why order square $I_0^2$ is connected?

Question

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 ...

Answer 1

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.