The Sorgenfrey topology on $\mathbb{R}$ is the topology whose basic open sets are of the form $[a,b)$ where $a < b \in \mathbb{R}$.

Does it have a countable base? (I suspect not.) Certainly it is separable and first countable.

The reals with the usual topology is second countable because open intervals with rational endpoints form a countable base. However this approach does not work for the Sorgenfrey topology, since open sets of the form $[i,b)$ where $i$ is irrational, cannot be written as the union of basic open sets $[p,q)$ with rational endpoints.


No, the Sorgenfrey line is not second countable.

A base for the topology would have to contain a neighborhood base for each point.

A neighborhood base for the point $a$ must contain some neighborhood whose least element is $a$.

Therefore, any base for the topology of the Sorgenfrey line must have the cardinality of the continuum.

Also see this old question.


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