Why the order square is not path-connected

Question

In the book of Topology by Munkres, at page 156, it is given and proven that

The ordered square $I_0^2$ (no idea why it is shown as the square of $I_0$) is connected but not path connected.

I understood the given proof, but cannot imagine which pair of point might causing the issue that this space is not path connected, so can you point me two elements in this space where there is no path connecting these two points ?

Answer 1

Intuitively, if two points have different $x$ coordinates then in order to construct a path that respects the lexicographic ordering you would need a path that either passes through the top of the square or bottom of the square and join up with the "next" number on the $x$-axis somehow. Since the concept of "next" number in the reals is nonsense so to is the possibility of path connecting the space.

Answer 2

The lexicographic square $X$ is connected, and so are its intervals (it fulfills the criteria of being a linear continuum as formulated in Munkres).

If $p=(x_1,y_1), q=(x_2,y_2)$ are points with $ x_1 < x_2$, and suppose $f: [0,1] \to X$ is a path from $p$ to $q$, then as $f[[0,1]]$ is connected it must be contain $[p,q]$. The continuous image of $[0,1]$ is metrisable and compact.

But $[p,q]$ is not metrisable, as it is compact but not separable (the family of open intervals $\{x\} \times (0,1)$ where $x_1 < x < x_2$, is uncountable and pairwise disjoint and any dense set must intersect them all). So no path can exist from $p$ to $q$. If $x_1 = x_2$ a linear path from $p$ to $q$ inside $\{x_1\} \times [0,1]$ does exist. So the path components are the vertical stalks. So $(0,0)$ and $(1,1)$ have no path between them, to be concrete. I believe Munkres also uses these points as an example.