The square of a Suslin line is not ccc (checked on audit)

A Suslin line is a linearly ordered set whose order topology satisfies the countable chain condition (every family of pairwise disjoint nonempty open intervals is countable) but which is not separable.

The result recorded here. If $L$ is a Suslin line then the product $L \times L$ fails the countable chain condition.

Status: settled, and outside this library's stack. This is Lemma 4.3 of Chapter II of Kunen, Set Theory: An Introduction to Independence Proofs, stated there as "if $X$ is a Suslin line, $X^2$ is not c.c.c.", with the definition of a Suslin line given in Definition 4.1 of the same section exactly as above. It is a plain ZFC theorem, and it asserts nothing about whether a Suslin line exists: only what follows if one does. It is not proved here. The library now develops order topologies and transfinite recursion through $\omega_1$, but it has not authored Kunen's specialised Suslin-line construction and square argument.

History of this entry, kept deliberately. Until the audit of 2026-07-26 this item recorded the statement as unverified: the claim had been encountered, flagged for checking against Kunen, and not checked, and the item accordingly asserted nothing. The check has now been carried out against exactly that reference, and the claim is a theorem, in the exact place the note said to look. The item id still contains the word "unverified" because ids in this library are immutable; the status above is what holds.

Not proved in this library. The result is recorded and cited, not derived, and no page here may present it as established from anything on these pages. Nothing in this library depends on it.

The proof, and why it is out of reach rather than hard. Kunen's argument is a recursion of length $\omega_1$: one picks $a_\alpha < b_\alpha < c_\alpha$ in $L$ with both intervals $(a_\alpha, b_\alpha)$ and $(b_\alpha, c_\alpha)$ nonempty and with $(a_\alpha, c_\alpha)$ avoiding every $b_\xi$ already chosen, which is possible because $L$ is not separable. The $\omega_1$ many open rectangles $V_\alpha = (a_\alpha, b_\alpha) \times (b_\alpha, c_\alpha)$ are then nonempty and pairwise disjoint, so $L^2$ is not ccc. The obstacle here is the setting, not the difficulty: separability is developed later in Separability: the existence of an at most countable dense subset ↗, but this item still records rather than proves Kunen's specialised $\omega_1$-length recursion.

Relation to the partial-order form. The Suslin hypothesis is independent of ZFC, in both directions ‡ records, with citations, that a Suslin line yields a ccc partial order whose square is not ccc, which is the Suslin tree viewed as a forcing. The statement here is the topological one about the square of the line itself, and the two agree: the nonempty open subsets of a ccc space, ordered by inclusion, form a ccc partial order, so the rectangles above give the partial-order statement too.

What is still not asserted. The existence of a Suslin line is independent of ZFC: the consistency of Suslin's Hypothesis, that no Suslin line exists, is due to Solovay and Tennenbaum (1971) by iterated ccc forcing, while $\diamondsuit$ implies one exists, so one exists in the constructible universe. Nothing here asserts that a Suslin line exists, and therefore nothing here asserts outright that the countable chain condition fails to be productive. That conclusion is conditional on there being a Suslin line, which is precisely why "the product of two ccc spaces is ccc" is not decided by ZFC.

Why it matters here. Ccc arguments will appear in the library's topology material, and productivity of the countable chain condition is exactly the point at which a plausible-sounding claim silently imports an independence result. Any page wanting an unconditional ZFC counterexample about the countable chain condition should use the Cantor cube $\{0,1\}^{\kappa}$ for $\kappa$ larger than the continuum, which is ccc and not separable and needs no independence result at all, and leave the Suslin line to this item.