[Math] One question on subspace topology and order topology

general-topology

everyone! I am learning topology with Munkres's topology book. Some examples of second chapter are very hard for me to understand.

The first question is the example 3 which is on the page 90, the section of subspace topology. This example says let $I=[0,1]$, the dictionary order on $I\times I$ is just the restriction to $I\times I$ of the dictionary order on the plane $\mathbb R\times \mathbb R$. However, the dictionary order topology on $I\times I$ is not the same as the subspace topology on $I\times I$ obtained from the dictionary order topology on $\mathbb R\times \mathbb R$! For example, the set $\{\frac{1}{2}\}\times(\frac{1}{2},1]$ is open in $I\times I$ in the subspace topology, but not in the order topology. I could not understand why the set $\{\frac{1}{2}\}\times(\frac{1}{2},1]$ is open in $I\times I$ in the subspace topology, but not in the order topology. Could someone give me a more detailed explanation? Thanks~

PS: ($I\times I$,$\mathbb R\times \mathbb R$, $\{\frac{1}{2}\}\times(\frac{1}{2},1]$ means Cartesian product).

Best Answer

It's open in the subspace topology because $\{\frac{1}{2}\} \times (\frac{1}{2},1] = \{\frac{1}{2}\} \times (\frac{1}{2},\frac{3}{2}) \cap I \times I$, and the set $\{\frac{1}{2}\} \times (\frac{1}{2},\frac{3}{2})$ is open in the dictionary topology on $\mathbb{R} \times \mathbb{R}$.

It is not open in the dictionary topology on $I \times I$ because any open interval containing the point $(\frac{1}{2}, 1)$ in $I \times I$ must also contain points of the form $(x, y)$ where $x > \frac{1}{2}$. This is because an interval around $(1/2,1)$ in $I \times I$ contains points larger than $(1/2,1)$ in the dictionary order. This means points either of the form $(1/2,y)$ where $y > 1$ (of which there are none in $I \times I$), or points of the form $(x,y)$ where $x > 1/2$.

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