Open set in order topology How the set ${x}×[0,1]$ where $x \in [0,1]$ is open in order topology of $\mathbb R^2$ defined by $(x_1,y_1)<(x_2,y_2)$ if EITHER $x_1<x_2$ OR  $x_1=x_2$ and $y_1<y_2$.
I know that this set is not open in usual topology of $\mathbb R^2$ using the fact that the neighborhood. of end points is not contained in the set. But what happens if it is an order topology of $\mathbb R^2$? Can I use the same neighborhood concept here too? I really want to visualize this difference.
 A: So we are considering the set $[0,1] \times [0,1]$, and especially subsets of the form $\{x\} \times [0,1]$, where $x \in [0,1]$, in the lexicographic order topology. 
As your previous question already showed, there is a difference here when we consider $[0,1] \times [0,1]$ as a subspace of $\mathbb{R}^2$ in the lexicographic order topology, and when we restrict the lexicographic order to $[0,1] \times [0,1]$ and use the order topology from that restricted order as the topology on $[0,1] \times [0,1]$.
In the former case, when we look at $\{x\} \times [0,1]$, we see that it is contained in an open interval $I = ((x,-1), (x,2))$, which is open in the lexicographic order on the plane, and so $I \cap ([0,1] \times [0,1]) = \{x\} \times [0,1])$ is open in $[0,1]$. So the neighbourhoods of $(x,0)$ in this subspace topology are essentially of the form $\{x\} \times [0,r)$ and those of $(x,1)$ are of the form $\{x\} \times (r,1]$ where $r \in (0,1)$.
In the latter case, where we consider the order on $[0,1] \times [0,1]$ (so there are no points outside to consider; note that we used the open intervals with endpoints outside of $[0,1] \times [0,1]$ to show that the subspace topology on $[0,1] \times [0,1]$ has $\{x\} \times [0,1]$ as an open set) and use the order topology directly on that space.
So now we only have to consider open intervals (or half-open intervals for neighbourhoods of the minimum or maximum) where endpoints are themselves in $[0,1] \times [0,1]$, and this makes all the difference. Because if we look at $(x,1), x 
\neq 1$, and we are looking at a neighbourhood of $(x,1)$, which is not the maximum, so we need some points $(p,q) < (x,1) < (r,s), (p,q),(r,x) \in [0,1] \times [0,1]$, and then $((p,q),(r,s)) = \{(u,v) \in [0,1] \times [0,1]: (p,q) < (u,v), (u,v) < (r,s) \}$ is a basic neighbourhood of $(x,1)$.
But what can $(r,s)$ be? We know that $(x,1) < (r,s)$. So either $x = r$ and $1 < x$, and this cannot be, as we have no points with coordinate $>1$ in the unit square, or (and this must be the case) $x < r$, and we have no information on the second coordinate. This is just the definition of the lexicographic order.    
Now, if $x < r' < r$, then $(r',y) \in ((p,q), (r,s))$ for any $y \in [0,1]$ (check the definition of the order). So we have shown that any open neighbourhood of $(x,1), x \neq 1$ in the lexicographic order contains some point of the form $(r',y)$ where $r' > r$,a nd so $\{x\} \times [0,1]$ is not open in this topology.
And can hold a symmetric argument for points of the form $(x,0), x > 0$ and points $(p',y)$ for some $p < x$. So also $(x,0)$ is not an interior point.
The cases $(1,1)$ and $(0,0)$ are slightly different: As $(1,1)$ is the max of the set, its basic neighbourhoods are of the form $((p,q),(1,1)]$
, where $(p,q) < (1,1)$, so there $(1,1)$ is an interior point of $\{1\} \times [0,1]$ (but $(1,0)$ still is not) and similarly, $(0,0)$ is an interior point of $\{0\} \times [0,1]$ but still $(0,1)$ is not, for the reasons above. 
In fact, one can show that $[0,1] \times [0,1]$ is compact in the order topology, and very non-compact (as the cover $\{x \} \times [0,1], x \in [0,1]$  shows) in the subspace topology induced from the order topology on the plane.
