proof of complete partial orders do not add small subsets

Take any x𝔐[G], xκ. Let x^ be a name for x. There is some pG such that

px^ is a subset of κ bounded by λ<κ


For any qp, we construct by inductionMathworldPlanetmath a series of elements qα stronger than p. Each qα will determine whether or not αx^. Since we know the subset is bounded below κ, we can use the fact that P is κ completePlanetmathPlanetmathPlanetmathPlanetmathPlanetmath to find a single element stronger than q which fixes the exact value of x^. Since the series is definable in 𝔐, so is x^, so we can conclude that above any element qp is an element which forces x^𝔐. Then p also forces x^𝔐, completing the proof.


Since forcingMathworldPlanetmath can be described within 𝔐, S={qPqx^V} is a set in 𝔐. Then, given any qp, we can define q0=q and for any qα (α<λ), qα+1 is an element of P stronger than qα such that either qα+1α+1x^ or qα+1α+1x^. For limit α, let qα be any upper bound of qβ for α<β (this exists since P is κ-complete and α<κ), and let qα be stronger than qα and satisfy either qα+1αx^ or qα+1αx^. Finally let q* be the upper bound of qα for α<λ. q*P since P is κ-complete.

Note that these elements all exist since for any pP and any (first-order) sentenceMathworldPlanetmath ϕ there is some qp such that q forces either ϕ or ¬ϕ.

q* not only forces that x^ is a bounded subset of κ, but for every ordinalMathworldPlanetmathPlanetmath it forces whether or not that ordinal is contained in x^. But the set {α<λq*αx^} is defineable in 𝔐, and is of course equal to x^[G*] in any generic G* containing q*. So q*x^𝔐.

Since this holds for any element stronger than p, it follows that px^𝔐, and therefore x^[G]𝔐.

Title proof of complete partial orders do not add small subsets
Canonical name ProofOfCompletePartialOrdersDoNotAddSmallSubsets
Date of creation 2013-03-22 12:53:35
Last modified on 2013-03-22 12:53:35
Owner Henry (455)
Last modified by Henry (455)
Numerical id 4
Author Henry (455)
Entry type Proof
Classification msc 03E40