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Let $\kappa = (2^{\aleph_0})^+$. Let $F : [\kappa]^2 \to \omega$ be a partition of $[\kappa]^2$ into $\omega$ pieces. Define $F_{a} : \kappa\setminus \{a\} \to \omega$ by $F_a (x) = F(\{a,x \})$. Construct a sequence $A_0 \subseteq A_1 \subseteq \ldots \subseteq A_{\alpha} \subseteq \ldots (\alpha \in \omega_1)$ of subsets of $\kappa$, where each set has size $2^{\aleph_0}$. To do this, define $A_0$ to be any old set in $\kappa$ which has cardinality $2^{{\aleph}_0}$, and for limit ordinals $\alpha$, let $A_{\alpha} = \bigcup_{\beta < \alpha} A_{\beta}$. Now I am getting stuck where we define the successor steps:

"Given $A_{\alpha}$, there exists a set $A_{\alpha+1} \supseteq A_{\alpha}$ of size $2^{\aleph_{0}}$ such that for each countable $C \subseteq A_{\alpha}$ and every $u ∈ κ \setminus C$ there exists $v ∈ A_{α+1} \setminus C$ such that $F_v$ agrees with $F_u$ on $C$."

I am unsure how to define $A_{\alpha + 1}$. Now since $C$ is countable, the number of functions from $C \to \omega$ is $2^{\aleph_0}$, so there must be $\le 2^{\aleph_0}$ functions on $C$ of the form $F_v$ for some $v \in \kappa$. So given any $u \in \kappa \setminus C$, there must exist some $v \in \kappa \setminus C$ where $F_u |_C = F_v|_C$, otherwise there would be too many functions on $C$. My idea was to define $A_{\alpha + 1}$ to be the collection of all such $v$, taken together with $A_{\alpha}$. But all I can work out is that $|A_{\alpha + 1}| \le |[A_{\alpha}^{< \omega_1}]| \cdot | \kappa \setminus C| = \kappa$ which is useless (multiplying each countable subset of $A_{\alpha}$ by every possible $v \in \kappa \setminus C$).

Can anyone shed some light on how to get $A_{\alpha + 1}$? Any help would be appreciated, thanks.

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up vote 4 down vote accepted

Enumerate $[A_\alpha]^{\aleph_0}$ as $\{C_\xi: \xi<2^{\aleph_0}\}$. For each $\xi$, we can find $D_\xi\subset \kappa\setminus C_\xi$ such that $|D_\xi|\leq 2^{\aleph_0}$ and for every $\nu \in \kappa\setminus C_\xi$ there is $\nu'\in D_\xi$ so that $F_\nu|_{C_\xi}=F_{\nu'}|_{C_\xi}$, the last part is possible since $|\{ F_\nu|_{C_\xi}: \nu\in \kappa\setminus C_\xi\}|\leq 2^{\aleph_0}$.

Finally, let $A_{\alpha+1}=\bigcup_{\xi<2^{\aleph_0}}D_\xi.$

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Is it true that $A_{\alpha} \subseteq \bigcup_{\xi < 2^{\aleph_0}} D_{\xi}$? I guess I can just unite this with $A_{\alpha}$ to ensure this. – Paul Slevin Mar 19 '12 at 17:49
@PaulSlevin you are right, we should set $A_{\alpha+1}=A_\alpha\cup\bigcup_{\xi<2^{\aleph_0}}D_\xi.$ – azarel Mar 19 '12 at 19:12

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