question about the basis of order topology recently I read the order topology and the basis element,I have a question that if (a,b)$\cap$(c,d)=$\emptyset $,but $\emptyset $is not belongs to any basis elements(ie. (a,b) for a&lt;b),why it can be basis?
thanks
You are misunderstanding the condition:
$$\forall B_1, B_2 \in \mathcal{B}: \forall x \in B_1 \cap B_2 : \exists B_3 \in \mathcal{B}: x \in B_3 \subseteq B_1 \cap B_2$$
If $B_1 \cap B_2$ are empty, the statement is already true because we have a quantor $\forall$ over the empty set! See vacuous truth.
You can formulate this condition equivalently as: the intersection of any two basic elements is a union of a family of basic elements, and then it works too as the empty set is the result of the empty union (the union of the empty subfamily of the base).
You have to throw the null set in, to make it a basis.  Any topology includes the null set.