Why is $V_{\frac{1}{2}} = \{ (\frac{1}{2}, y) \in X : y \in [0,1] \}$ not open in the lexicographical topology on unit square Where X is the unit square with the lexicographical topology. 
I think its because taking any open set around $V_{\frac{1}{2}}$ say $(a \times b, c\times d)$ we can always find some x s.t $a < x < \frac{1}{2}$ and so  $(x \times b, c\times d)$ will be a strict subset of $(a \times b, c\times d)$ and still be in the nbhd of $V_{\frac{1}{2}}$, so $V_{\frac{1}{2}}$  is not open. Basically I'm trying to say is that it doen't have any immediate predecessors or immediate successors. Do I have the right idea and is there a better way to phrase this argument . 
 A: $V_{\frac12}$ is not open because the points $p_0=\left\langle\frac12,0\right\rangle$ and $p_1=\left\langle\frac12,1\right\rangle$ are both in $V_{\frac12}$, and neither has an open nbhd contained in $V_{\frac12}$. 
Suppose, for instance, that $U$ is an open nbhd of $p_0$. Then by the definition of the order topology there must be points $q_0=\langle x_0,y_0\rangle$ and $q_1=\langle x_1,y_1\rangle$ such that $q_0\prec p_0\prec q_1$, and $q\in U$ whenever $q_0\prec q\prec q_1$. (Here $\prec$ is the strict lexicographic order.) Since $q_0\prec p_0$, i.e., 
$$\langle x_0,y_0\rangle\prec\left\langle\frac12,0\right\rangle\;,$$
we know that $x_0<\frac12$. Pick any $z\in\left(x_0,\frac12\right)$; then 
$$q_0=\langle x_0,y_0\rangle\prec\langle z,0\rangle\prec\left\langle\frac12,0\right\rangle\;,$$
so $\langle z,0\rangle\in U$. On the other hand, clearly $\langle z,0\rangle\notin V_{\frac12}$, so $U\nsubseteq V_{\frac12}$. $U$ was an arbitrary open nbhd of $p_0$, so $p_0$ has no open nbhd contained in $V_{\frac12}$.
