# Question Regarding Ordinals

Where $$\mu, \beta, \gamma, \alpha,$$ and $$\kappa$$ are ordinals and $$\alpha, \kappa > 0$$ does there exist a function $$\phi$$ such that $$\phi(\alpha) = (\mu, \beta, \gamma)$$ where:

1) $$\mu, \beta, \gamma < \alpha$$, and

2) $$\alpha \neq \kappa \implies \phi(\alpha) \neq \phi(\kappa)$$

We can restrict all variables to being less than $$\omega_1$$ (or any arbitrary regular uncountable cardinal) so as to apply Fodor's lemma if that helps, though we are able to consider $$\phi$$ as a class function too.

Where there are $$\omega_1^3$$ triplets $$(\mu, \beta, \gamma)$$ that can be comprised from the elements of $$\omega_1$$ given $$\mu, \beta, \gamma < \omega_1$$, we generally have $$\phi : \omega_1 \rightarrow \omega_1^3$$ (or, if a class function, $$\phi: Ord \rightarrow Ord^3$$). This is a little different than Fodor's lemma, which considers only $$\phi : \omega_1 \rightarrow \omega_1$$. That is not to say Fodor's lemma may not prove valuable in answering the question, as the cardinality of $$\omega_1$$ and $$\omega_1^3$$ are equal, etc.

I personally am asking because I am working with different models for what I have been calling $$T$$ sequences (sequences generated by starting with some initial finite segment of ordinals and then considering all the triplets, quadruplets, quintuplets, and so on, that can be made from the initial segment so as to generate rules that add additional ordinals to the sequence based on the availability of the triplets, quadruplets, etc., over $$\omega$$ initial segments generated using an iterative process).

• Do you know Fodor's lemma? – Andrés E. Caicedo Oct 20 '19 at 13:16
• Thank you. I looked it up, but didn’t know it off the top of my head. The wiki article uses club sets to show proof of the lemma so I’m trying to grasp those too. I may need to pause for a sec and think about how Fodor’s lemma helps answer the question, etc. – AplanisTophet Oct 20 '19 at 17:51
• From Wiki for Fodor’s lemma: $\omega_1$ is a stationary subset of itself, so $S = \omega_1$. Fodor’s lemma considers a regressive function $f : \omega_1 \rightarrow \omega_1$. My $\phi$ is regressive, but since there are $\omega_1^3$ triplets $(\mu,\beta,\gamma)$ that can be made from the elements of $\omega_1$, I’m not sure there is a direct connection. I.e, $\phi : \omega_1 \rightarrow \omega_1^3$. I'm also still trying to grapple with wiki's definition of club sets, but I gather that any set of limit ordinals that is unbounded in $\omega_1$ is club with respect to $\omega_1$, correct? – AplanisTophet Oct 21 '19 at 12:34
• No, not necessarily. The set should also be closed. – Andrés E. Caicedo Oct 21 '19 at 13:43
• Thank you, I have made sense of Fodor's lemma, stationary sets, club sets, etc. It is ok to re-open the question from my perspective if you like. I believe the answer will end up being that no $\phi$ can exist as I've defined it because it would contradict Fodor's lemma, but since we're going from $\omega_1$ to $\omega_1^3$ here as opposed to Fodor's going from $\omega_1$ to $\omega_1$, I am thinking that maybe this changes things? – AplanisTophet Oct 23 '19 at 0:06

1. First, there is such a function $$\phi$$, as long as we restrict its domain to $$\delta\smallsetminus\{0\}$$ for some countable ordinal $$\delta$$. Indeed, we can define $$\phi(n+1)=(0,n,0)$$ for $$n\in\omega$$ and if $$\delta$$ is infinite, fix a bijection $$g$$ between $$\delta$$ and $$\omega\times\omega$$, and let $$\phi(\alpha)=(1,g(\alpha))$$ for $$\omega\le\alpha<\delta$$.
2. There is no such function if we want its domain to contain $$\omega_1\smallsetminus\{0\}$$. Indeed, the result follows from Fodor's lemma for $$\omega_1$$, which states that if $$f$$ is a regressive function defined on a stationary subset of $$\omega_1$$, then $$f$$ is constant on a stationary subset of its domain. Apply Fodor's lemma to the first coordinate of $$\phi\upharpoonright\omega_1\smallsetminus\{0\}$$. There is a stationary set $$S_1$$ where this first coordinate is fixed, with value (say) $$\alpha$$. Now apply Fodor's lemma to the second coordinate of $$\phi\upharpoonright S_1$$. Again, there is a stationary subset $$S_2$$ of $$S_1$$ where this second coordinate is fixed, say equal to $$\beta$$. One last application, now to the last coordinate of $$\phi\upharpoonright S_2$$ completes the proof: There is a stationary set $$S_3\subset\omega_1$$, and countable ordinals $$\alpha,\beta,\gamma$$, such that $$\phi(\rho)=(\alpha,\beta,\gamma)$$ for all $$\rho\in S_3$$.