6
$\begingroup$

I've been trying to prove a property that apparently all functions $g: \omega_1 \rightarrow \omega_1$ have, where $\omega_1$ is the least uncountable ordinal. For $\alpha \in \omega_1$, define $g^\rightarrow(\alpha) = \{g(\beta) : \beta \in \alpha\}$; the claim is then that for all functions $g: \omega_1 \rightarrow \omega_1$, there exists $0< \alpha \in \omega_1$ such that $g^\rightarrow(\alpha) \subseteq \alpha$.

This fact seems strange, and I just don't see why it should even be true. Any thoughts would be appreciated!

$\endgroup$
7
$\begingroup$

There are actually "club-many" $\alpha < \omega_1$ for which $g[\alpha] = g^{\to} (\alpha) \subseteq \alpha$. Here club here means "closed" (the limit of every increasing sequence of ordinals with the property also has the property) and "unbounded" (with respect to the usual order). Set-theoretically, this is a notion of "largeness;" not only do such points exists, but there are lots of them.

One important fact about $\omega_1$ that is useful here is that it has "uncountable cofinality" which means that any countable subset $A$ of $\omega_1$ is bounded: there is a $\gamma < \omega_1$ such that $A \subseteq \gamma$.

  • For the "unbounded" part, note first that for any $\beta < \omega_1$, as $g[\beta]$ is countable there is a $\gamma \geq \beta$ such that $g[\beta] \subseteq \gamma$. Now, given any $\beta_0 < \omega_1$ we can use this to inductively construct a non-decreasing sequence $\langle \beta_n \rangle_{n \in \omega}$ of countably ordinals such that $g[\beta_n] \subseteq \beta_{n+1}$ for all $n$. Letting $\alpha = \lim_n \beta_n = \bigcup_n \beta_n$ it follows that $$g[\alpha] = g [ {\textstyle \bigcup_n} \beta_n ] = {\textstyle \bigcup_n} g[\beta_n] \subseteq {\textstyle \bigcup_n} \beta_{n+1} = \alpha.$$ (The uncountable cofinality of $\omega_1$ is also used here to show that $\alpha < \omega_1$.)

  • For the "closed" part, note that if $\langle \alpha_n \rangle_{n \in \omega}$ is an increasing sequence of countable ordinals with this property, then letting $\alpha = \lim_n \alpha_n = \bigcup_n \alpha_n$ it follows that $$g[\alpha] = g[ {\textstyle \bigcup_n} \alpha_n ] = {\textstyle \bigcup_n} g[\alpha_n] \subseteq {\textstyle \bigcup_n} \alpha_n = \alpha,$$ and so $\alpha$ also has this property.

$\endgroup$
  • $\begingroup$ This is a perfect answer, thank you! The inductive construction is very nice. $\endgroup$ – Anthony Nov 10 '14 at 6:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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