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.