2
$\begingroup$

Let $(X,\leq)$ be a pospace: a topological poset such that uppersets and downsets are closed in the topology. Let $A\subseteq X$ be a closed (connected) subspace and suppose $\phi: A\rightarrow A$ is an order isomorphism: a monotone bijection whose inverse is also monotone. Does $\phi$ extend to an order isomorphism of $X$? If not, what kind of conditions on $X$ do guarantee this property?

I'm specifically interested in the case where $X=[0,1]^n$ in which case things like Tietze's extension theorem might be helpful (or just the general fact that it is a compact metric space).

$\endgroup$
  • $\begingroup$ Can you clarify your definition of "pospace"? By "uppersets" and "downsets" do you just mean the sets $\{x:x\geq a\}$ and $\{x:x\leq a\}$ for fixed $a$? I would think the natural definition of "topological poset" is that the order relation is closed as a subset of $X\times X$. $\endgroup$ – Eric Wofsey Dec 18 '16 at 22:17
  • $\begingroup$ You are completely right, that is the correct definition. I'm used to working with compact metric spaces in which these two notions of "closedness" coincide. $\endgroup$ – John Dec 18 '16 at 22:32
0
$\begingroup$

I found a counter-example to the case of $X=[0,1]^n$:

Pick the subspace $A=\frac{1}{2}+[0,\frac{1}{2}]^n$. So it is an upper corner that is order-isomorphic to $X$. One of the corners of $A$ is an interior point of $X$, while the other corners of $A$ are also corners of $X$. If we permute the coordinates of $A$ such that one of the corner points of $A$ is mapped to the "interior" corner point, then we get an order isomorphism of $A$ that can't be extended to an order isomorphism of $X$, since any order isomorphism of $X$ must map corner points to corner points.

Since this applies to a connected compact subset of a compact metric space, it is unlikely to hold in general for any other general class of pospaces.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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