A totally order-disconnected space (TOD) is a tuple $(P, \leq, \tau)$ where $(P, \leq)$ is a poset and $(P,\tau)$ is a topological space such that for $x\not\leq y$ in $P$ there is a clopen down-set that contains $y$, but not $x$. If $P, Q$ are TODs then a map $f:P\to Q$ is a morphism if $f$ is continuous and order-preserving. The set of all morphisms from $P$ to $Q$ will be denoted by $\text{Mor}(P,Q)$.

The "Stone-Cech" compactification for totally order-disconnected spaces is constructed as follows. Let $\mathbf{2}$ be the space $\{0,1\}$ endowed with the discrete topology and ordered by $0<1$. For any TOD $P$ we set $C^*(P) = \text{Mor}(P,\mathbf{2})$. Note that any product of $\mathbf{2}$ is a TOD when endowed with the componentwise ordering and the product topology. We define the evaluation map $$e_P : P \to \mathbf{2}^{C^*(P)}$$ by $e(p):f\in C^*(P) \mapsto f(p) \in \mathbf{2}$ for all $p\in P$ and set $$\beta_\mathbf{2}(P) := \text{cl}(\text{im}(e_P)),$$ as it is done with the usual Stone-Cech compactification. (It turns out that this construction is indeed a compactification in the category of TODs.)

We say that a TOD $P$ is well-behaved if the following is true: if $U$ is a clopen up-set of $P$ then $\downarrow U := \{p \in P: p\leq u \text{ for some } u\in U\}$ is clopen.

Question: If $P$ is well-behaved, is the same true for $\beta_\mathbf{2}(P)$?

  • $\begingroup$ What about discrete $P$? Is $\beta_2(P)$ always well-behaved? $\endgroup$ – Taras Banakh Oct 21 '17 at 8:25
  • $\begingroup$ You mean - if the order of $P$ is an antichain? $\endgroup$ – Dominic van der Zypen Oct 21 '17 at 11:25
  • $\begingroup$ No, I mean a poset $P$, endowed with the discrete topology. It will be automatically well-behaved. $\endgroup$ – Taras Banakh Oct 21 '17 at 14:20
  • $\begingroup$ OK thanks for the clarification - so you have a proof for this, or do I understand that this is a question? (Like a sub-question of mine) $\endgroup$ – Dominic van der Zypen Oct 21 '17 at 16:09
  • $\begingroup$ No, I do not have a proof yet. So, you can consider this as a partial case of your question. $\endgroup$ – Taras Banakh Oct 21 '17 at 17:13

It seems that the answer to this problem is affirmative:

Take a well-behaved pospace $P$. To show that $\beta_2(P)$ is well-behaved, take any clopen upper set $U\subset \beta_2(P)$. We should prove that its lower set ${\downarrow}U$ is clopen in $\beta_2(P)$.

For this consider the upper clopen set $U\cap P$ in $P$ and observe that its lower set $${\downarrow}_P(U\cap P):=\{x\in P:\exists u\in U\cap P\mbox{ with }x\le u\}$$ in $P$ is clopen (as $P$ is well-behaved). Then the characteristic function $\chi:P\to\{0,1\}$ of the set $P\setminus {\downarrow}_P(U\cap P)$ is monotone and continuous and hence it admits a continuous monotone extension $\bar \chi:\beta_2(P)\to\{0,1\}$. Then the preimage $W=\bar\chi^{-1}(0)$ is a clopen subset of $\beta_2(X)$.

It remains to prove that $W={\downarrow}U$.

The inclusion ${\downarrow}U\subset W$ is clear as for any $v\in{\downarrow}U$ there exists $u\in U$ such that $v\le u$. Taking into account that $U\cap P$ is dense in $U$, we conclude that $\bar\chi(u)\in \bar\chi(U)=\overline{\chi(U\cap P)}=\{0\}$ and hence $\bar\chi(v)\le\bar\chi(u)=0$ implies $\bar\chi(v)=0$ and $v\in W$.

To see that $W\subset{\downarrow}U$, observe that $W$ (being clopen), coincides with the closure of the set ${\downarrow}_P(U\cap P)=\chi^{-1}(0)$ in $\beta_2(P)$. Since the partial order of the compact pospace $\beta_2(P)$ is closed, the set ${\downarrow}U$ is closed in $\beta_2(P)$. Consequently, $$W=\overline{\chi^{-1}(0)}=\overline{{\downarrow}_P(U\cap P)}\subset\overline{{\downarrow}U}={\downarrow}U.$$

  • $\begingroup$ Great, thanks! The problem was unanswered for quite some time, I am delighted you solved it $\endgroup$ – Dominic van der Zypen Oct 22 '17 at 19:14

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