proof of is equivalent to and continuum hypothesis

The proof that S implies both S and that for every λ<κ, 2λκ are given in the entries for S and S.

Let A=AααS be a sequence which satisfies S.

Since there are only κ boundedPlanetmathPlanetmathPlanetmath subsets of κ, there is a surjective function f:κBounded(κ)×κ where Bounded(κ) is the bounded subsets of κ. Define a sequence B=Bαα<κ by Bα=f(α) if sup(Bα)<α and otherwise. Since the set of (Bα,λ)Bounded(κ)×κ such that Bα=T is unboundedPlanetmathPlanetmath for any bounded subset T, it follow that every bounded subset of κ occurs κ times in B.

We can define a new sequence, D=DααS such that xDαxBβ for some βAα. We can show that D satisfies S.

First, for any α, xDα means that xBβ for some βAα, and since BββAαα, we have Dαα.

Next take any Tκ. We consider two cases:

T is bounded

The set of α such that T=Bα forms an unbounded sequence T, so there is a stationary SS such that αSAαT. For each such α, xDαxBi for some iAαT. But each such Bi is equal to T, so Dα=T.

T is unbounded

We define a function j:κκ as follows:

  • j(0)=0

  • To find j(α), take X{j(β)β<α}. This is a bounded subset of κ, so is equal to an unbounded series of elements of B. Take j(α)=γ, where γ is the least number greater than any element of {α}{j(β)β<α} such that Bγ=X{j(β)β<α}.

Let T=range(j). This is obviously unbounded, and so there is a stationary SS such that αSAαT.

Next, consider C, the set of ordinalsMathworldPlanetmathPlanetmath less than κ closed under j. Clearly it is unbounded, since if λ<κ then j(λ) includes j(α) for α<λ, and so inductionMathworldPlanetmath gives an ordinal greater than λ closed under j (essentially the result of applying j an infiniteMathworldPlanetmath number of times). Also, C is closed: take any cC and suppose sup(cα)=α. Then for any β<α, there is some γc such that β<γ<α and therefore j(β)<γ. So α is closed under j, and therefore contained in C.

Since C is a club, C=CS is stationary. Suppose αC. Then xDαxBβ where βAα. Since αS, βrange(j), and therefore BβT. Next take any xTα. Since αC, it is closed under j, hence there is some γα such that j(x)γ. Since sup(Aα)=α, there is some ηAα such that γ<η, so j(x)η. Since ηAα, BηDα, and since ηrange(j), j(δ)Bη for any δ<j-1(η), and in particular xBη. Since we showed above that Dαα, we have Dα=Tα for any αC.

Title proof of is equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath to and continuum hypothesisMathworldPlanetmath
Canonical name ProofOfDiamondIsEquivalentToclubsuitAndContinuumHypothesis
Date of creation 2013-03-22 12:53:57
Last modified on 2013-03-22 12:53:57
Owner Henry (455)
Last modified by Henry (455)
Numerical id 6
Author Henry (455)
Entry type Proof
Classification msc 03E45
Synonym proof that diamond is equivalent to club and continuum hypothesis
Related topic DiamondMathworldPlanetmath
Related topic Clubsuit