# $\{1/2\}\times (1/2,1]$ is not open in the ordered square [duplicate]

Consider the square $[0,1]^2$ with the dictionary order (the restriction of the dictionary order on $\mathbb R^2$ to the subset $[0,1]^2$. I'm trying to understand why the set $A=\{1/2\}\times (1/2,1]$ is not open in the order topology on $[0,1]^2$.

Munkres give this picture that should clarify things:

The order topology on the square is given by the basis that consists of all elements of the form $[a_0,b)$ where $a_0=(-1,1)$, $(a,b_0]$ where $b_0=[1,1]$, and $(a,b)$ (w.r.t. the dictionary order). The definition of $A$ being open is this: for every every $t\in A$ there is a basis element $U$ s.t. $t\in U\subset A$. Its negation is: there is $t\in A$ such that for all basis elements $U$ it is not true that $t\in U\subset A$. But I couldn't find such $t$ and unravel "it is not true that $t\in U\subset A$".

Suppose $$p=(\frac12, 1)$$ is an interior point of $$\{\frac12\} \times (\frac12, 1]$$. As $$p$$ is not an endpoint of $$[0,1]\times[0,1]$$ in the lexicographic order, a local base for its neighbourhoods are the open intervals $$((a,b), (c,d))$$ that contain $$p$$, where $$(a,b) ,(c,d) \in [0,1] \times [0,1]$$ (end points of the intervals lie in the set). This is by definition of the order topology. So we must have such an interval $$((a,b), (c,d))$$ such that $$p \in ((a,b), (c,d)) \subseteq \{\frac12\} \times (\frac12, 1]$$ if $$p$$ is to be an interior point of $$\{\frac12\} \times (\frac12, 1]$$.
As $$(\frac12, 1] < (c,d)$$ in the lexicographic order, either $$\frac12 = c$$ and $$1 < d$$ or $$\frac12 < c$$. The former cannot happen as $$(c,d) \in [0,1] \times [0,1]$$ so $$d \le 1$$, so we conclude that $$\frac12 < c$$. That's why the end point of the open interval drawn in your right hand picture is stricly to the right of $$p$$.
Now take any $$c'$$ with $$\frac12 < c' < c$$, and note that by virtue of the first coordinate alone we can already say that $$p=(\frac12, 1) < (c',y) < (c,d)$$ for any choice of $$y$$, so the open interval $$((a,b), (c,d))$$ contains all points of the form $$(c',y)$$ and so cannot be a subset of $$\{\frac12\} \times (\frac12, 1]$$ as we supposed.
This shows that $$p$$ is indeed not an interior point of $$\{\frac12\} \times (\frac12, 1]$$ even though that set is open when we consider $$[0,1] \times [0,1]$$ as a subspace of the plane in its lexicographical order topology: $$\{\frac12\} \times (\frac12, 1] = \left((\frac12, \frac12), (\frac12, 2)\right) \cap [0,1] \times [0,1]$$
Every open set containing the point $\left(\frac{1}{2},1\right)$ contains a point of the open set $\left(\frac{1}{2},1\right]\times[0,1]$ so $\left(\frac{1}{2},1\right)$ is not an interior point of $\left\{\dfrac{1}{2}\right\}\times \left(\dfrac{1}{2},1\right]$.