My question relates to group actions on manifolds. If $G$ is a group and $M$ is a manifold, then the set of orbits $M/G$ is equipped with the quotient topology. I consider the following example:
The natural action of $\mathbb{R}^+$ on $\mathbb{R}$ by multiplication. This has three orbits; $\lbrace \mathbb{R}^-, 0, \mathbb{R}^+ \rbrace$
This I understand. I happen to know that the open sets with respect to the quotient topology are $\lbrace \mathbb{R}^+ \rbrace, \lbrace \mathbb{R}^- \rbrace, \lbrace \mathbb{R}^+, \mathbb{R}^- \rbrace, M/G$ and $\emptyset$. How can I show this?
Thanks in advance.
For each $x\in\mathbb R$, let $\pi(x)$ be its orbit.
The set $\{\mathbb{R}^+\}$ is an open set because $\pi^{-1}\bigl(\{\mathbb{R}^+\}\bigr)=\mathbb{R}^+$, which is an open subcet of $\mathbb{R}$. For the same reason, $\{\mathbb{R}^-\}$ is open and therefore $\{\mathbb{R}^+,\mathbb{R}^-\}$ is open.
On the other hand, $\{0\}$ is not open, since $\pi^{-1}\bigl(\{0\}\bigr)$ isn't open. For the same reason $\{0,\mathbb{R}^+\}$ and $\{0,\mathbb{R}^-\}$ aren't open.
Finally, $\emptyset$ and the whole space are always open.