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Let $\kappa$ be an infinite cardinal, and consider $\{ 0, 1 \}^{\kappa}$ with the lexicographic order (I have this defined to mean $f < g \iff f(\alpha) < g(\alpha)$ for the least value $\alpha$ that causes $f(\alpha) \not= g(\alpha)$ ). Let $W = \{ f_{\alpha} : \alpha < \kappa^+ \}$, a strictly increasing sequence of functions in $\{0,1\}^{\kappa}$. Let $\gamma \le \kappa$ be the least $\gamma$ such that $\{ f_{\alpha}|_{\gamma} : \alpha < \kappa^+ \}$ has size $\kappa^+$.

For each $\alpha < \kappa^+$, let $\xi_{\alpha}$ be such that $f_{\alpha}|_{\xi_{\alpha}} = f_{\alpha + 1}|_{\xi_{\alpha}}$, and $f_{\alpha}(\xi_{\alpha})=0,\ f_{\alpha+1}(\xi_{\alpha}) = 1$. It's obvious that $\xi_{\alpha} < \gamma$.

My question is, how do I prove the following statement: "Hence there exists $\xi < \gamma$ such that $\xi = \xi_{\alpha}$ for $\kappa^+$ elements $f_{\alpha}$ of $W$."

I'm a bit stumped here. I don't know how at successor steps we can pick such a $\xi$ because we need $f_{\alpha+1}(\xi) = 0$ but $f_{\alpha+1}(\xi_{\alpha}) = 1$. I think it must have something to do with $\kappa^+$ because we haven't really used it in the argument so far. Any help would be appreciated.

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You’re making it too hard. You don’t really pick $\xi$ at all, in the sense of looking in detail at the individual $\xi_\alpha$’s.

Consider the function $f:\kappa^+\to\gamma:\alpha\mapsto\xi_\alpha$; $\gamma\le\kappa$, and $\kappa^+$ is regular, so $f$ must be constant on a set of cardinality $\kappa^+$.

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Because $\kappa^+ \rightarrow (\kappa^+)_{\gamma'}^1 $? (Where $\gamma '$ is possibly smaller than $\gamma$). Of course, because we have $\operatorname{cf}{\kappa^+} > \gamma > \gamma'$ so that relation holds. –  Paul Slevin Mar 17 '12 at 23:33
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@Paul: No need to get bogged down in arrows. For each $\alpha<\gamma$ let $A_\alpha=\{\beta<\kappa^+:f(\beta)=\alpha\}$; then $\kappa^+=\bigcup_{\alpha<\gamma}A_\alpha$, so if $|A_\alpha|\le\kappa$ for all $\alpha<\gamma$, then $\kappa^+\le\gamma\cdot\kappa=\kappa$, which is impossible. –  Brian M. Scott Mar 17 '12 at 23:37
    
Brilliant, thanks. –  Paul Slevin Mar 17 '12 at 23:38

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