Worked solution: Independence of Suslin's Hypothesis from ZFC (Jensen, Solovay–Tennenbaum, 1971)

Step 1 of 8: Suslin's question: can "separable" be weakened to "ccc"?

In plain words

The real line $\mathbb{R}$ has several familiar order properties: it is dense (between any two points there is another), complete (no gaps), has no first or last point, and it is separable — the rationals $\mathbb{Q}$ sit inside it as a countable set that gets arbitrarily close to every point. Cantor proved these four properties, taken together, uniquely characterize $\mathbb{R}$ up to relabeling.

Separability has an easy consequence: any collection of pairwise disjoint open intervals must be countable, since each one has to contain a distinct rational number. Suslin asked, in 1920, whether this weaker consequence alone — called the countable chain condition — could replace separability in Cantor's characterization, or whether some mysterious "impostor" line could satisfy it without secretly having a countable dense subset.

R \text{ dense, complete, no endpoints, ccc} \ \overset{?}{\Longrightarrow}\ R \cong \mathbb{R}

Detailed analysis

Cantor's characterization theorem states that a linearly ordered set which is dense (between any two points lies a third), complete (every bounded set has a supremum), without endpoints, and separable (has a countable dense subset) is order-isomorphic to $\mathbb{R}$ . Each hypothesis is needed; the question Suslin posed in 1920 concerns weakening the last one.

Separability implies the countable chain condition (ccc): any family of pairwise disjoint nonempty open intervals is countable, since each interval must contain a distinct point of the countable dense set. The converse is not obvious at all — disjointness of intervals is a much weaker demand than density of a fixed countable set, and a priori a ccc continuum could 'hide' its density behind uncountably many mutually avoiding regions.

Suslin asked whether ccc can replace separability in Cantor's theorem: is every dense complete endpoint-free ccc order isomorphic to $\mathbb{R}$ ? A counterexample — a ccc non-separable linear continuum — is called a Suslin line, and Suslin's Hypothesis ( $\mathrm{SH}$ ) asserts none exists. The question resisted direct attack for decades because the order-theoretic hypothesis ccc is hard to exploit in constructions; the next step's equivalence with a tree problem is what made it tractable.

Terms in this step

Separable order: A linearly ordered set that contains a countable subset getting arbitrarily close to every point, the way $\mathbb{Q}$ sits densely inside $\mathbb{R}$ .
Countable chain condition (ccc): An order (or more generally a poset) satisfies ccc if every family of pairwise disjoint nonempty open intervals (or pairwise incompatible elements) is countable.

Knowledge used in this step

The ZFC axioms