Willard (in the book General Topology) defines something called the “radial plane”: the set of points is and a set is declared open if it meets the following property: for all and each unit vector there is some such that
In words: a set is open if, for every point in the set, there is an open line segment in every direction from the point that stays with in the set; note the line segments do NOT have to be of the same length in every direction.
Of course, a set that is open in the usual topology for is open in the radial topology.
It turns out that the radial topology is strictly finer than the usual topology.
I am not going to prove that here but I am going to show a very curious closed set.
Consider the following set . In the usual topology, this set is neither closed (it lacks the limit point ) nor open. But in the radial topology, is a closed set.
To see this we need only show that there is an open set that misses and contains the origin (it is easy to find an open set that shields other points in the complement from . )
First note that the line contains and is disjoint from , as is the line . Now what about the line ? and so the set meets only at and at no other points; hence, by definition, is an open set which contains .
Of course, we can do that at ANY point on the usual graph of ; the graphs of such “curvy” functions have no limit points.
Therefore such a graph, in the subspace topology…has the discrete topology.
On the other hand, the lattice of rational points in the plane form a countable, dense set (a line segment from a rational lattice point with a rational slope will intercept another rational lattice point).
So we have a separable topological space that lacks a countable basis: with the radial topology is not metric. Therefore it is strictly finer than the usual topology.
PS: I haven’t checked the above carefully, but I am reasonably sure it is right; a reader who spots an error is encouraged to point it out in the comments. I’ll have to think about this a bit.