Martin’s axiom is consistent

If κ is an uncountable strong limit cardinal such that for any λ<κ, κλ=κ then it is consistentPlanetmathPlanetmath that 20=κ and MA. This is shown by using finite support iterated forcing to construct a model of ZFC in which this is true. Historically, this proof was the motivation for developing iterated forcing.


The proof uses the convenient fact that MAκ holds as long as it holds for all partial orders smaller than κ. Given the conditions on κ, there are at most κ names for these partial orders. At each step in the forcingMathworldPlanetmath, we force with one of these names. The result is that the actual generic subset we add intersects every dense subset of every partial order.

Construction of Pκ

Q^α will be constructed by inductionMathworldPlanetmath with three conditions: |Pα|κ for all ακ, PαQ^α, and Pα satisfies the ccc. Note that a partial ordering on a cardinal λ<κ is a functionMathworldPlanetmath from λ×λ to {0,1}, so there are at most 2λ<κ of them. Since a canonical name for a partial ordering of a cardinal is just a function from Pα to that cardinal, there are at most κ2λκ of them.

At each of the κ steps, we want to deal with one of these possible partial orderings, so we need to partitionMathworldPlanetmathPlanetmath the κ steps in to κ steps for each of the κ cardinals less than κ. In addition, we need to include every Pα name for any level. Therefore, we partion κ into Sγ,δγ,δ<κ for each cardinal δ, with each Sγ,δ having cardinality κ and the added condition that ηSγ,δ implies ηγ. Then each Pγ name for a partial ordering of δ is assigned some index ηSγ,δ, and that partial order will be dealt with at stage Qη.

Formally, given Q^β for β<α, Pα can be constructed and the Pα names for partial orderings of each cardinal δ enumerated by the elements of Sα,δ. αSγ,δ for some γα and δα, and αγα so some canonical Pγα name for a partial order ^α of δα has already been assigned to α.

Since ^α is a Pγα name, it is also a Pα name, so Q^α can be defined as δα,^α if Pαδα,^α satisfies the ccc and by the trivial partial order 1,{1,1} otherwise. Obviously this satisfies the ccc, and so Pα+1 does as well. Since Q^α is either trivial or a cardinal together with a canonical name, PαQ^α. Finally, |Pα+q|n|α|n(supi|Q^i|)nκ.

Proof that MAλ holds for λ<κ

Lemma: It suffices to show that MAλ holds for partial order with size λ


Suppose P is a partial order with |P|>κ and let Dαα<λ be dense subsets of P. Define functions fi:PDα for ακ with fα(p)p (obviously such elements exist since Dα is dense). Let g:P×PP be a function such that g(p,q)p,q whenever p and q are compatibleMathworldPlanetmath. Then pick some element qP and let Q be the closure of {q} under fα and g with the same ordering as P (restricted to Q).

Since there are only κ functions being used, it must be that |Q|κ. If pQ then fα(p)p and clearly fα(p)QDα, so each DαQ is dense in Q. In addition, Q is ccc: if A is an antichainMathworldPlanetmath in Q and p1,p2A then p1,p2 are incompatible in Q. But if they were compatible in P then g(p1,p2)p1,p2 would be an element of Q, so they must be incompatible in P. Therefore A is an antichain in P, and therefore must have countableMathworldPlanetmath cardinality, since P satisfies the ccc.

By assumptionPlanetmathPlanetmath, there is a directed GQ such that G(DαQ) for each α<κ, and therefore MAλ holds in full. ∎

Now we must prove that, if G is a generic subset of Pκ, R some partial order with |R|λ and Dαα<λ are dense subsets of R then there is some directed subset of R intersecting each Dα.

If |R|<λ then λ additional elements can be added greater than any other element of R to make |R|=λ, and then since there is an order isomorphism into some partial order of λ, assume R is a partial ordering of λ. Then let D={α,βαDβ}.

Take canonical names so that R=R^[G], D=D^[G] and Di=D^i[G] for each i<λ and:

PκR^ is a partial ordering satisfying ccc andD^λ×λ and Dα^ is dense in R^

For any α,β there is a maximal antichain Dα,βPκ such that if pDα,β then either pPκαR^β or pPκαR^β and another maximal antichain Eα,βPκ such that if pEα,β then either pPκα,βD^ or pPκα,βD^. These antichains determine the value of those two formulasMathworldPlanetmathPlanetmath.

Then, since κcfκ>κ and κμ=κ for μ<κ, it must be that cfκ=κ, so κ is regularPlanetmathPlanetmath. Then γ=sup({α+1αdom(p),pα,β<λDα,βEα,β)<κ, so Dα,β,Eα,βPγ, and therefore the Pκ names R^ and D^ are also Pγ names.

Lemma: For any γ, Gγ={pγpG} is a generic subset of Pγ


First, it is directed, since if p1γ,p2γGγ then there is some pG such that pp1,p2, and therefore pγGγ and pp1γ,p2γ.

Also, it is generic. If D is a dense subset of Pγ then Dκ={pPκpqD} is dense in Pκ, since if pPκ then there is some dp, but then d is compatible with p, so dpDκ. Therefore there is some pDκGκ, and so pDGγ. ∎

Since R^ and D^ are Pγ names, R^[G]=R^[Gγ]=R and D^[G]=D^[Gγ]=D, so

V[Gγ]R^ is a partial ordering of λ satisfying the ccc andDα^ is dense in R^

Then there must be some pGγ such that

pPγR^ is a partial ordering of λ satisfying the ccc

Let Ap be a maximal antichain of Pγ such that pAp, and define ^* as a Pγ name with (p,m)^* for each mR^ and (a,n)^* if n=(α,β) where α<β<λ and paAp. That is, ^*[G]=R when pG and ^*[G]=λ otherwise. Then this is the name for a partial ordering of λ, and therefore there is some ηSγ,λ such that ^*=^η, and ηγ. Since pGγGη, Q^η[Gη]=^η[Gη]=R.

Since Pη+1=Pη*Qη, we know that GQηQη is generic since with the compositionMathworldPlanetmath is equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmath to successive forcing. Since DiV[Gγ]V[Gη] and is dense, it follows that DiGQη and since GQη is a subset of R in Pκ, MAλ holds.

Proof that 20=κ

The relationship between Martin’s axiom and the continuum hypothesisMathworldPlanetmath tells us that 20κ. Since 20 was less than κ in V, and since |Pκ|=κ adds at most κ elements, it must be that 20=κ.

Title Martin’s axiom is consistent
Canonical name MartinsAxiomIsConsistent
Date of creation 2013-03-22 13:21:59
Last modified on 2013-03-22 13:21:59
Owner mathcam (2727)
Last modified by mathcam (2727)
Numerical id 7
Author mathcam (2727)
Entry type Result
Classification msc 03E50