Topology Between the Radial Topology and Standard One on $\mathbb{R}^2$? A subset $U\subseteq \mathbb{R}^2$ is radially open if for every $x∈U$ and every $v∈\mathbb{R}^2$, there exists $ϵ>0$ such that $x+sv∈U$ for every $s∈(−ϵ,ϵ)$. Then the collection of radially open sets defines the radial topology on $\mathbb{R}^2$. See the following article for more on this topology:
familly of Radially open sets in $\mathbb{R}^2$
My question: is there a topology properly between the standard one and the radial one that is translation-invariant (i.e. such that $x \mapsto x + y$ for any fixed $y \in \mathbb{R}^2$ is a homeomorphism)? It seems like there should be infinitely many because the standard topology is second countable and the radial one is not, but I can't think of one.
 A: For one natural example in this context, one may consider the semi-regularization of the radial topology. 
In any topological space, a set $U$ is called regularly open if $U=\mathrm{Int}(\overline{U})$ (i.e. it equals the interior of its closure).
The regularly open sets form a base for a topology which is called the 
semi-regularization of the original topology. 
For details about semi-regularization (in particular, it has the same regularly open sets as the original topology) see for example: Regular open sets and semi-regularization.
The semi-regularization of the radial topology is not regular (but is semi-regular, which just means that its regular open sets form a basis). On the other hand there are open sets in the radial topology that are not open in the semi-regularization. This shows that the semi-regularization of the radial topology is strictly in between the radial topology and the usual topology. 
You need to prove two things: 


*

*The radial topology is not semi-regular. 

*The semi-regularization of the radial topology is not regular. 
Both are proved (as far as I remember) in:
S. G. Popvassilev,  Baire property versus non-regularity in some topologies on $\Bbb R^n$. C. R. Acad. Bulgare Sci. 49 (1996), no. 5, 1-14. 
This is a rather difficultly accessible journal (I am the author but I won't easily locate my own copy). Fortunately, it is also available online (though perhaps in a slightly different version) at topology atlas (moved there after it was originally posted at Beverly Brechner's topology eprints, I believe), at:
http://at.yorku.ca/p/a/b/b/12.htm
Corollary 3.1.1 there shows that the radial topology is not semi-regular. 
This corollary is only stated for the cross topology (defined like the radial topology but only considering "vertical" and "horizontal" vectors $v$), but it is also valid for the radial topology. The paragraph after 3.1.1 (till the end of the paper) gives a rather sketchy description of a certain self-similar set in the plane which shows that the semi-regularization of the cross topology is not regular (and same for semi-regularization of the radial topology). I am unaware if this result is available anywhere else, and I have occasionally contemplated writing it up in a better form, including pictures (I had generated some with a computer long ago and lost them by now) - since I believe this result is interesting and the short description in one paragraph at the end of the paper doesn't do it justice - but I never got around to do that.  I would hope that with some effort and persistence that short description could be deciphered (in absence of a better option, as far as I am aware). For that matter, there is also a picture online, from a 2003 conference talk, thanks to Jack Brown who organized the conference, took the picture (after my talk, and I insisted that this particular self-similar set is on the background), and he posted the pictures online, see:
http://topo.math.auburn.edu/pub/photos/AuburnMini03/pres0061.html 
I am aware that one may want a bit more details, but on the other hand I hope this answer would give a good start at taking a look at some of these proofs. 
A: Here is one more example, easier to describe, and easier to verify its properties. 
Note that a set is radially open if and only if its intersection with each straight line is open in the usual topology of that line. (I have first encountered the radial topology under a different name via this definition before I saw the definition involving vectors and epsilons.) 
Let me call a set circularly open if its intersection with every circle is open in the usual topology of the circle. Let us also count straight lines as circles (of infinite radius), so under this convention we have that every circularly open set is radially open.
Define the circular topology as the topology of circularly open sets. 
The radial topology induces the discrete topology on every circle (as discussed in the link provided by OP). On the other hand every circle has its usual topology induced from the circular topology. Therefore the radial topology is strictly stronger than the circular topology. 
One could show that the circular topology is strictly stronger than the usual topology as follows. 
Use that every two points determine a line, and every three points determine a circle (or a line). One may construct a countable set $F$ that is: (a) dense in the usual topology, and (b) intersects every straight line in at most two points and every circle in at most three points. This could be done by starting with a countable basis $\{B_n:n\in\Bbb N\}$ of the usual topology and picking a point $x_n\in B_n$ that does not belong to any of the lines and circles determined by points $x_k$ with $k<n$. 
Note that the set $F$ constructed as above is closed in the circular (as well as in the radial) topologies, but not in the usual topology (as its closure is $\Bbb R^2$ but  $F$ is countable, hence not equal to $\Bbb R^2$). This shows that the circular topology is strictly stronger than the usual topology. Translation-invariance of the circular topology is obvious. 
One would obtain another example if, in addition to lines and circles, one also uses ellipses. As yet another example, one may consider lines, circles, ellipses, as well as all algebraic curves. 
One would be able to construct a countable set $F$ that is: (a) dense in the usual topology, and (b) intersects every algebraic curve in at most finitely many points. Such an $F$ would be closed in this topology, but not closed in the usual topology. 
