Prove the following.
Theorem. Let \(X\) be a locally compact Hausdorff space; let \(G\) be a group of homeomorphisms of \(X\) such that the action of \(G\) is fixed-point free. Suppose that for each compact subspace \(C\) of \(X\), there are only finitely many elements \(g\) of \(G\) such that the intersection \(C \cap g(C)\) is nonempty. Then the action of \(G\) is properly discontinuous, and \(X/G\) is locally compact Hausdorff.
For each compact subspace \(C\) of \(X\), show that the union of the sets \(g(C)\), for \(g \in G\), is closed in \(X\). [Hint: If \(U\) is a neighborhood of \(x\) with \(\overline{U}\) compact, then \(\overline{U} \cup C\) intersects \(g(\overline{U} \cup C)\) for only finitely many \(g\).]
Show that \(X/G\) is Hausdorff.
Show that the action of \(G\) is properly discontinuous.
Show that \(X/G\) is locally compact.
Solution
Goal: Prove that a free, properly discontinuous group action with compact interference on a locally compact Hausdorff space \(X\) yields a properly discontinuous action and a locally compact Hausdorff quotient \(X/G\).
1. Part (a): Orbit union \(G \cdot C = \bigcup_{g \in G} g(C)\) is closed in \(X\) for any compact \(C \subseteq X\).Proof:
1.1. Let \(x \in X \setminus \bigcup_{g \in G} g(C)\).
1.2. Since \(X\) is locally compact Hausdorff, choose an open neighborhood \(U\) of \(x\) whose closure \(K = \overline{U}\) is compact.
1.3. The union \(K' = K \cup C\) is compact, so by hypothesis the set \(F = \{g \in G \mid K' \cap g(K') \neq \varnothing\}\) is finite.
1.4. For any \(g \in G \setminus F\), \(K \cap g(C) \subseteq K' \cap g(K') = \varnothing\), which implies \(U \cap g(C) = \varnothing\).
1.5. For each \(g \in F\), \(x \notin g(C)\). Because \(g(C)\) is compact and \(X\) is Hausdorff, \(g(C)\) is closed, so \(X \setminus g(C)\) is open.
1.6. Define \(V = U \cap \bigcap_{g \in F} (X \setminus g(C))\). Then \(V\) is an open neighborhood of \(x\).
1.7. For \(g \in F\), \(V \cap g(C) = \varnothing\) by construction; for \(g \notin F\), \(V \cap g(C) \subseteq U \cap g(C) = \varnothing\).
1.8. Thus \(V \cap (\bigcup_{g \in G} g(C)) = \varnothing\), proving \(X \setminus \bigcup_{g \in G} g(C)\) is open, so \(\bigcup_{g \in G} g(C)\) is closed.
2. Part (b): \(X/G\) is Hausdorff.Proof:
2.1. The quotient map \(p: X \to X/G\) is open because \(p^{-1}(p(W)) = \bigcup_{g \in G} g(W)\) is open for every open \(W \subseteq X\).
2.2. Let \(p(x) \neq p(y)\) in \(X/G\), so \(G \cdot x \cap G \cdot y = \varnothing\).
2.3. Choose open neighborhoods \(U_x\) of \(x\) and \(U_y\) of \(y\) with compact closures \(K_x = \overline{U_x}\) and \(K_y = \overline{U_y}\).
2.4. The compact set \(K = K_x \cup K_y\) has only finitely many \(g \in G\) with \(K \cap g(K) \neq \varnothing\).
2.5. Let \(F = \{g \in G \mid K_x \cap g(K_y) \neq \varnothing\}\). For each \(g \in F\), since \(x \neq g(y)\), choose disjoint open sets \(V_g \ni x\) and \(W_g \ni g(y)\).
2.6. Setting \(U = U_x \cap \bigcap_{g \in F} V_g\) and \(V = U_y \cap \bigcap_{g \in F} g^{-1}(W_g)\) gives open neighborhoods of \(x\) and \(y\) such that \(g(U) \cap V = \varnothing\) for all \(g \in G\).
2.7. Then \(p(U)\) and \(p(V)\) are disjoint open neighborhoods of \(p(x)\) and \(p(y)\) in \(X/G\). Thus \(X/G\) is Hausdorff.
3. Part (c): The action of \(G\) is properly discontinuous.Proof:
3.1. Let \(x \in X\). Choose an open neighborhood \(U_0\) of \(x\) with \(C = \overline{U_0}\) compact.
3.2. The set \(F = \{g \in G \setminus \{e\} \mid C \cap g(C) \neq \varnothing\}\) is finite.
3.3. For each \(g \in F\), since the action is fixed-point free, \(g(x) \neq x\).
3.4. By Hausdorffness, choose disjoint open neighborhoods \(V_g \ni x\) and \(W_g \ni g(x)\).
3.5. Define \(U = U_0 \cap \bigcap_{g \in F} (V_g \cap g^{-1}(W_g))\).
3.6. For each \(g \in F\), \(g(U) \cap U \subseteq W_g \cap V_g = \varnothing\).
3.7. For each \(g \in G \setminus (F \cup \{e\})\), \(g(U) \cap U \subseteq g(C) \cap C = \varnothing\).
3.8. Thus \(g(U) \cap U = \varnothing\) for all \(g \neq e\), so the action is properly discontinuous.
4. Part (d): \(X/G\) is locally compact.Proof:
4.1. For any point \(p(x) \in X/G\), choose a compact neighborhood \(K \subseteq X\) containing an open neighborhood \(W \subseteq K\) of \(x\).
4.2. Since \(p\) is continuous, \(p(K)\) is compact in \(X/G\).
4.3. Since \(p\) is an open map, \(p(W)\) is open in \(X/G\), and \(p(x) \in p(W) \subseteq p(K)\).
4.4. Thus \(p(K)\) is a compact neighborhood of \(p(x)\), so \(X/G\) is locally compact. Q.E.D.