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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.