# Do order isomorphisms of sub-pospaces extend to the whole space?

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).

• 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 $X\times X$. – Eric Wofsey Dec 18 '16 at 22:17
• 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. – John Dec 18 '16 at 22:32

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.