What is the Basis of an Ordered Square? The dictionary order topology on $\Bbb R\times \Bbb R$ is generated by a basis having elements $(a\times b,c\times d)$ for $a<c$ or $a=c$ and $b<d$.
Let $I=[0,1]$ and consider $I\times I$. The restriction of dictionary order topology on $I\times I$ defines a topology on $I\times I.$
I'm trying to figure out what would be its basis. The following is what I have;
For $a,b,c,d\in I$
Its basis will contain elements of the form $(a\times b,c\times d)$ for $a<c$ or $a=c$ and $b<d$
The basis contains elements  $[0\times0,a\times b)$ with $0<a$ or $0=a$ and $0<b$
Also $(a\times b,1\times1]$ with $a<1$ or $a=1$ and $b<1$ belongs to the basis.
Is that correct? Suggestions Please!

The general way to contruct a base for an ordered space is the following:
If $(X,<)$ is an ordered space, a subbase for its order topology is given by all sets of the form $L(a) = \{x \in X: x < a\}, a \in X$ (the lower sets) together with all sets $U(a) = \{x: x > a\}$, where $a \in X$ (the upper sets).
The base derived from it (i.e. all finite intersections of subbase elements) depends on whether $X$ has a minimum $m$ or maximum $M$, as $m$ cannot be in any upper set, nor $M$ in any lower set. All others can.
$L(a) \cap L(a') = L(\min(a,a'))$ and $U(a) \cap U(a') = U(\max(a,a')$ so both types of sets among themselves are closed under finite intersections. So we only need to consider the intersections of one $L(a)$ and one $U(a')$ which are exactly the open intervals $(a',a)$ in this case, and we only have possibly non-empty intersections when $a' < a$. If we have $m$ we also need to include all $L(a)$ in the base (where $a>m$, and if we have $M$, all $U(a)$(where $a < M$ as well.
So for $I \times I$ in the lexicographic order $<_l$ we do have $m= 0 \times 0$, and $M = 1 \times 1$, so the standard base is indeed by the above general procedure:


*

*all sets $(a \times b, c \times d)$ with $a \times b <_l c \times d$.

*all sets $L(a \times b) = [0 \times 0, a \times b)$ where $a \times b \neq 0\times 0$.

*all sets $U(a \times b) = (a \times b, 1 \times 1]$ where $a \times b \neq 1 \times 1$.


Which is the same as the basis you describe.
