Dictionary order topology and subspace topology Compare $(0,1) \times (0,1)$ with the dictionary order topology to the same set with the subspace topology given by the dictionary order on $\mathbb{R} \times \mathbb{R}$. 
This is an exercise in my intro topology book that I am reading. Can anyone explain this? I've been thinking about it for awhile but haven't really gotten anywhere. 
The subspace topology of $\mathbb{R} \times \mathbb{R}$ is just the topology made up of intersections of some subset $Y$ and basis elements of the topology we have. 
 A: The solution is explained well in the comments, but I thought I would draw a few pictures to clarify things further!
Consider the dictionary order topology on $(0, \ 1) \ \times \ (0, \ 1)$, which consists of all sets of the form $(a \ \times \ b, \ c \ \times \ d)$, with $0 \ < \ x \ < \ 1$ for $x \ = \ a, \ b, \ c, \ d$ and $a \ < \ c$ or $a \ = \ c$ and $b \ < \ d$.
Now, the dictionary order topology on $\mathbb{R} \ \times \ \mathbb{R}$ behaves the same way, but consists of dictionary ordered open intervals of all real numbers. We then take the intersection of all these sets with $(0, \ 1) \ \times \ (0, \ 1)$ to get the subspace topology.
The best way to prove this is to simply visualize it. Consider the two forms of open set for the first topology we described:

Then consider sets of this same form in $\mathbb{R} \ \times \ \mathbb{R}$, intersecting $(0, \ 1) \ \times \ (0, \ 1)$:

Proving this is just a matter of taking these pictures, and writing down what is visually true (the topologies are the same), in more rigorous language, which is fairly straightforward once the intuition is gained with the pictures.
