# Constant functions in set-theory

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 $$\|c_\lambda^\lambda\|_S >\lambda$$?

• What have you tried? Where are you stuck? – John Coleman Jan 8 at 13:39
• You might want to provide a definition of $\|f\|_S$ as well. – Asaf Karagila Jan 8 at 13:44
• What precisely is difficult here? – Andrés E. Caicedo Jan 8 at 14:47
• $||Ψ||_S= \bigcup\ \{ ||φ||_S +1 | φ<_S Ψ\}$ 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. – N. Leveling Jan 8 at 16:02
• 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 $\alpha>\beta$? – Andrés E. Caicedo Jan 8 at 18:27

## 1 Answer

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 $$f. This means that $$\{\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 $$c_\lambda^\lambda$$.

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