Let \(G\) be a group of homeomorphisms of \(X\). The action of \(G\) on \(X\) is said to be fixed-point free if no element of \(G\) other than the identity \(e\) has a fixed point. Show that if \(X\) is Hausdorff, and if \(G\) is a finite group of homeomorphisms of \(X\) whose action is fixed-point free, then the action of \(G\) is properly discontinuous.
Solution
Goal: Prove that every fixed-point free (free) action of a finite group \(G\) of homeomorphisms on a Hausdorff space \(X\) is properly discontinuous.
1. Definition of proper discontinuity:The action of \(G\) on \(X\) is properly discontinuous if for every point \(x \in X\), there exists an open neighborhood \(U \subseteq X\) of \(x\) such that:\[g(U) \cap U = \varnothing \quad \text{for all } g \in G \setminus \{e\}.\]
2. Separation for individual group elements:For each \(g \in G \setminus \{e\}\), there exists an open neighborhood \(U_g \subseteq X\) of \(x\) such that \(g(U_g) \cap U_g = \varnothing\).Proof:
2.1. Because the action is fixed-point free, \(g(x) \neq x\).
2.2. Since \(X\) is Hausdorff, there exist disjoint open sets \(V_g, W_g \subseteq X\) such that \(x \in V_g\), \(g(x) \in W_g\), and \(V_g \cap W_g = \varnothing\).
2.3. Since \(g: X \to X\) is continuous and \(g(x) \in W_g\), the preimage \(g^{-1}(W_g)\) is an open neighborhood of \(x\).
2.4. Define \(U_g = V_g \cap g^{-1}(W_g)\). Then \(U_g\) is an open neighborhood of \(x\).
2.5. If \(y \in U_g\), then \(y \in V_g\) and \(g(y) \in W_g\).
2.6. Since \(V_g \cap W_g = \varnothing\), \(g(y) \notin V_g\), which implies \(g(y) \notin U_g\).
2.7. Thus \(g(U_g) \cap U_g = \varnothing\).
3. Construction of the simultaneous neighborhood:Proof:
3.1. Write \(G \setminus \{e\} = \{g_1, \dots, g_n\}\). Since \(G\) is finite, this is a finite set of elements.
3.2. Define \(U = \bigcap_{i=1}^n U_{g_i}\).
3.3. As a finite intersection of open neighborhoods containing \(x\), \(U\) is an open neighborhood of \(x\) in \(X\).
3.4. For every \(g_k \in G \setminus \{e\}\), since \(U \subseteq U_{g_k}\):\[g_k(U) \cap U \subseteq g_k(U_{g_k}) \cap U_{g_k} = \varnothing.\]
3.5. Therefore, \(g(U) \cap U = \varnothing\) for all \(g \neq e\).
4. Conclusion:The action of \(G\) on \(X\) is properly discontinuous. Q.E.D.