# Order-preserving map of regressive functions on $\omega_1$

This questions has now been published in a journal, see update at the bottom.

I posted the following question in March 2014 on MO. It did receive some attention, but the answer there remains incomplete. It was motivated by some paracompactness-type properties as discussed at the end, but the question is purely order-theoretic, about $$\omega_1$$.

Let $$\omega_1$$ be the first uncountable ordinal, same as the set of all countable ordinals.
Let $$\mathcal F$$ be the set of all functions $$f:\omega_1\to\omega_1$$ that are:
(a) regressive i.e. $$f(\alpha) < \alpha$$ for all $$0 < \alpha < \omega_1$$, and
(b) non-decreasing (same as $$\le$$-order-preserving), i.e.,
if $$0\le\alpha \leq \beta<\omega_1$$ then $$f(\alpha)\leq f(\beta)$$ .
Define a partial order $$\sqsubseteq$$ on $$\mathcal F$$ by $$f \sqsubseteq g$$ if $$f(\alpha) \leq g(\alpha)$$ for all $$\alpha < \omega_1$$.
Let $$\mathcal K$$ be the subset of $$\mathcal F$$, consisting of functions with a finite range.
Formally $$\mathcal K=\{f\in\mathcal F: |\{f(\alpha):\alpha<\omega_1\}|<\aleph_0\}$$.

Question:
Is there a $$\sqsubseteq$$-order-preserving map (same as a $$\sqsubseteq$$-non-decreasing map)
$$\psi : \mathcal F \to \mathcal K$$, i.e if $$f \sqsubseteq g$$ then $$\psi(f) \sqsubseteq \psi(g)$$, and with
the additional property that $$\psi(f) \sqsupseteq f$$ for all $$f\in \mathcal F$$ ?

Let me summarize some comments made at MO, clarifying certain partial answers.

Partial answer (A). Since every $$f\in\mathcal F$$ is regressive and non-decreasing, it must be eventually constant and reach its maximal value $$\mu_f=\max\{f(\alpha):\alpha < \omega_1\}$$. One is tempted to define $$\psi(f)(\alpha)=\mu_f$$ for all $$\alpha$$. The problem is that this is not regressive: We have $$\psi(f)(\alpha)<\alpha$$ only when $$\alpha>\mu_f$$, but I insist that $$\psi(f)(\alpha)<\alpha$$ whenever $$0<\alpha<\omega_1$$.

Partial answer (B). If we drop the requirement that $$\psi$$ be a $$\sqsubseteq$$-non-decreasing map then the answer by @NoahS at MO works, as well as one of my comments there, which I include below. As above let $$\mu_f=\max\{f(\alpha):\alpha < \omega_1\}$$ and let $$\gamma_f=\min\{\alpha:f(\alpha)=\mu_f\}$$. (Then $$f(\alpha)=\mu_f$$ for $$\alpha\ge\gamma_f$$, and $$f(\alpha)<\mu_f$$ for $$\alpha<\gamma_f$$. Usually $$\mu_f<\gamma_f$$ unless $$\mu_f=0=\gamma_f$$.) Let $$\alpha_{0,f}=\mu_f$$. If $$\mu_f\ge1$$ then let $$\alpha_{1,f}=f(\alpha_{0,f})<\alpha_{0,f}$$. There is a non-negative integer $$n_f$$ such that $$\alpha_{k+1,f}=f(\alpha_{k,f})<\alpha_{k,f}$$ for $$k, and $$\alpha_{n_f,f}=0$$. Define $$\psi(f)$$ as follows. If $$\alpha>\alpha_{0,f}$$ then let $$\psi(f)(\alpha)=\alpha_{0,f}=\mu_f$$. If $$\alpha_{k+1,f}<\alpha\le\alpha_{k,f}$$ then let $$\psi(f)(\alpha)=\alpha_{k+1,f}$$. (Formally also $$\psi(f)(0)=0$$, but in general each function in $$\mathcal F$$ being regressive must take value $$0$$ at $$1$$, and being non-decreasing must take value $$0$$ at $$0$$ as well.) Then $$\psi(f)\in\mathcal K$$ and $$\psi(f)\sqsupseteq f$$.

So partial answer (A) above achieves that $$\psi(f)$$ has a finite range, and $$\psi(f) \sqsubseteq \psi(g)$$ whenever $$f \sqsubseteq g$$, and also $$\psi(f) \sqsupseteq f$$. It almost achieves that $$\psi(f)$$ is regressive, but not quite, and it follows that $$\psi(f)$$ is not in $$\mathcal K$$ unless $$\mu_f=0$$. (One could perhaps say that $$\psi(f)$$ is "regressive on a tail" only, which might in a different context be good enough, but the requirement in my question is that $$\psi(f)(\alpha)<\alpha$$ whenever $$0<\alpha<\omega_1$$.) On the other hand, partial answer $$B$$ achieves that $$\psi(f)\in\mathcal K$$ (in particular both that $$\psi(f)$$ is regressive and has a finite range), and $$\psi(f) \sqsupseteq f$$ for all $$f\in \mathcal F$$, but not necessarily that $$\psi(f) \sqsubseteq \psi(g)$$ whenever $$f\sqsubseteq g$$. It is not clear to me if we could achieve all conditions simultaneously. Edit. Following a comment (on MO), let me clarify why in partial answer $$B$$ we need not have $$\psi(f) \sqsubseteq \psi(g)$$ whenever $$f\sqsubseteq g$$. Fix any ordinals $$0<\beta<\delta<\nu<\omega_1$$. Let $$f(\alpha)=g(\alpha)=0$$ if $$0\le\alpha<\nu$$. Let $$f(\alpha)=\beta$$ and $$g(\alpha)=\delta$$ if $$\alpha\ge\nu$$. Clearly $$f\sqsubseteq g$$. Then $$\psi(f)(\alpha)=\beta$$ if $$\alpha>\beta$$, and $$\psi(f)(\alpha)=0$$ if $$0\le\alpha\le\beta$$ (where $$\psi$$ is as in partial answer $$B$$). While $$\psi(g)(\alpha)=\delta$$ if $$\alpha>\delta$$, and $$\psi(g)(\alpha)=0$$ if $$0\le\alpha\le\delta$$. In particular, if $$\beta<\alpha\le\delta$$ then $$\psi(g)(\alpha)=0<\beta=\psi(f)(\alpha)$$, so $$\psi(f)\not\sqsubseteq \psi(g)$$.

If I were to make a guess, I would say the answer is no. This question is an order-theoretic restatement of a question from general topology that a co-author and I considered: Whether $$\omega_1$$ has a monotone interior-preserving open operator $$r$$, that is, if $$\mathcal U$$ is any open cover of $$\omega_1$$, with the order topology, then $$r(\mathcal U)$$ is an interior-preserving open refinement that covers $$\omega_1$$, and if $$\mathcal U$$ refines $$\mathcal V$$ then $$r(\mathcal U)$$ refines $$r(\mathcal V)$$. As usual we would write $$\mathcal U\preceq \mathcal V$$ if $$\mathcal U$$ refines $$\mathcal V$$. In this context $$f$$ is intended to encode an open cover $$\mathcal U(f)=\{0\}\cup\{(f(\alpha),\alpha]:\alpha<\omega_1\}$$. Note that if $$f\sqsubseteq g$$ then $$\mathcal U(g)\preceq \mathcal U(f)$$.

Update Oct 19, 2018 (and May 21, 2019):
This questions has been included in the following paper:
Serdica Math. J. 44 (2018) (dedicated to the memory
of Professor Stoyan Nedev (1942−2015))
ON MONOTONE ORTHOCOMPACTNESS
S.G. Popvassilev, J.E. Porter
Here is a temporary link from the editors:
http://www.math.bas.bg/serdica/2018/2018-177-186.pdf

Thank you!

• Oh, my; it’s been a long time since I thought about orthocompactness! – Brian M. Scott Apr 12 '15 at 15:00
• "$\mathcal{K}$ consists of functions with a finite range" doesn't mean that $k \in \mathcal{K}$ implies $\mathrm{range}(k) \subseteq \omega$, but that such a $k$ takes only finitely many values (some of which may be infinite), yes? – HTFB Apr 12 '15 at 21:21
• @Brian: Ornithocompactness is the study of small birds! – Asaf Karagila Apr 12 '15 at 23:23
• @HTFB Ha! You tried to hide that link but I will make it overt. But guys, please focus! Don't function so regressively. My question is not about cardinals It is all about ordinals. – Mirko Apr 13 '15 at 12:48
• repeat not about cardinals, not about cardinals! – Mirko Apr 13 '15 at 12:57