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show that the ordered square is locally connected but not locally path connected. what are the path components of this space?

as i got the answer here,,but i have some confusion in my mind..(underline in red line)

show that the ordered square is locally connected~

enter image description here

im confusion that that how can $[0,1] $ is homeormorphics to $[0,1] \times [0,1]$..??..as if we remove one point from [0,1] it will not connected but if we removed one point from order square it will remain connected.....now my confusion is that how to verify that if $X$ is the lexicographically ordered square, then the map

$$\varphi:R\to X:\langle x,y\rangle\mapsto\langle h(x),y\rangle$$

is a homeomorphism.

Pliz help me,,,

jasmine
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1 Answers1

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Note that it doesn't claim that $[0,1]$ is homeomorphic to $([0,1]\times [0,1], \mathcal{T}(<_{\text{lex}}))$. What is true is that $[0,1]$ is homeomorphic to a "stalk" $\{t\} \times [0,1]$ in that square, for any fixed $t$; the obvious map $x \to (t,x)$ is an order preserving bijection between these sets and hence a homeomorphism.

The crucial fact about the square is that there can be no continuous path from $(a,x)$ to $(b,y)$ if $a \neq b$ and so the stalks (being path-connected as homeomorphic copies of $[0,1]$) are the path-components.

Henno Brandsma
  • 242,131