Is there a rigid dense linear order?

Question

Does there exist a dense linear order with at least two points that is rigid, in other words, has no nontrivial automorphisms?

Ooh, I like this question. – Noah Schweber Apr 27 '21 at 17:22 — Noah Schweber, Apr 27 '21 at 17:22

score 6 · Accepted Answer · answered Apr 27 '21 at 22:06

In this old paper I showed inter alia that $[0,1]$ has dense subspaces $X$ of cardinality $2^\omega$ with the property that $X\setminus\{x\}$ and $X\setminus\{y\}$ are not homeomorphic whenever $x,y\in X$ and $x\ne y$. Such an $X$ must be a densely ordered subset of $[0,1]$, so its subspace and order topologies are identical, and therefore it cannot have a non-trivial order-automorphism. (In fact there are $2^\omega$ of them that are pairwise disjoint and pairwise non-homeomorphic.)

Very nifty, I didn't know this result! – Noah Schweber May 01 '21 at 20:54 — Noah Schweber, May 01 '21 at 20:54

Is there a rigid dense linear order?

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