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Let $(P,\leq)$ be a poset with more than $1$ point such that the interval topology $\tau_i(P)$ is path-connected and $T_2$. Does this imply that $[0,1]$ order-embeds into $P$?

(This is a follow-up question to Path-connected interval topologies to which Ramiro de la Vega gave a concise argument that I had overlooked.)

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  • $\begingroup$ If $P$ is Hausdorff, does it follow that $P$ is linearly ordered? $\endgroup$ – user1688 Aug 27 '15 at 7:34
  • $\begingroup$ No, it doesn't. First, any finite poset has discrete (therefore Hausdorff) interval topology. Moreover the power-set ${\cal P}(X)$ of any set $X$ has Hausdorff interval topology - it is isomorphic to $\{0,1\}^X$ with the product topology (where $\{0,1\}$ is given the discrete topology). $\endgroup$ – Dominic van der Zypen Aug 27 '15 at 7:40
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There is a counterexample as a subset of the real plane endowed with the natural partial order.

Take the graph $\Gamma$ of the standard continuous monotone map $f:C\to[0,1]$ of the Cantor set $C\subset[0,1]$ onto $[0,1]$. The map $f$ is defined anlytically by $f:\sum_{n=1}^\infty \frac{2x_n}{3^n}\mapsto \sum_{n=1}^\infty \frac{x_n}{2^n}$ where $x_n\in\{0,1\}$.

Let $Q_2$ be the set of binary-rational number in the open interval $(0,1)$. By the construction of $f$, for any $y\in Q_2$ the preimage $f^{-1}(y)$ consists of two points and for $y\in[0,1]\setminus Q_2$ the preimage is a singleton. It follows that $\Gamma_0=\Gamma\cap (\mathbb R\times Q_2)$ is a countable dense subset of $\Gamma$.

Now let $\rho$ be the clockwise rotation of the real plane on $\pi/4$. Consider the line $L=\{(x,y):x+y=0\}$ and the subset $X=\rho(\Gamma)\cup(L+\rho(\Gamma_0))$. It can be shown that $X$ is a required path-connected ordered pospace containing no ordered arc. (To prove this property of $X$ use the argument of Baire category).

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