Alphabeta Math
Remark‡ sources checked 2026-07-26‡ not proved here
‡ Recorded, not proved here. This statement is included so the library can refer to it honestly, with a citation to the literature. It is not proved anywhere in this library: the track that would prove it has not been developed here yet.

Martin's Axiom

Statement

A partial order P has the countable chain condition (ccc) when every family of pairwise incompatible elements of P is countable. For a cardinal κ, MA(κ) asserts:

for every ccc partial order P and every family D of at most κ dense subsets of P, there is a filter on P meeting every member of D.

Martin's Axiom (MA) is the assertion that MA(κ) holds for every κ<2ℵ0.

Three facts fix its status.

(a) MA(ℵ0) is a theorem of ZFC (the Rasiowa-Sikorski lemma), so MA is not vacuous but its content is entirely in the uncountable cases.

(b) CH implies MA, trivially, since under CH there is no κ with ℵ0<κ<2ℵ0. So MA alone decides nothing that CH does not.

(c) If ZFC is consistent, then so is ZFC + MA + (not CH). This is Solovay and Tennenbaum (1971), and it is where finite-support iterated ccc forcing was invented: one iterates ccc posets ℵ2 times, catching every ccc poset of size less than the continuum along the way, and the iteration is itself ccc so no cardinal is collapsed.

What MA + (not CH) buys. Every set of reals of cardinality less than 2ℵ0 is Lebesgue null and meagre; the union of fewer than 2ℵ0 meagre sets is meagre; 2κ=2ℵ0 for every infinite κ<2ℵ0; and the product of two ccc spaces is ccc, so ccc is productive.

Remarks

Depends on

Used by

Dependency tree · two levels

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Sources