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 $\mathcal K=\{f\in\mathcal F: |\{f(\alpha):\alpha<\omega_1\}|<\aleph_0\}$.

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 $k<n_f$, 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 (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))
S.G. Popvassilev, J.E. Porter
Here is a temporary link from the editors:

(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!

  • 2
    $\begingroup$ Oh, my; it’s been a long time since I thought about orthocompactness! $\endgroup$ Apr 12, 2015 at 15:00
  • 3
    $\begingroup$ @Brian: Ornithocompactness is the study of small birds! $\endgroup$
    – Asaf Karagila
    Apr 12, 2015 at 23:23
  • 1
    $\begingroup$ @Asaf: That deserves some sort of emuticon! $\endgroup$ Apr 12, 2015 at 23:34
  • 3
    $\begingroup$ @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. $\endgroup$
    – Mirko
    Apr 13, 2015 at 12:48
  • 3
    $\begingroup$ repeat not about cardinals, not about cardinals! $\endgroup$
    – Mirko
    Apr 13, 2015 at 12:57

1 Answer 1


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:

S.G. Popvassilev, J.E. Porter
Serdica Math. J. 44 (2018) (dedicated to the memory
of Professor Stoyan Nedev (1942−2015))

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.

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)


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