# College Math Teaching

## March 12, 2015

### Radial plane: interesting topology for the plane

Filed under: advanced mathematics, topology — Tags: , — collegemathteaching @ 4:14 pm

Willard (in the book General Topology) defines something called the “radial plane”: the set of points is $R^2$ and a set $U$ is declared open if it meets the following property: for all $\vec{x} \in U$ and each unit vector $\vec{u}_{\theta} = \langle cos(\theta), sin(\theta) \rangle$ there is some $\epsilon_{\theta} > 0$ such that $\vec{x} + \epsilon_{\theta} \vec{u}_{\theta} \subset U$

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 $R^2$ 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 $C = \{(x, x^4), x > 0 \}$. In the usual topology, this set is neither closed (it lacks the limit point $(0,0)$ ) nor open. But in the radial topology, $C$ is a closed set.

To see this we need only show that there is an open set $U$ that misses $C$ and contains the origin (it is easy to find an open set that shields other points in the complement from $C$. )

First note that the line $x = 0$ contains $(0,0)$ and is disjoint from $C$, as is the line $y = 0$. Now what about the line $y = mx$? $mx = x^4 \rightarrow x^4-mx = (x^3-m)x = 0$ and so the set $\{(x, mx) \}$ meets $C$ only at $x = m^{\frac{1}{3}}, y = m^{\frac{4}{3}}$ and at no other points; hence, by definition, $R^2 - C$ is an open set which contains $(0,0)$.

Of course, we can do that at ANY point on the usual graph of $f(x) = x^4$; 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: $R^2$ 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.