$\{1/2\}\times (1/2,1]$ is not open in the ordered square 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 $t\in U\subset A$".
 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: $$\{\frac12\} \times (\frac12, 1] = \left((\frac12, \frac12), (\frac12, 2)\right) \cap [0,1] \times [0,1]$$
A: 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 $\left\{\dfrac{1}{2}\right\}\times \left(\dfrac{1}{2},1\right]$.
