I posted the following question more than two years ago on MO (and then reposted on MSE), but the answer remains incomplete, so I thought I would rephrase it a bit (to make the statement clearer) and try again.


Let $\omega\_1$ be the first uncountable ordinal,
same as the set of all countable ordinals.  

$\omega\_1=\{\alpha:0\le\alpha<\omega\_1\} = \{\alpha:\alpha {\mathrm{\ is\ a\ countable\ ordinal}}\}$.  

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 below works, as well as one of my comments below, which I move here. 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 $k<n\_f$, 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, 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](http://at.yorku.ca/cgi-bin/abstract/cbjp-06): 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](http://en.wikipedia.org/wiki/Orthocompact_space) 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!


Original version of this post:


Let $\omega\_1$ be the first uncountable ordinal,
same as the set of all countable ordinals.
Let $F$ be the set of all functions
$f$ from $\omega\_1$ minus singleton $0$ into $\omega\_1$ that
are (a) regressive i.e. $f(\alpha) < \alpha$ for $0 < \alpha < \omega\_1$,
and (b) order-preserving (same as non-decreasing)
i.e. $f(\alpha)\leq f(\beta)$ if $\alpha \leq \beta$.
Define a partial order on $F$ by $f \leq g$ if
$f(\alpha) \leq g(\alpha)$ for all $0<\alpha < \omega\_1$.
Let $K$ be the subset of $F$, of functions with
a finite range.


Question:
Is there an order-preserving homomorphism
$h : F \to K$, i.e if $f \le g$ then $h(f) \le h(g)$,
and with the additional property that $f \le h(f)$ ?


(I had dropped (b) in my first post, but comments below
show that it is essential. Also, I did mean the functions $f$ must be regressive, when I had imprecisely said decreasing, in the original statement. I added the gn tag since the question stated is an order-theoretic translation of a question from general topology: Whether $\omega\_1$ has a monotone interior-preserving open operator.)


Edit April 28, 2014: The answer by Noah S below is correct but *incomplete* (for each $f\in F$ it finds $h(f)\in K$ with $f \le h(f)$ but does not consider whether $h(f) \le h(g)$ when $f\le g \in F$). The question is open. Thank you


Edit April 12, 2015. I [reposted at MSE](https://math.stackexchange.com/questions/1231431/order-preserving-map-of-regressive-functions-on-omega-1)


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>



Order homomorphism functions on $\omega_1$
*OK, third time's the charm, hopefully:*


First, a lemma:



> 
> Fix an arbitrary *successor* ordinal $\beta$ and a nondecreasing regressive map $f$ with domain $\beta-\{0\}$. Then there is a nondecreasing regressive map $d\_f$ with the same domain such that $f\le d\_f$ and $d\_f$ takes on only finitely many values.
> 
> 
> 


The proof is by induction on $\beta$. The base case $\beta=2$ is trivial, as is the successor-of-a-successor case.


Now suppose the proof holds for all non-limit $\gamma<\lambda+1$ for $\lambda$ a limit, and take $\beta=\lambda+1$. Fix some successor ordinal $\chi$ with $\chi<\lambda$ and $\chi>f(\lambda)$; such a $\chi$ must exist since $\lambda$ is limit and $f$ is regressive.


But now consider the map $F=f\upharpoonright \chi$. Since $\lambda$ is limit and $\chi<\lambda$ is a successor ordinal, we may apply the induction hypothesis to get a $d\_{F}$. But now consider the map $d\_f$ with domain $\lambda-\{0\}$ extending $d\_{F}$ and taking on the value $max\{f(\lambda), d\_F(\epsilon)\}$ (where $\epsilon+1=\chi$) on all inputs $\theta$ with $\chi<\theta \le\lambda$. This function has the desired properties - in particular, it is regressive by choice of $\chi$ - so we are done. QED




---


Now, to answer the actual question: fix a nondecreasing regressive $f$ defined on $\omega\_1-\{0\}$. By Fodor's lemma + nondecreasingness, $f$ is eventually constant; fix some countable successor ordinal $\alpha$ such that for all $\beta>\alpha$, $f(\beta)=f(\alpha)=\gamma$.


Apply the lemma to $f\upharpoonright \alpha$ to get a function $d\_0$; now extend $d\_0$ to a function $d$ defined on all of $\omega\_1-\{0\}$ by setting $d(\theta)=max\{f(\alpha), d\_0(\epsilon)\}$ (where $\epsilon+1=\alpha$) for all $\theta\ge\alpha$.


