Let us consider $\Bbb R^2$ with the topology $\tau$ consisting of all subsets which contain an open line segment in each direction about each of its points.
Question. Show that the topology $\tau$ on $\Bbb R^2$ is not second countable.
In fact, we know that the Euclidean topology is properly contained in $\tau$ by here .
Could anyone give me a hint? Thanks!
Best Answer
Hint: Following up on the answer to your previous question, what is the subspace topology on $S^1$ induced by $\tau$?