13
$\begingroup$

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 $k<n_f$, 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))
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!

$\endgroup$
9
  • 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

0
$\begingroup$

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)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .