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

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 question has now been published in a journal.
It is Question 3.2 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

(Update as of August 21, 2020.)
This question has been answered in the negative by Gary Gruenhage. I will post a complete answer some time in the future. Here is a sketch of the proof. The existence of an order-preserving map $$\psi$$ as in the question is equivalent to $$\omega_1$$ being monotonically orthocompact via open refinements, abbrevaited MO$$_o$$ (this is Theorem 3.1 in the paper, a link to which is enclosed at the end of this question). What Gary proved is that MO$$_o$$ implies a certain property called (A$$_o$$) (defined in terms of certain neignborhoods), and that $$\omega_1$$ does not have this property (A$$_o$$).

(Update April 25, 2021.)
I am about to publish an answer here with details of Gary Gruenhage's proof (thus answering the above question is the negative).

Thank you!

• Oh, my; it’s been a long time since I thought about orthocompactness! Apr 12 '15 at 15:00
• @Brian: Ornithocompactness is the study of small birds! Apr 12 '15 at 23:23
• @Asaf: That deserves some sort of emuticon! Apr 12 '15 at 23:34
• @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. Apr 13 '15 at 12:48
• repeat not about cardinals, not about cardinals! Apr 13 '15 at 12:57

## 1 Answer

The answer to the above question is no (there is no such mapping $$\psi$$).

The question asked above is an order-theoretic restatement of a question from general topology:

Question. Does $$\omega_1$$ (with its usual order topology) have a monotone interior-preserving open operator $$r$$ (defined below)?

Definition. A topological space $$X$$ in monotonically orthocompact via open refinements (abbreviated MO$$_o$$) if it has a monotone interior preserving open operator $$r$$, that is:
(i) if $$\mathcal U$$ is any open cover then $$r(\mathcal U)$$ is an interior-preserving open refinement (that covers $$X$$), and
(ii) if $$\mathcal U$$ refines $$\mathcal V$$ then $$r(\mathcal U)$$ refines $$r(\mathcal V)$$.
The operator $$r$$ will also be called an MO$$_o$$ operator (for $$X$$).

The proof that the two questions are equivalent is in Theorem 3.1 in the following paper:

ON MONOTONE ORTHOCOMPACTNESS
S.G. Popvassilev, J.E. Porter
Serdica Math. J. 44 (2018) (dedicated to the memory
of Professor Stoyan Nedev (1942−2015))
http://www.math.bas.bg/serdica/2018/2018-177-186.pdf

As the above paper seems to be reliably available online, I will only include here a negative answer (due to Gary Gruenhage) of the topological version. Namely:

Theorem. $$\omega_1$$ is not MO$$_o$$.

This result is included in the following paper.

MONOTONE ORTHOCOMPACTNESS AND PROPERTY (A$$_o$$)
Gary Gruenhage, Strashimir G. Popvassilev, and John E. Porter
(accepted for publication in the journal Topology Proceedings)

The proof consists of two steps (both due to Gary Gruenhage):
(1) Every regular Hausdorff MO$$_o$$ space has property (A$$_o$$) (defined below), and
(2) $$\omega_1$$ does not have property (A$$_o$$).

I just copy these results from our paper and paste them here.

Definition 2.1. Let (A$$_o$$) be the following property (of a topological space $$X$$):
One can assign to each pair $$(x,U)$$, where $$U$$ is open and $$x \in U$$, an open set $$V(x,U)$$ such that
(1) $$x\in V(x,U)\subset U$$ ;
(2) whenever $$y\in \bigcap_{\alpha<\kappa} V(x_\alpha, U_\alpha)$$, there is $$A \subset \kappa$$ such that $$y \in \textrm{Int}(\bigcap_{\beta \in A}U_\beta)$$ and for each $$\alpha \in \kappa$$ there is $$\beta \in A$$ with $$V(x_\alpha, U_\alpha) \subset U_\beta$$.

Theorem 2.2. For a regular Hausdorff space $$X$$, MO$$_o$$ implies property (A$$_o$$).

Proof. Let $$r$$ be an MO$$_o$$ operator for $$X$$. Let $$U$$ be open and $$x \in U$$. Choose open $$W(x,U)$$ with $$x \in W(x,U) \subset \overline{W(x,U)} \subset U$$, and let $$\mathcal{O}_{x,U} = \{U, X \setminus \overline{W(x,U)}\}.$$ Then choose $$P(x,U) \in r(\mathcal{O}_{x,U})$$ such that $$x \in P(x,U)$$ and let $$V(x,U)= W(x,U) \cap P(x,U)$$.

Suppose $$y \in \bigcap_{\alpha<\kappa} V(x_\alpha, U_\alpha)$$. Let $$\mathcal{O}^*= \bigcup_{\alpha<\kappa} \mathcal{O}_{x_\alpha, U_\alpha}$$. For each $$Q \in r(\mathcal{O}^*)$$ with $$y \in Q$$, there is $$O \in \mathcal{O}^*$$ with $$Q \subset O$$. Since $$y \in V(x_\alpha, U_\alpha) \subset W(x_\alpha, U_\alpha)$$ for all $$\alpha$$, it must be that $$O=U_\alpha$$ for some $$\alpha$$. Choose such an $$\alpha$$ and denote it $$\alpha(Q)$$. So $$Q \subset U_{\alpha(Q)}$$. Finally let $$A=\{\alpha(Q): y \in Q \in r(\mathcal{O}^*)\}.$$

We claim that
(1) $$y \in \textrm{Int}(\bigcap_{\beta \in A}U_\beta)$$ and
(2) for each $$\alpha \in \kappa$$ there is $$\beta \in A$$ with $$V(x_\alpha, U_\alpha) \subset U_\beta.$$

Note that (1) holds because each $$U_\beta$$ for $$\beta \in A$$ contains some $$Q$$ with $$y \in Q \in r(\mathcal{O}^*)$$, and $$r(\mathcal{O}^*)$$ is interior preserving.

To see (2), suppose $$\alpha <\kappa$$. Note that $$\mathcal{O}_{x_\alpha, U_\alpha}$$ refines $$\mathcal{O}^*$$ and so $$r(\mathcal{O}_{x_\alpha,U_\alpha})$$ refines $$r(\mathcal{O}^*)$$. Hence $$P(x_\alpha,U_\alpha) \subset Q$$ for some $$Q \in r(\mathcal{O}^*).$$ Note that $$y \in Q$$ because $$y \in V(x_\alpha, U_\alpha) \subset P(x_\alpha,U_\alpha).$$ So now we have $$V(x_\alpha, U_\alpha) \subset P(x_\alpha, U_\alpha) \subset Q \subset U_{\alpha(Q)},$$ and (2) follows.
(End of proof of Theorem 2.2.}

Theorem 3.1. Let $$S$$ be a stationary subset of a regular uncountable cardinal $$\kappa$$. Then $$S$$ does not have property (A$$_o$$) (and hence is not MO$$_o$$; in particular $$\omega_1$$ is not MO$$_o$$).

Proof. Suppose by way of contradiction that the operator $$V$$ witnesses (A$$_o$$). Let $$\alpha_0=0$$. For each $$x \in S$$, $$x>1$$, we may assume $$V(x, (\alpha_0+1,x]) =(\beta_x, x]$$ for some $$0<\beta_x < x$$. By the pressing down lemma, there is $$\alpha_1 > \alpha_0$$ such that $$\beta_x = \alpha_1$$ for $$\kappa$$-many $$x$$ in $$S$$.

Similarly, there is $$\alpha_2> \alpha_1$$ such that $$V(x, (\alpha_1 +1,x]) =(\alpha_2, x]$$ for $$\kappa$$-many $$x$$. And so on. Continue in this way to define a strictly increasing $$\kappa$$-sequence $$\alpha_\gamma$$, $$\gamma<\kappa$$, of elements of $$\kappa$$ such that
(i) $$\forall \gamma<\kappa ( |\{x>\alpha_\gamma: V(x, (\alpha_\gamma+1,x])=(\alpha_{\gamma+1},x]\}|=\kappa$$);
(ii) if $$\beta$$ is a limit, then $$\alpha_\beta=\sup\{\alpha_\gamma: \gamma<\beta\}$$.

Then $$\gamma\mapsto \alpha_\gamma$$ is an increasing continuous mapping of $$\kappa$$ onto a club subset of $$\kappa$$; this is also the case if $$\gamma$$ is restricted to limit ordinals. It follows that there is some $$\delta$$ in $$S$$ such that $$\delta=\alpha_\beta$$ for some limit ordinal $$\beta$$. Then $$\delta=\sup\{\alpha_\gamma: \gamma <\beta\}.$$

Now we may inductively choose a strictly increasing sequence $$x_\gamma$$, $$\gamma<\beta$$, in $$S$$ such that $$x_0 > \delta$$ and $$V(x_\gamma, (\alpha_\gamma+1, x_\gamma]) = (\alpha_{\gamma+1}, x_\gamma]$$.

For $$\gamma<\beta$$, let $$U_\gamma = (\alpha_\gamma +1, x_\gamma]$$ and $$V_\gamma= (\alpha_{\gamma+1}, x_\gamma]$$. Then $$V(x_\gamma, U_\gamma) = V_\gamma$$.

Note that $$\delta$$ is in every $$V_\gamma$$ for $$\gamma < \beta$$. Since $$V$$ witnesses property (A$$_o$$), there is a subset $$A$$ of $$\beta$$ such that (I) $$\delta \in \textrm{Int}(\bigcap_{\gamma \in A}U_\gamma)$$, and (II) for each $$\gamma \in \beta$$ there is $$\eta \in A$$ with $$V_\gamma \subset U_\eta$$.

Since the $$\alpha_\gamma$$'s increase to $$\alpha_\beta=\delta$$, for (I) to hold the set $$A$$ cannot be cofinal in $$\beta$$. But since the $$x_\gamma$$'s are strictly increasing, $$A$$ must be cofinal in $$\beta$$ for (II) to hold, so we have a contradiction.
(End of proof or Theorem 3.1)