Connected sets in GO-spaces

Question

I'd like to understand what connected subsets and connected components of a GO-space look like.

Some definitions:

A linearly ordered topological space (LOTS) is a totally ordered set $(X,<)$ together with the corresponding order topology generated by all the open rays $(\leftarrow,x)$ and $(x,\rightarrow)$ for $x\in X$. For a point $x$ that is not the minimum or the maximum of $X$, the open intervals $(a,b)$ with $a<x<b$ form a local base of open nbhds at $x$. Let's write $\lambda(<)$ for this order topology.

A GO-space (generalized ordered space) is a totally ordered set with a more general topology, not determined by the order. Specifically and formally, it's a triple $(X,<,\tau)$ where $<$ is a total order on $X$ and $\tau$ is a topology on $X$ containing $\lambda(<)$ with a base of open sets that are order-convex. (A set $A\subseteq X$ is order-convex if for any two points $x<y$ in $A$ the whole interval $[x,y]$ is contained in $A$.) So, a subbase for $\tau$ is given by the collection of sets consisting of:

• (1) all sets of the form $(\leftarrow,x)$ and $(x,\to)$ for $x\in X$;

• (2) all sets of the form $[x,\to)$ such that $[x,\to)$ is $\tau$-open, $x$ is not the minimum element of $X$ and $x$ has no immediate predecessor in $(X,<)$;

• (3) all sets of the form $(\leftarrow,x]$ such that $(\leftarrow,x]$ is $\tau$-open, $x$ is not the maximum element of $X$ and $x$ has no immediate successor in $(X,<)$.

See Equivalent definitions for GO-spaces (generalized ordered spaces) for some equivalent characterizations.

As an example, the Sorgenfrey line is a GO-space with $(\mathbb R,<)$ as underlying ordered set. The topology contains the usual open rays, as well as all closed rays $[x,\rightarrow)$ (which are not open in the Euclidean topology). Another example is the Michael line.

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First the situation with LOTS (then I'll get to the question for GO-spaces). Suppose $X$ is a LOTS with order $<$ and $A$ is a nonempty subset of $X$.

Let $<_A$ denote the restriction of the order $<$ to the set $A$. The order $<_A$ is a total order on $A$ with associated order topology $\lambda(<_A)$. In general, the topology $\lambda(<_A)$ on $A$ can be strictly coarser than the subspace topology induced by the topology $\lambda(<)$ on $X$. (Example: $A=\{0\}\cup(1,2)\subseteq\mathbb R$) But if $A$ is order-convex in $X$, the two topologies coincide (easy to check).

On the other hand, being order-convex is a necessary condition for $A$ to be connected. Indeed, if $a<c<b$ with $a,b\in A$ and $c\notin A$, then the disjoint open sets $(\leftarrow,c)$ and $(c,\rightarrow)$ cover $A$ and give a non-trivial separation of $A$.

Therefore, the subset $A\subseteq X$ is connected iff $A$ is order-convex in $X$ and the ordered set $(A,<_A)$ itself is connected in its order topology. (As is well known, that condition can be expressed entirely by properties of the order $<_A$, namely the order is Dedekind-complete and order-dense.)

Aso, every locally connected LOTS is locally compact. See for example here.

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What of these results still hold for GO-spaces?

Even though the order does not determine the topology for a GO-space, is there a way to detect that a subset is connected just by looking at the order?

Answer 1

It turns out things are a lot simpler than I was making it to be. We can use this other characterization of GO-spaces:

$(X,<,\tau)$ is a GO-space iff (a) $\lambda(<)\subseteq\tau$ and (b) there is a base for the topology $\tau$ consisting of order-convex sets.

where $\lambda(<)$ is the order topology induced by the order $<$.

Also a basic fact for LOTS: Suppose $(X,<)$ is a totally ordered set with order topology $\lambda(<)$ and let $A$ be an arbitrary subset of $X$. Let $<_A$ be the order $<$ restricted to $A$. There are two topologies on $A$: the order topology $\lambda(<_A)$ induced by $<_A$ and the subspace topology $\lambda(<)_A$ induced by the topology $\lambda(<)$ on $X$. In general, the two topologies are not equal. But it is easy to check that $\lambda(<_A)\subseteq\lambda(<)_A$. And if $A$ is order-convex, the two topologies are equal: $\lambda(<_A)=\lambda(<)_A$.

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Now suppose $(X,<,\tau)$ is a GO-space and let $A\subseteq X$ be a connected set for the topology $\tau$. Let $\tau_A$ be the subspace topology on $A$ induced by $\tau$. I'll show that $(A,<_A,\tau_A)$ is also a GO-space.

The set $A$ has to be order-convex in $(X,<)$. (Just as in the LOTS case: if $a<c<b$ with $a,b\in A$ and $c\notin A$, then $A=(A\cap(\leftarrow,c))\cup(A\cap(c,\to)$ is a partition of $A$ into two disjoint nonempty $\tau_A$-open sets.)

We have $\lambda(<)\subseteq\tau$ as topologies on $X$. Therefore restricting the topologies to $A$ gives $\lambda(<)_A\subseteq\tau_A$. And since $A$ is order-convex, $\lambda(<_A)=\lambda(<)_A$. Thus $\lambda(<_A)\subseteq\tau_A$.

And on the other hand, there is a base for the topology $\tau$ on $X$ consisting of order-convex sets in $(X,<)$. Intersecting each of these basic sets with $A$ gives a base for the topology $\tau_A$. And since $A$ is order-convex in $X$, the intersections will also be order-convex in $(X,<)$, and in $(A,<_A)$.

By the characterization of GO-spaces at the top, $(A,<_A,\tau_A)$ is a GO-space. We can now use the known result that a connected GO-space is actually a LOTS with the same order. That is, $\tau_A=\lambda(<_A)$.

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Summary: The connected sets in a GO-space $(X,<,\tau)$ can be determined just by looking at the order $<$. They are exactly the order-convex subsets $A$ of $X$ that satisfy the usual condition for being connected as a LOTS when restricting attention to the order $<_A$ on $A$. Namely, $(A,<_A)$ is Dedekind-complete (= least upper bound condition) and order-dense.

And from that, some local results for LOTS imply local results for GO-spaces. In particular, any interval $[x,y]$ in a connected LOTS is compact (since it is Dedekind complete and has a minimum element and a maximum element). Hence if a GO-space is locally connected, each point $x$ has a connected nbhd $V$. That nbhd $V$ with the restriction of the order from $X$ is a LOTS and the point $x$ has a compact nbhd within $V$. That shows:

A locally connected GO-space is locally compact.