order topology on singleton set

Question

Let $E$ be a singleton, then obviously there is only one total order on $E.$ Notice that every open interval of $E$ is empty, we can see that the "order topology" of $E$ is $\left\{\emptyset\right\},$ which is not a topology.

Question: Should we add the assumption "$\mathrm{Card}(E)>1$" to the definition of order topology?

Answer 1

No. The open rays define a subbase of the topology, so the whole space is automatically open in the induced topology.