The elements are orbits, but how do we find the neighbourhoods?
In particular, let $G= ( \mathbb R^+ , \cdot )$ and $X=[0, \infty )$.
Let $G$ act on $X$ via the usual multiplication.
Then $X/G = {\bigg \{ } \{0 \} \ , \ (0, \infty ) {\bigg \} }$ has two elements.
- I guess the topology on $X/G$ is not discrete. What is it?
- Can I consider $G$ only as a group to define a topology on $X/G$? Or does $G$ have to be a topological group?
Thanks in advance.
Might be helpful to consider the quotient topology. Let $\pi: X \to X/G$ be the projection mapping. Then $U \subseteq X/G$ is open if and only if $\pi^{-1}(U) \subseteq X$ is open. Using this idea we then have only a few subsets to check. Clearly $\pi^{-1}(\emptyset) = \emptyset$ and $\pi^{-1}(X/G) = X$. We then have that $\pi^{-1}(\{\{0\}\}) = \{0\}$ which is not open in $[0,\infty)$. We do however have that $\pi^{-1}(\{(0,\infty)\}) = (0,\infty)$ which is open in $[0,\infty)$ so this subset is open. Therefore the topology on $X/G$ is given by
$$ \mathcal{T} \;\; =\;\; \Big\{\emptyset, X/G, \{(0,\infty)\} \Big\}. $$