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Is there a name for an order topology $\tau$ over a totally ordered set $X$ that satisfies the following condition: for every $a, b \in X\cup\{\pm\infty\}$ if $a < b$, then $(a,b) \neq \emptyset$?

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An order dense LOTS (linearly ordered topological space, see Wikipedia, e.g. ) without maximum or mininum. $\mathbb{Q}$ is the only such countable linear order, up to isomorphism. The absence of a minimum or maximum could be called "unbounded", I suppose. So "an unbounded order-dense LOTS"?

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  • $\begingroup$ @RobArthan I’ll use LOTS then ( linearly ordered topological space), as is common. $\endgroup$ – Henno Brandsma May 31 '18 at 22:27
  • $\begingroup$ Thanks. I will leave this link (which I found thanks to your answer) here for future reference. $\endgroup$ – Evan Aad May 31 '18 at 22:33
  • $\begingroup$ order-dense to distinguish it from topologically dense, though it does imply that the space is so-called dense-in-itself (dii). $\endgroup$ – Henno Brandsma May 31 '18 at 22:34
  • $\begingroup$ Would you mind clarifying the acronym LOTS inside your answer, for future reference? $\endgroup$ – Evan Aad May 31 '18 at 22:40
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    $\begingroup$ @EvanAad I did some edits for clarification. $\endgroup$ – Henno Brandsma May 31 '18 at 22:46

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