Locally countable orderings

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Lemma Suppose X ⊂ 2 ω which only contains ∆ 1 2-random reals, if µ(X) = 0, then µ(U h (X)) = 0. Proof. Define R = {x|x is ∆ 1 2-random.}. Take a maximal set X ⊂ R so that ∀x ∈ X∀y ∈ X(x ≠ y =⇒ ∀z(z ≤ h x & z ≤ h y =⇒ z is not ∆ 1 2 -random)). Lemma There exists a nonmeasurable antichain in the h-degrees.
Solving the question Theorem For any locally <strong>countable</strong> Σ 1 1 partial order P = 〈2ω , ≤ P 〉, there exists a nonmeasurable antichain in P.
Page 1 and 2: Locally countable orderings Liang Y
Page 3 and 4: The basic Refs are Downey and Hirsc
Page 5 and 6: Motivation Counting the antichains.
Page 7 and 8: Higher recursion theory I x ≤ h y
Page 9 and 10: Proposition 1 Assume ZFC + V = L, t
Page 11 and 12: Definition Given a string σ ∈ 2
Page 13 and 14: 0-law Proposition If X ⊆ 2 ω and
Page 15 and 16: Calculating the complexity I Propos
Page 17 and 18: Attacking the question I Lemma For
Page 19 and 20: continued. Then µ({z ∈ 2 ω |
Page 21: continued. Define a ∆ 1 2 (z)-seq
Page 25 and 26: Applications to some specific order
Page 27 and 28: Possible generalizations II We say
Page 29: Thanks

Locally countable orderings

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