Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 2 down vote accepted

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 $\kappa^+$.

share|cite|improve this answer
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
@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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.