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why order square $I_0^2$ is connected ?

My thinking is that $ I_0^2 = [0,1] \times [0,1]$ as $ [0,1] $is connected ,by theorem product of connected set is connected,,

My confusion: Munkre said that $I_0^2$ is connected by linear continuum.

as i don't know how to proved that it is connected by linear continuum

Pliz help me ...

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    $\begingroup$ This is not the product topology. This notation means that you're using the dictionary order topology on $[0,1]\times [0,1]$. $\endgroup$ May 18 '18 at 22:54
  • $\begingroup$ why downvotes ?? $\endgroup$
    – user525416
    May 18 '18 at 22:55
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    $\begingroup$ For starters, you need to read your textbook and make clear in your question what $I_0^2$ actually means. I did not downvote you, but I should. $\endgroup$ May 18 '18 at 23:14
  • $\begingroup$ im very sorry@TedShifrin,,my concept is very weak in topology $\endgroup$
    – user525416
    May 18 '18 at 23:20
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    $\begingroup$ The lexicographic-order-topology on $[0,1^2$ is NOT the product-topology. $\endgroup$ May 19 '18 at 5:07
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With questions like that, it always helps to go back to the definitions and read them carefully. Breaking down the concepts in the definition, here are the properties that make a set $L$ a linear continuum:

  1. $L$ has at least two elements.
  2. $L$ is simply ordered:
    • if $x, y \in L$ with $x \neq y$, then either $x < y$ or $y < x$.
    • if $x < y$, then $x \neq y$
    • if $x < y$ and $y < z$, then $x < z$
  3. $L$ has the least upper bound property.
  4. If $x < y$, there exists $z$ such that $x < z < y$.

That's just the definition. Now Theorem 24.1 (page 153) shows any linear continuum in the order topology is connected, so if you show the ordered square is a linear continuum, the Theorem gives you the connectedness.

That gives you the setup. You need to verify the four properties listed above for $I_0 ^2$. The first two are given by the definition of the beast, the third property is given to you by Munkres on page 155. The last density property you should verify yourself.

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  • $\begingroup$ thanks u @Andrey Portnoy $\endgroup$
    – user525416
    May 18 '18 at 23:27
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    $\begingroup$ You're welcome. I swear, 95% of the work is reading the definitions, so do that. $\endgroup$ May 18 '18 at 23:29
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    $\begingroup$ "it always helps to go back to the definitions" +1 $\endgroup$ May 18 '18 at 23:29

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