Given a compact metric space $X$ and a preorder $R$ on $X$ (a reflexive and transitive relation) which is continuous in the sense that the upper and lower contour set of any point, \begin{equation*} U(x):=\{y\in X: yRx\} ,\; \; L(x):=\{y\in X:xRy\} \end{equation*} are closed, is there a coherent completion of $X$ which is continuous in the above sense? Any help appreciated.

  • $\begingroup$ What is a coherent completetion of $X$ (maybe $R$ is meant?) $\endgroup$ – Henno Brandsma Jul 21 '17 at 15:17
  • $\begingroup$ A completion of $R$ is a binary relation $S$ such that $S\supseteq R$ and $S$ is complete (i.e. total) . It is coherent if $xRy$ and $\lnot yRx$ imply $xSy$ and $\lnot ySx$. I know coherent completions exist as an application of Szilprajn theorem, but I cannot adapt the machinery of the theorem to guarantee the existence of a coherent completion which is also continuous $\endgroup$ – Enrico Jul 21 '17 at 15:23

If such an extension $R'$ would exist, then consider the order topology $\mathcal{T}_<$ induced by $R'$, which I'll denote by $\le$ (and its strict version by $<$), which has $\{x \in X: x < a\}$ as a subbasic open (for $a \in X$). This subbasic set is exactly the complement of $U(a)$ for this $R'$. So this subbasic element is open in $X$. Similarly, all other subbasic sets of the form $\{x : x > a\}$ for $a \in X$ are also open in $X$ when $R'$ is continuous in your sense.

So $f: X \to (X, \mathcal{T}_<) \, f(x) = x$ is continuous and as all order topologies are Hausdorff, we get by a standard result that the topology of $X$ is exactly the order topology induced by this extension.

But not all compact spaces are orderable, e.g. take $X= [0,1]^2$ and this already fails: in a connected orderable space, there are at most $2$ non-cutpoints. And $[0,1]^2$ has more.


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