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.


  1. You have, in fact proved that its strictly finer. Because you have found a radially open set (your space without C) which is not a usual open set. Hence it’s finer.

    Comment by Daniel — June 6, 2016 @ 6:24 pm

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