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I need some help with an exercise in set theory, which is about certain constant functions.

Let $S$ be a stationary subset of a regular uncountable cardinal $\lambda$. Given an ordinal $\alpha$, let $c_\alpha^\lambda$ denote the constant function with domain $\lambda$ and range $\{\alpha\}$.

Letting $\psi,\varphi$ range over all ordinal-valued functions with domain $\lambda$, define $$\varphi<_S\psi\mbox{ if and only if }\{\delta\in S\mid \varphi(\delta)≥\psi(\delta)\}\mbox{ is non-stationary}.$$ The relation $<_S$ is well-founded, so we can use it to define a rank $\|\cdot\|_S$ by recursion as $$ \|\psi\|_S=\bigcup\{\|\varphi\|_S+1\mid \varphi<_S\psi\}. $$

How can we prove that, for all $\alpha\in{\rm Ord}$, $\|c_\alpha^\lambda\|_S \ge\alpha$ holds?

How can we determine the value of $\|c_\alpha^\lambda\|_S$ for all $\alpha<\lambda$?

Can we prove that $\|c_\lambda^\lambda\|_S >\lambda$?

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  • $\begingroup$ What have you tried? Where are you stuck? $\endgroup$ – John Coleman Jan 8 at 13:39
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    $\begingroup$ You might want to provide a definition of $\|f\|_S$ as well. $\endgroup$ – Asaf Karagila Jan 8 at 13:44
  • $\begingroup$ What precisely is difficult here? $\endgroup$ – Andrés E. Caicedo Jan 8 at 14:47
  • $\begingroup$ $||Ψ||_S= \bigcup\ \{ ||φ||_S +1 | φ<_S Ψ\}$ is the definition of the S-rank, with S being stationary and $φ<_S ψ $ iff $ \{δ∈S|φ(δ)> ψ(δ)\} ∈ NS_λ$ (NS is the Non-stationary ideal on λ). I just can't seem to figure out how I can determine the rank of the constant function, and thus don't know how to prove anything about it. $\endgroup$ – N. Leveling Jan 8 at 16:02
  • $\begingroup$ Do you see that $\|c_\alpha^\lambda\|_S$ is an ordinal for all $\alpha$, and can you prove that $\|c_\alpha^\lambda\|_S>\|c_\beta^\lambda\|_S$ whenever $\alpha>\beta$? $\endgroup$ – Andrés E. Caicedo Jan 8 at 18:27
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Note that $\alpha\mapsto\|c_\alpha^\lambda\|_S$ is strictly increasing (trivially): After all, $$\{\delta\in S\mid c_\beta^\lambda(\delta)\ge c_\alpha^\lambda(\delta)\}=\{\delta\in S\mid\beta\ge \alpha\}=\emptyset$$ if $\beta<\alpha$. This immediately gives that $\|c_\alpha^\lambda\|_S\ge\alpha$ for all $\alpha$.

Suppose now that $f<c_\alpha^\lambda$. This means that $\{\delta\in S\mid f(\delta)\ge \alpha\}$ is non-stationary, or, what is the same, $f(\delta)<\alpha$ for almost every $\delta\in S$. If, in addition, $\alpha<\lambda$, then in fact $f(\delta)<\delta$ for almost every $\delta\in S$. Use Fodor's lemma to conclude that $f$ coincides with some $c_\beta^\lambda$ for some $\beta<\alpha$ ("coincides" in the sense of $=_S$, where $f=_S g$ implies in particular that $\|f\|_S=\|g\|_S$). This should give you that $\|c_\alpha^\lambda\|_S=\alpha$ for all $\alpha<\lambda$.

Finally, check that the identity map is above all $c_\alpha^\lambda$, $\alpha<\lambda$, and below $c_\lambda^\lambda$.

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  • $\begingroup$ Thank you very much! Just one more question: Why is the identity map below $c_λ^λ$? I don't understand this point $\endgroup$ – N. Leveling Jan 8 at 19:55
  • $\begingroup$ Use the definitions, it is just as immediate as the other properties we discussed in the comments above. $\endgroup$ – Andrés E. Caicedo Jan 8 at 19:57

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