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I want to show that the ordered square is connected but not path connected.

It's connected because it's a linear continuum. But its not path connected. This is where I'm stuck. So I took two points $(x_0,y_0)$ and $(x_1,y_1)$. The lines are the open neighborhood $U$ that contains both of those points.

Just by looking at the picture below, I'm stuck as to why its not path connected.

Sorry if the picture is poorly drawn, but I just learned how to use GIMP to sketch stuff out.

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  • $\begingroup$ It’s actually the lexicographically ordered square; we can tell from the picture, but you really ought to specify the ordering. $\endgroup$ – Brian M. Scott Apr 12 '13 at 7:30
  • $\begingroup$ Ah, sorry about that. I'm just trying to work out problems that deal with these problems because of a conceptual hurdle I am experiencing. $\endgroup$ – emka Apr 12 '13 at 7:31
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Note that the interval from $\langle x_0,y_0\rangle$ to $\langle x_1,y_1\rangle$ contains uncountably many pairwise disjoint open sets, so it’s not separable. But a path is a continuous image of $[0,1]$, so ... ?

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  • $\begingroup$ It it because we have an uncountable collection of disjoint sets contained in a connected set? $\endgroup$ – emka Apr 12 '13 at 7:39
  • $\begingroup$ @jdla: No, it’s because the continuous image of a separable set is ... ? $\endgroup$ – Brian M. Scott Apr 12 '13 at 7:43
  • $\begingroup$ a seperable set. $\endgroup$ – emka Apr 12 '13 at 7:44
  • $\begingroup$ @jdla: There you go. And any space with uncountably many pairwise disjoint open sets must be non-separable: if $D$ is a dense set in that space, there has to be at least one member of $D$ in each of those open sets. $\endgroup$ – Brian M. Scott Apr 12 '13 at 7:45

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