Must a sequentially discrete subspace of a LOTS itself be a LOTS

examples-counterexamplesgeneral-topology

By LOTS I mean linearly ordered topological spaces.

Suppose that $X$ is a LOTS, and $Y$ is a sequentially discrete subspace of $Y$, meaning that any sequence contained in $Y$ that has a limit in $Y$ is eventually constant. Must the subspace topology of $Y$ induced by some order on $Y$? Note that such order may not coincide with the order inherited from $X$.

As a trivial example, if $Y$ is a discrete subspace of $X$, then such order exists because every discrete space is a LOTS. On the other hand, I have constructed an example of sequentially discrete LOTS that is not discrete in this MSE question.

Note that $Y$ is anticompact, meaning that every compact subset $Y$ is finite: Suppose that $Z$ is a compact subset of $Y$ (in the subspace topology), then the subspace topology of $Z$ coincides with its topology induced by the order on $X$ (a coarser Hausdorff topology and a finer compact topology must coincide); in particuler, $Z$ is a compact LOTS, so it is sequentially compact (as a countably compact LOTS is: Every sequence has a monotone subsequence and it converges), then sequential discreteness of $Z$ implies finiteness.

The problem is that I do not know many examples of a subspace of LOTS that is not itself a LOTS. An such example is the disjoint union of $\mathbb{R}$ and a singleton. You can see here for a proof that it is not a LOTS, but such a proof uses the fact that $\mathbb{R}$ is connected. Here we have instead a totally disconnected space: any nontrivial closed inteval in a connected LOTS must have cardinality $\ge\mathfrak{c}$ as any nontrivial connected $T_4$ space does.

Thank you in advance for any help.

Best Answer

No:

Recall that a space is called non-Archimedean, if it is $T_1$ and has a base such that every two elements of this base are either disjoint or comparable by inclusion.

It is well-known that non-Archimedean spaces are GO spaces (= subspaces of LOTS). For instance, this is mentioned here. Unfortunately, I couldn't find a reference for a proof, and do not have time now to write it down here. Perhaps someone could give a reference?

Let $X$ be a set. Following (1) we call a function $\rho: X \times X \rightarrow \omega_1+1$ $\space$ special $\omega_1$-metric, if for all $x,y,z \in X, \xi < \omega_1$ the following hold:

  1. $\rho(x,y)= \omega_1$, iff $x=y$
  2. $\rho(x,y)= \rho(y,x)$
  3. $\rho(x,y), \rho(y,z) > \xi$ implies $\rho(x,z) > \xi$

In this case define $B_\xi(x) = \{y \in X: \rho(x,y) > \xi \}$ for $x \in X, \xi < \omega_1$.

Then $\{B_\xi(x): x \in X, \xi < \omega_1 \} $ is a base for a topology on $X$, which witnesses that the space is non-Archimedean, hence a GO space. Moreover, it is easy to see that $X$ is sequentially discrete.

By theorem A in (1) there is such a space, which is not a LOTS. (Note that in the construction the map $\Phi: \omega_1 \rightarrow \omega_1 = id$, hence this is a special $\omega_1$-metric.)

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