Exercise 13, § 8

Let $X$ be a set and let $\Gamma$ be a group of permutations of $X$. a) Let $\mathfrak{F}$ be a set of subsets of $X$. Let $\mathfrak{S}$ be the set of all topologies on $X$ in which all the sets of $\mathfrak{F}$ are closed and every $u \in \Gamma$ is a homeomorphism. Show that the set $\mathfrak{S}$ has a least element $\mathfrak{T}(\Gamma, \mathfrak{F})$. If $X$ is infinite, show that there exist sets $\mathfrak{F}$ of subsets of $X$ such that $\mathfrak{T}(\Gamma, \mathfrak{F})$ is accessible but not discrete (§ 2, Exercise 8). b) Let $J(\Gamma)$ be the set of subsets $A$ of $X$ which have the following property: for each $x \notin A$ there is a mapping $f \in \Gamma$ such that $f(x) \neq x$ which leaves fixed every element of $A$. Show that every Hausdorff topology on $X$, in which every mapping $u \in \Gamma$ is a homeomorphism of $X$ onto itself, is finer than $\mathfrak{T}_J(\Gamma) = \mathfrak{T}(\Gamma, J(\Gamma))$. Show that the topology $\mathfrak{T}_J(\Gamma)$ is Hausdorff if and only if the group $\Gamma$ satisfies the following condition: given any two distinct elements $x, y$ of $X$ there is a finite number of elements $u_k \in \Gamma$ such that, if $F(u_k)$ is the set of elements of $X$ which are invariant under $u_k$, the $F(u_k)$ cover $X$ and none of the $F(u_k)$ contains both $x$ and $y$. c) Let $\mathfrak{T}_0$ denote the topology of the rational line $\mathbf{Q}$ (resp. the real line $\mathbf{R}_*$) and let $\Gamma$ be the group of all homeomorphisms of $\mathbf{Q}$ (resp. $\mathbf{R}_*$) onto itself. Show that the topology $\mathfrak{T}_J(\Gamma)$ is the same as $\mathfrak{T}_0$. d) Let $X$ be the subspace of the rational line consisting of 0 and the points $1/n$, where $n$ is an integer $\geq 1$, and let $\Gamma$ be the group of homeomorphisms of $X$. Show that the topology $\mathfrak{T}_J(\Gamma)$ is not Hausdorff, and that the group of homeomorphisms of $X$ with respect to this topology is $\Gamma$. e) Let $X$ be a topological space such that for each pair of distinct points $x, y$ of $X$ there exists a homeomorphism $u$ of $X$ onto itself such that $u(x) = x$ and $u(y) \neq y$(*for example, the spaces $\mathbf{R}^{n_*}$). Let $G$ be the group of homeomorphisms of $X$ onto itself, and for each $x \in X$ let $S_x$ be the subgroup of $G$ which leaves $x$ fixed. Show that $S_x$ is equal to its normalizer in $G$ and that the intersection of all the subgroups

$S_x$ consists of the identity element alone; in particular, the centre of $G$ consists of the identity element alone.