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Show that the dictionary order topology on the set $\mathbb{R}\times\mathbb{R}$ is the same as the product topology on $\mathbb{R}_d\times\mathbb{R}$, where $\mathbb{R}_d$ denoted $\mathbb{R}$ in the discrete topology. Compare this topology with the standard topology on $\mathbb{R}^2$.

I am trying to answer the question but I am confused about what it is even asking. Could someone explain the the dictionary order topology perhaps by giving a basis. I understand what a product topology generally is, but what does Munkres mean when he writes the plain $\mathbb{R}$ in the statement $\mathbb{R}_d\times\mathbb{R}$. Which topology does he mean to be used on the space $\mathbb{R}$?

Sorry I'm very new to this subject. Thanks for the help.

Santana Afton
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Zachary Isaacson
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    I have this book, and $\Bbb R$ means the set of the real numbers with its usual topology (intervals, order, absolute value based metric...) – ajotatxe Sep 16 '19 at 19:00
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    Given a linearly ordered set $(X,\leq)$ the order topology has as a basis the open intervals $(x,y)={z\mid x<z<y}.$ The dictionary order is $(\mathbb R\times\mathbb R, \leq_d)$ where $(x_1,x_2)\leq_d (y_1,y_2)$ if and only if $x_1<y_1$ or $x_1=y_1$ and $x_2\leq y_2.$ – Thomas Andrews Sep 16 '19 at 19:15

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