open sets in the order topological space I have  a question.  I am really confused about determining if a set is open.  First, the idea of  a set being closed has nothing to do with homomorphic ideas of closure: if $x,y \in F$  then $x+ y$  in $F$ .  (This is not the idea of closure.)
My question is from an example from the book:
Consider $I \times I  =[0,1] \times [0,1]$, where $I \times I$ has the dictionary order.
Then, consider the set $Y :=\{\frac{1}{2}\} \times (1/2, 1]$.
For $Y$ is open in the subspace topology of $I \times I$ since for $a \in [0,1]$ and $b \in [0,1]$  we have  $(a,b) \bigcap \{\frac{1}{2}\} \times (1/2, 1] = \left(a \bigcap  \{\frac{1}{2}\}\right) \times \left(b \bigcap (\frac{1}{2},1]\right) = \{\frac{1}{2}\} \times (\frac{1}{2},1]$.  So this is in $Y$ as a subspace.
But $I \times I$ is not in the order topology.  Why?  I just know in the subspace we have an open interval.  But in the order topology, we have not an open interval, but a subset... so to speak.  Need some clarification here... Thank you so much.
 A: Your question isn’t entirely clear, but if you’re starting with $\Bbb R\times\Bbb R$ with the dictionary order and then giving $I\times I$ the subspace topology that it inherits from $\Bbb R\times\Bbb R$, then it’s true that $Y$ is open in $I\times I$. To see this, let $p=\left\langle\frac12,\frac12\right\rangle$ and $q=\left\langle\frac12,2\right\rangle$. Let $\preceq$ denote the dictionary order on $\Bbb R\times\Bbb R$. Then the open interval $(p,q)$ in $\langle\Bbb R\times\Bbb R,\preceq\rangle$ is by definition
$$\begin{align*}
(p,q)&=\{\langle x,y\rangle\in\Bbb R\times\Bbb R:p\precneqq\langle x,y\rangle\precneqq q\}\\
&=\left\{\left\langle\frac12,y\right\rangle:\frac12<y<2\right\}\\
&=\left\{\frac12\right\}\times\left(\frac12,2\right)\;,
\end{align*}$$
and
$$\begin{align*}
(p,q)\cap(I\times I)&=\left(\left\{\frac12\right\}\times\left(\frac12,2\right)\right)\cap\left(\left\{\frac12\right\}\times\left(\frac12,1\right]\right)\\
&=\left\{\frac12\right\}\times\left(\left(\frac12,2\right)\cap\left(\frac12,1\right]\right)\\
&=\left\{\frac12\right\}\times\left(\frac12,1\right]\\
&=Y\;.
\end{align*}$$
Thus, $Y$ is the intersection with $I\times I$ of an open set — in fact an open interval — in $\Bbb R\times\Bbb R$, so $Y$ is relatively open in the subspace $I\times I$.
(In the definition of $q$ you can replace the second coordinate by any real number greater than $1$.)
