This is a question from Munkres which has been answered a lot of times but my approach has been a bit different:
I have been able to prove in the previous problem that $\mathbb{R}_d \times \mathbb{R}$ where $\mathbb{R}_d$ is the discrete topology has the same topology as $\mathbb{R} \times \mathbb{R}$ in the dictionary order topology.
Then shouldn't $I \times I$ in the dictionary order topology and $I_d \times I$ be same?
If they are same then we start with the basis of $I_d \times I$ Where $I_d$ denotes the discrete topology. So the basis of $I_d \times I$ will be $\{x\} \cup (a, b) $.
I am trying to show that the product topology is finer than the dictionary order topology.
let $(x_0,y_0) \in \{x_0\} \times (a, b) $.Then, $ a < y_0 < b$.
We see that, $(x_0,y_0) \in (x_0,x_0)
\times (a, b) \subset \{x_0\} \times (a, b) $(**I am not sure about this because I think $ \{1/4\} \times (0,1)$ will not be open in the product topology but it is a basis of $I \times I$ in the dictionary order topology **)
Now, the dictionary order topology is finer than the product topology :
We see that, $(x_0,y_0) \in \{x_0\} \times (a, b) \subset (x_0-\epsilon,x_0+\epsilon)
\times (a, b) $
Where am I going wrong?
Best Answer
First of all the base for $I_d \times I$ is all sets of the form $\{x\} \times (a,b)$ with $x,a,b \in I$ but also all sets of the form $\{x\} \times [0,a)$ and $\{x\} \times (a,1]$, because for a base of $[0,1]$ (in it usual (=order) topology) we also need the "special" sets for $0$ and $1$.
A base for $I \times I$ in the lex-order is all sets of the form $[(0,0), (a,b))$ for $(0,0)$, which is the minimum in this order and all sets of the form $((a,b), (1,1)]$ for the maximum $(1,1)$, and $(a,b) \in I \times I$ in either case. These sets are also open in $I_d \times I$ (of we take into account the remark of the first paragraph!) as you can easily see. Most basic open sets will be of the form $((a,b), (c,d))$ for two points $(a,b), (c,d) \in I \times I$.
A product open set like $\{\frac12\} \times [0, \frac12)$ will not be open in the order topology because any order neighbourhood of $(\frac12,0)$ will "stick out" to the left ( consider the left boundary point in any open interval containing it). So $\mathcal{T}_{<} \subsetneq \mathcal{T}_{\text{prod}}$ holds and not equality.