Does $[0,1]^\omega$ have only trivial path components under the dictionary order topology? Let $I = [0,1]$ and consider $I^{\omega}$ as a topological space under the dictionary order. Are all path components of $I^\omega$ trivial? Otherwise I expect $I^{\omega}$ to be path connected since I think it, without its endpoints, is $2$-transitive. This seems odd since $I^n$ are not path connected for finite $n > 1$.
I proved (Do linear continua contain $\mathbb{R}$? Can a nontrivial connected space have only trivial path components?) that this is true for $I^{\omega_1}$ but that proof uses the fact that $\omega_1$ has uncountably many elements.
I realized by cardinal arithmetic that the set of points between any two points of $I^{\omega_1}$ has cardinality equal to that of $2^{\mathfrak c}$ and so there cannot be a path between them since the cardinalities don't agree. Closed intervals of $I^{\omega}$ do however have the right cardinality so this argument doesn't work.
 A: The fact that $I^\omega$ has no nontrivial path components can be deduced from the fact that no nontrivial subcontinuum is second countable.
Note that for every $x \in I$ the sequences $x, 0, 0, 0, ...$ and $x, 1, 1,1, ...$ are the bounds of an open interval in $I^\omega$, and these intervals are pairwise disjoint. Hence $I^\omega$ itself is not second countable.
Every subcontinuum of a linear continuum is a closed interval, and we will show that a nondegenerate interval in $I^\omega$ contains a
smaller interval that is order-isomorphic to $I^\omega$.
Let $(a_n)_{n<\omega}$ and $(d_n)_{n<\omega}$ be elements of $I^\omega$
with $(a_n) < (d_n)$ and let $m = \min\{n < \omega \mid a_n < d_n \}$. Define $b_n = c_n = a_n$ for $n < m$, $b_m = c_m = (a_m + d_m)/2$ and $b_n = 0, c_n = 1$ for  $n > m$. The closed interval $[(b_n), (c_n)]$ is contained in $[(a_n), (d_n)]$ and it is easy to verify that $(x_n)_{n<\omega} \mapsto (x_{n+m+1})_{n<\omega}$ defines an order-isomorphism from  $[(b_n), (c_n)]$ to $I^\omega$, which must also be
a homeomorphism. 
It follows that $[(a_n), (d_n)]$ is not second countable and therefore cannot be a continuous image
of a compact metric space.
A: This paper shows that we can order embed an ordered space $X$ into a countable lexicographic product of subsets of the reals, iff $X$ only has at most countable well-ordered or reverse well-ordered subsets. The problem is that this need not be a topological embedding. It's just an injective continuous map.
