Is the interval $((a,b), (a,d))$ is open in $\mathbb{R}\times \mathbb{R}$ which is equipped with the dictionary order topology

Question

Is the interval $((a,b), (a,d))$ is open in $\mathbb{R}\times \mathbb{R}$ which is equipped with the dictionary order topology.

For me, since $(a,b)$ and $(a,c)$ are points in $\mathbb{R^2}$ with $(a,b) \not = (a,c)$, by the definition of order topology, the interval $((a,b), (a,d))$ should be open in $\mathbb{R^2}$.

However, is it really the case ? I just want to make sure that I'm not missing anything.

Answer 1

Let $(X,<)$ be any linearly ordered set, while $(a,b)=\{x\in X:a<x<b\}$, $(a,\rightarrow )=\{x \in X:a<x\}$, $(\leftarrow,a)=\{x\in X: x<a\}$. Now we can easly see that the familly $\mathcal{B}=\{(a,b), (\leftarrow,a), (a\rightarrow)\}$ is a base for a topology. The topology is called order topology.

So if you take $X=\mathbb{R}^2$ and "$<$" is dictionary order on $X$. Therefore it is obvious that the interval $((a,b),(a,c))$ is an element of the base of the order topology. So it is open.