By nontrivial LOTS I mean a linearly ordered space that contains more then one point. Being totally path disconnected means that every path in the space is constant.
A connected linearly ordered topological space (LOTS) may not be path-connected, the ordered square being an standard counterexample. But can it be totally path disconnected?
I don't have any idea. Any help appreciated.
Let $X = [0,1]^\omega$ lexicographically ordered.
To show that $X$ is connected, one has to show that the order on $X$ is dense, which is clear, and that it is Dedekind complete (= any subset with an upper bound has a least upper bound). Since $X$ has a maximum and a minimum, it is equivalent to check that every nonempty $A\subseteq X$ has a least upper bound.
Let $\pi_n$ be the projection onto the $n$-th coordinate. Let $a_0=\sup\pi_0(A)\in[0,1]$. It this supremum is not attained (i.e., $\sup\ne\max$), $\sup A=(a_0,0,0,\dots)$. Otherwise, let $A_0=\{z\in A:\pi_0(z)=a_0\}\ne\emptyset$ and let $a_1=\sup\pi_1(A_0)$. If this supremum is not attained, $\sup A=(a_0,a_1,0,0,\dots)$. Otherwise, let $A_1=\{z\in A_0:\pi_1(z)=a_1\}\ne\emptyset$, etc. Continue recursively. If the supremum at step $n$ is not attained, $\sup A=(a_0,\dots,a_n,0,0,\dots)$. If the supremum is attained at every step, $\sup A=(a_0,a_1,a_2,\dots)$.
To see that $X$ is totally path disconnected, any non-trivial interval of $X$ contains an uncountable pairwise disjoint family $(\;](x_0, ..., x_n, x, 0, 0, ...), (x_0, ..., x_n, x, 1, 1, ...)[\;)_{0 \le x \le 1}$ of open, non-empty intervals, hence cannot be homeomorphic to a subset of the reals. Therefore any path in X is constant.
