There is a problem which I am given to solve by one of my friend.The problem is "Prove that $[0,1]*[0,1]$ is not compact in Order Topology".We have defined the Basis of order Topology the collection of all open intervals of the form $(a*b,c*d)$ for $a$ $<$ $c$ ,and for $a=c$ and $b$ $<$ $d$.We consider the usual Topology in $R$ and then constract the Product Topology on $R*R$ has the basis of the form $U*V$;$U$ and $V$ are open in $R$ w.r.t. Usual Topology.
I don't belive that $X=$ $[0,1]*[0,1]$ is not compact in Ordered Topology.Since any basis element of Order Topology is also in the Basis of product Topology;Product Topology is finner than Order Topology.Let ${C}$ be any open cover of $X$ in Ordered Topology then as Product Topology is finner than it ;$C$ is a subcover of $X$ in Product Topology.$X$ is compact in Product Topology and hence $C$ has a finite subcollection $C^*$ (say) which covers $X$.By the construction of $C$ each member of $C^*$ are open in Ordered Topology also.So for any arbitary open cover of $X$ has finite subcover in Ordered Topology and hence $X$ is compact.
Please correct me if I am wrong and provide any useful hint to solve that problem.
$[0,1]\times [0,1]$ is compact in the lexicographic ordering. This follows from the theorem that a linearly ordered topological space $(X,<)$ is compact iff every subset of $X$ has a supremum (BTW including that $\sup(\emptyset) = \min(X)$ must exist).
A minor correction: the order topology also includes all intervals of the form $[\min(X), a)$ (whenever $\min(X)$ exists) and $(b,\max(X)]$ (whenever $\max(X)$ exists, not just the open intervals; the open segments form a subbase).
It has nothing to do with the topology induced by the product topology, e.g. $\{\frac{1}{2}\} \times (0,1)$ is lexicographically open (an open interval) but not product open. Moreover, a set like $(0,1) \times (\frac{1}{2},1]$ is product open but not lexicographically open (as e.g. $(\frac{1}{2},1)$ is not an interior point), so the topologies aren't even comparable, neitehr is finer than the other.