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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.

Jianing Song
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1 Answers1

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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.)

Ulli
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  • That's an interesting construction. Thanks! – Jianing Song Jun 01 '24 at 14:17
  • If I understood well, for $\kappa>\aleph_0$ a regular cardinal, the author defines $\Omega={\delta<\kappa:\operatorname{cf}(\delta)=\omega}$, and for each $\delta\in\Omega$ one chooses a countable subset $p_\delta$ cofinal to $\delta$. For $S\subset\Omega$ being a stationary subset of $\kappa$, the auther proves that the space $X(S):={p_\delta:\delta\in S}\cup{\text{finite subset of }\kappa}$ is not a LOTS with respect to the topology induced by the "metric" $\rho(x,x):=\kappa$, $\rho(x,y):=\min(x\Delta y)$ for $x\neq y$. Just out of curiosity: Can this $X(S)$ be extremally disconnected? – Jianing Song Jun 01 '24 at 15:05
  • @Jianing Song No, an extremally disconnected GO space is discrete. This can be proven analogously to the LOTS case, considering the equivalent definition of GO spaces referenced here. – Ulli Jun 01 '24 at 16:17
  • ... or: I'm pretty sure that non-Archimedean extremally disconneted spaces are discrete as well. – Ulli Jun 01 '24 at 16:21
  • That's very kind of you. Thanks a lot! – Jianing Song Jun 01 '24 at 16:28
  • ... for my second comment, it's probably enough to assume that each point has a linearly ordered neighbourhood base, which is, of course, the case for non-Archimedean spaces. – Ulli Jun 01 '24 at 16:37
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    "Each point has a linearly ordered neighbourhood base": This is called well-based in pi-base. Well in general just well-basedness, extremal disconnectedness and being $T_1$ do not imply discreteness; we need stronger separation axioms (perhaps just a little bit stronger: well-based + extremally disconnected + US + ~discrete yields no examples in pi-base). – Jianing Song Jun 01 '24 at 19:47
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    The second comment is a bit strange because non-archimedean implies GO-space, so if extremally disconnected GO spaces are discrete then of course the same holds for non-archimedean spaces. – Jianing Song Jun 01 '24 at 19:49
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    @Jianing Song: yes, of course, the non-Archimedean case is redundant, didn't think about it. Thanks for reminding me ;-). For the well-based case: yes, true, there are first countable counterexamples (cofinite topology). But T2 should suffice! – Ulli Jun 01 '24 at 21:02