Cardinality of order topology?

Question

Just out of interest: The cardinality of the Euclidean topology on the real line is $c$. In general, if $X$ is totally ordered of cardinality $\alpha$, the order topology on $X$ must have cardinality $\geq\alpha$. When is it precisely $\alpha$?

Answer 1

We can calculate.

Why is the topology on $\Bbb R$ have the same cardinality as $\Bbb R$? Because we have a basis of size $\aleph_0$ and each open set is the union of $\aleph_0$ basic open sets.

So if $(X,\leq)$ is a linear order, then its order topology has size $\leq\kappa$ if (and only if) there exists a basis of size $\mu$ such that every open set is the union of at most $\lambda$ basic open sets, and $\mu^\lambda\leq\kappa$.

Some observations:

1. $|X|\leq\mu^\lambda$.
2. We can assume $\lambda\leq\mu$, since any union of more than $\mu$ elements of a set of size $\mu$, is really just the same union of at most $\mu$ elements.

So if $|X|=\mu^\lambda$ then the topology is exactly the wanted size.