Is $(\mathbb{R}^2, lexicographic) \cong (\mathbb{R}, discrete)\times (\mathbb{R}, usual)$? I have the following question:
Is $(\mathbb{R}^2, lexicographic) \cong (\mathbb{R}, discrete)\times (\mathbb{R}, usual)$?
Note: the symbol $\cong$ denotes a homeomorphism between topological spaces
What I think:
I think that in fact there is a homeomorphism between those spaces, because is possible to express the lexicographic order as union of vertical lines with open extremes, I mean this is a basis for the lexicographic order. If I consider the Identical function I have the homeomorphism. Am I right?
 A: Yes, they are the same. One could argue as follows:
If $O$ is open in the product topology you mention, and $(x,y) \in O$, then there is a basic open set $U \times V$ with $(x,y) \in U \times V \subset O$. Because open intervals form a base for $\mathbb{R}$ in the usual topology, and singletons for the discrete topology, there is some open interval $(a,b)$ such that $(x,y) \in \{x\} \times (a,b) \subset U \times V \subset O$. But $\{x\} \times (a,b) = ((x,a), (x,b))$, where the latter interval is taken in the lexicographic order, so this is an open set in that topology. This shows that (an arbitrary point) $(x,y)$ is an interior point of $O$ in the lexicographic order topology, so $O$ is also open in that topology.  
The reverse: suppose $O$ is open in the lexicographic order topology, and let $(x,y) \in O$. There is some open interval $((a,b),(c,d))$ that contains $(x,y)$ and is contained in $O$. So we know that $a \le x$. If $a < x$, set $e = y-1$, otherwise set $e = b$. Also $x \le c$. If $x < c$, set $f = y+1$, otherwise $f = d$. In all cases, $y \in (e,f)$ and $\{x\} \times (e,f) \subset ((a,b), (c,d)) \subset O$, which shows that $(x,y)$ is an interior point of $O$ in the product topology.
The key observation is, as you already saw, that any lexicographic interval can be shrunk to essentially a singleton times an open interval (i.e. these form a local base at all points). This does use that $\mathbb{R}$ is unbounded on two sides, the analogous result does not hold for the lexicographically ordered square $[0,1] \times [0,1]$. The latter space is not metrisable (it contains a copy of the Sorgenfrey line, e.g.) while the plane (the case at hand) is metrisable, as a product of two metrisable spaces, as we just saw. So restricting the order to $[0,1] \times [0,1]$ gives a new order topology which is not the subspace topology w.r.t. the original order topology.
