On ADS (Part 4)

February 10, 2012 at 14:40 | Posted in Uncategorized | Leave a comment

So what does it mean when we say {k_\beta} “almost works”? We’ll start with a lemma:

Lemma 1 For a fixed {\nu\in^{<\omega}\mu}, the set {\{\alpha<\beta:\nu=\eta_\alpha\upharpoonright k_\beta(\alpha)+1\}} contains at most one element from each set {A_{n+1}^\beta\setminus A^\beta_n}, and hence is at most countable.

Proof: Suppose by way of contradiction what {\alpha} and {\gamma} are distinct members of {A^\beta_{n+1}\setminus A^\beta_n} for which

\displaystyle  \nu=\eta_\alpha\upharpoonright (k_\beta(\alpha)+1)=\eta_\gamma\upharpoonright (k_\beta(\gamma)+1). \ \ \ \ \ (1)


\displaystyle  f^\beta_{n+1}(\alpha)=\eta_\alpha(k_\beta(\alpha))=\nu(k_\beta(\alpha))=\nu(k_\beta(\gamma))=\eta_\gamma(k_\beta(\gamma))=f^\beta_{n+1}(\gamma), \ \ \ \ \ (2)

and this contradicts the fact that {f^\beta_{n+1}} is a transversal for {\{x_\epsilon:\epsilon\in A^\beta_{n+1}\}}. \Box

Now given {\alpha<\beta}, we are going to define {E(\alpha)} to be those {\gamma<\beta} for which {k_\beta} has failed to work, that is,

\displaystyle  E(\alpha)=\{\gamma<\beta:\max\{k_\beta(\alpha),k_\beta(\gamma)\}<\Delta(\alpha,\gamma)\}. \ \ \ \ \ (3)

Lemma 2 The set {E(\alpha)} is at most countable.

Proof: If not, find {k^*<\omega} for which the set {B:=\{\gamma\in E(\alpha):k_\beta(\gamma)=k^*\}} is uncountable, and set

\displaystyle  \nu=\eta_\alpha\upharpoonright k^*+1. \ \ \ \ \ (4)

Then for each {\gamma\in B}, we have

\displaystyle  \eta_\gamma\upharpoonright k_\beta(\gamma)+1=\eta_\alpha\upharpoonright k^*+1=\nu, \ \ \ \ \ (5)

which contradicts the preceding lemma. \Box

So for a given {\alpha<\beta}, the function {k_\beta} will “disjointify” {A_\alpha} from all but countably many {A_\gamma} with {\gamma<\beta}: the function {k_\beta} “almost works”.

Notice that {\gamma\in E(\alpha)} if and only if {\alpha\in E(\gamma)}, so that we can define a graph {\Gamma} on {\beta} by connecting {\alpha} and {\gamma} if and only if {\gamma\in E(\alpha)}. By the preceding lemma, every vertex of the graph has at most countably many edges coming out of it. An easy argument tells us that the connected components of our graph {\Gamma} are at most countable as well.

Now if {\alpha} and {\gamma} are members of different connected components of {\Gamma} then {k_\beta} will disjointify {A_\alpha} and {A_\gamma}. But each connected component of {\Gamma} can be disjointified easily (by induction) as it is at most countable.

Thus, it is straightforward now to “correct” {k_\beta} to a function {h_\beta} which will work everywhere.


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