Discrete Group Acts Continuously iff Acts by Homeomorphisms
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Theorem
Let $G$ be a discrete group acting on a topological space $X$.
Then the following are equivalent:
Proof
If $G$ acts continuously, then by Continuous Group Action is by Homeomorphisms, $G$ acts by homeomorphisms
Let $G$ act by homeomorphisms
Let $\phi: G \times X \to X$ denote the group action.
For $g \in G$, denote $\phi_g : X \to X : x \mapsto \map \phi {g, x}$
Let $U \subset X$ be open.
Then:
| \(\ds \phi^{-1} \sqbrk U\) | \(=\) | \(\ds \set { \tuple {g, x} \in G \times X : \map \phi {g, x} \in U }\) | Definition of Preimage of Mapping with respect to $\phi$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \set { \tuple {g, x} \in G \times X : \map {\phi _g} x \in U }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \set { \tuple {g, x} \in G \times X : x \in \phi _g^{-1} \sqbrk U }\) | Definition of Preimage of Mapping with respect to $\phi _g$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \bigcup _{g \in G} \paren { \set g \times {\phi _g^{-1} } \sqbrk U }\) |
Since each $\set g \times {\phi _g^{-1} \sqbrk U }$ is open by Definition of Product Topology, $\phi^{-1} \sqbrk U$ is open.
Thus $\phi$ is continuous.
$\blacksquare$