A subset $U\in\mathbb{R}^2$ is called radially open if for every point $x\in U$, $U$ contains an open segment through $x$ in every direction, that is for every $v\in\mathbb{R}^2$, $|v|=1$, there exists $\epsilon>0$ such that $x+sv\in U$ for every $s\in (-\epsilon,\epsilon)$. I proved that the familly of radially open sets is a topology on $\mathbb{R}^2$.

What are the induced topologies of this topology on a straight line of $\mathbb{R}^2$ and on a circle ?

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    $\begingroup$ Would you say that this topology is finer or coarser than the standard topology? $\endgroup$ – Arthur Sep 14 '15 at 12:25
  • $\begingroup$ No, I am asking about the topology induced by this topology on a straight line of $\mathbb{R}^2$ and on a circle. $\endgroup$ – Omega Sep 14 '15 at 12:28
  • $\begingroup$ And a way to get to the induced topology is by getting a feel for how the original topology works. Hence my question: have you looked at what this topology looks like? More specifically, how does it compare to the standard topology? $\endgroup$ – Arthur Sep 14 '15 at 12:36
  • $\begingroup$ I proved that this topology contains the usual topology. But I can not find an example to show that it is not contained in the usual topology $\endgroup$ – Omega Sep 14 '15 at 12:40
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    $\begingroup$ Let $U$ be the open ball of radius $1$ centred at the origin. Let $C$ be the circle of radius $1$ centred at $\langle 0,1\rangle$, and let $A$ be all of $C$ except the origin. Then $U\setminus A$ is open in the radial topology but not in the usual one. (This example is even specifically relevant to the problem.) $\endgroup$ – Brian M. Scott Sep 14 '15 at 12:51

Once you prove that it is a topology $\tau$ on $\mathbb{R}^{2}$, it is useful to see that it is finer than the usual one $\tau_{u}$. A radially open set which is not open in the usual topology would be for example:

$A:=\{(0,0)\}\cup B_{1}((1,0))\cup B_{1}((-1,0))\cup ((-2,2)^{2}\setminus (B_{2}((2,0))\cup B_{2}((-2,0)))$

Check that for any $r>0$ the point $(\frac{r^{2}}{12}, \sqrt{\frac{3r^{2}}{12}-\frac{r⁴}{144}})$ is in $B_{r}((0,0))$ but not in $A$, because it lies in the curve $(x-\frac{3}{2})^{2}+y^{2}=(\frac{3}{2})^{2}$.

Straight lines

Consider the vectorial line $L$ generated by $\overrightarrow{v} \in \mathbb{R}^{2}$ with $\vert \overrightarrow{v} \vert = 1$ and take $x \in L$. Let $G \in \tau$ a radially open set such that $x\in G$. Then $\exists r>0$ such that $\{x + t\overrightarrow{v} \vert t\in (-r,r)\}\subset (L\cap G) \subset L$. Hence, the topology induced by $\tau$ on $L$ is the usual topology generated by the euclidean metric.


Consider the circumference $C=\{(x,y)\in \mathbb{R}^{2} \vert (x-1)^{2}+y^{2}=1\}$ and $p=(0,0)$. Take the previously defined $A\in \tau$. Then $A \cap C=\{p\}$ is open in the induced topology on $C$.

This seems to be a very particular example, but it isn't. By isometries of the euclidean plane we can do the same with every other point in any circumference. Hence, the topology induced by $\tau$ on circumferences is the discrete topology.

(Sorry for not being very formal in the last part).


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