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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!

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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.

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  • $\begingroup$ This is a perfect answer, thank you! The inductive construction is very nice. $\endgroup$
    – Anthony
    Commented Nov 10, 2014 at 6:35

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