsequence. Otherwise the sequence has a subsequence of distinct terms; write its set of values as \(A=\{a_1,a_2,\ldots\}\). If \(A\) has no limit point, then for each \(x\in X\) there is an open neighborhood meeting \(A\) in no point other than possibly \(x\). For \(x\notin A\), this neighborhood misses \(A\), so \(A\) is closed. For each \(j\), choose an open neighborhood \(V_j\) of \(a_j\) with \(V_j\cap A=\{a_j\}\). The countable family consisting of \(X\setminus A\) and all the \(V_j\) covers \(X\). A finite subfamily covers only finitely many points of \(A\), since \(X\setminus A\) contains none and each \(V_j\) contains only \(a_j\) from \(A\). This contradicts countable compactness. Therefore \(A\) has a limit point \(p\).

Every ball about \(p\) contains infinitely many points of \(A\) other than \(p\): if some ball contained only finitely many, a smaller ball could exclude those finitely many points, contradicting that \(p\) is a limit point. We can therefore choose distinct terms of the subsequence, with increasing original indices, whose distances from \(p\) are less than \(1,1/2,1/3,\ldots\), respectively. These terms converge to \(p\). This proves sequential compactness.

Next, for each \(n\in\mathbb N\), there is a finite set \(F_n\subseteq X\) such that the balls \(B_d(x,1/n)\), for \(x\in F_n\), cover \(X\). Otherwise we could choose \(x_1,x_2,\ldots\) recursively so that each new point lies outside the balls of radius \(1/n\) around all earlier choices. Then \(d(x_i,x_j)\ge 1/n\) whenever \(i\ne j\). No subsequence could converge, contradicting sequential compactness.

Set \(D=\bigcup_{n\in\mathbb N}F_n\). It is countable, being a countable union of finite sets, and dense: given \(x\in X\) and \(\varepsilon>0\), choose \(n\) with \(1/n<\varepsilon\). Some \(q\in F_n\) satisfies \(d(x,q)<1/n<\varepsilon\).

Consider all balls \(B_d(q,r)\) with \(q\in D\) and positive rational \(r\). This is a countable family. To verify it is a basis, let \(U\) be open and \(x\in U\). Choose \(\varepsilon>0\) with \(B_d(x,\varepsilon)\subseteq U\), and choose \(q\in D\) with \(d(x,q)<\varepsilon/3\). There is a positive rational \(r\) satisfying \(d(x,q)<r<\varepsilon-d(x,q)\). Then \(x\in B_d(q,r)\). If \(y\in B_d(q,r)\), the triangle inequality gives \[ d(y,x)\le d(y,q)+d(q,x)<r+d(q,x)<\varepsilon, \] so \(B_d(q,r)\subseteq U\). The countable family is a basis, proving the theorem. \(\square\)

The Nonmetrizability Conclusion

Proof. Suppose, for contradiction, that \(\mathbb L\) has a topology induced by a metric. Countable compactness is a property of the topology, so the metric space \(\mathbb L\) is countably compact. Theorem 2 then implies that \(\mathbb L\) is second countable. But the intervals \(U_\alpha\) constructed above form an uncountable family of pairwise disjoint nonempty open sets, and Theorem 1 shows that a second-countable space cannot have such a family. This contradiction proves that \(\mathbb L\) is not metrizable. \(\square\)

The roles of the two ingredients are distinct. The long intervals rule out second countability directly. Countable compactness supplies the bridge from a hypothetical metric to second countability. Without that bridge, the existence of many disjoint open sets is not a metrizability obstruction by itself.