If \(A\) and \(B\) are subsets of the topological group \(G\), let \(A \cdot B\) denote the set of all points \(a \cdot b\) for \(a \in A\) and \(b \in B\). Let \(A^{-1}\) denote the set of all points \(a^{-1}\), for \(a \in A\).
A neighborhood \(V\) of the identity element \(e\) is said to be symmetric if \(V = V^{-1}\). If \(U\) is a neighborhood of \(e\), show there is a symmetric neighborhood \(V\) of \(e\) such that \(V \cdot V \subset U\). [Hint: If \(W\) is a neighborhood of \(e\), then \(W \cdot W^{-1}\) is symmetric.]
Show that \(G\) is Hausdorff. In fact, show that if \(x \neq y\), there is a neighborhood \(V\) of \(e\) such that \(V \cdot x\) and \(V \cdot y\) are disjoint.
Show that \(G\) satisfies the following separation axiom, which is called the regularity axiom: given a closed set \(A\) and a point \(x\) not in \(A\), there exist disjoint open sets containing \(A\) and \(x\), respectively. [Hint: There is a neighborhood \(V\) of \(e\) such that \(V \cdot x\) and \(V \cdot A\) are disjoint.]
Let \(H\) be a subgroup of \(G\) that is closed in the topology of \(G\); let \(p: G \to G/H\) be the quotient map. Show that \(G/H\) satisfies the regularity axiom. [Hint: Examine the proof of (c) when \(A\) is saturated.]
Solution
Continuity of multiplication at \((e,e)\) gives a neighborhood \(W\) of \(e\) such that \[ W\cdot W\subseteq U. \] Set \[ V=W\cap W^{-1}. \] Since inversion is a homeomorphism, \(V\) is an open neighborhood of \(e\); it is symmetric and \[ V\cdot V\subseteq W\cdot W\subseteq U. \]
Let \(x\ne y\). One-point sets are closed, since a topological group is assumed to satisfy the \(T_1\) axiom, so \[ U=G-\{xy^{-1}\} \] is a neighborhood of \(e\). Choose symmetric \(V\) with \(VV\subseteq U\). If \[ v_1x=v_2y \] with \(v_1,v_2\in V\), then \[ xy^{-1}=v_1^{-1}v_2\in VV\subseteq U, \] a contradiction. Hence \(Vx\) and \(Vy\) are disjoint open neighborhoods, so \(G\) is Hausdorff.
Let \(A\) be closed and \(x\notin A\). The right translate \(Ax^{-1}\) is closed and does not contain \(e\). Choose a symmetric neighborhood \(V\) of \(e\) with \[ VV\subseteq G-Ax^{-1}. \] The sets \[ Vx,\qquad VA=\bigcup_{a\in A}Va \] are open and contain \(x\) and \(A\). If \(v_1x=v_2a\), then \[ ax^{-1}=v_2^{-1}v_1\in VV, \] contradicting the choice of \(V\). Thus they are disjoint, proving regularity.
Let \(C\subseteq G/H\) be closed and let \(xH\notin C\). Put \[ A=p^{-1}(C). \] Then \(A\) is closed, saturated, and \(x\notin A\); saturation means \(AH=A\). Apply (c) to choose symmetric \(V\) with \[ VA\cap Vx=\varnothing. \] The quotient map \(p:G\to G/H\) is open, since for open \(U\subseteq G\) the saturation \(p^{-1}(p(U))=UH=\bigcup_{h\in H}Uh\) is a union of translates of \(U\). So \(p(VA)\) and \(p(Vx)\) are open neighborhoods of \(C\) and \(xH\). They are disjoint: if a coset lay in both, there would be \(v_1,v_2\in V\), \(a\in A\), and \(h\in H\) with \[ v_1a=v_2xh. \] Then \[ v_1ah^{-1}=v_2x, \] and \(ah^{-1}\in AH=A\), contradicting \(VA\cap Vx=\varnothing\). Hence \(G/H\) satisfies the regularity axiom.