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This question is discussed in Steenrod's "The Topology of Fibre Bundles", part I, sections 7.3 and 7.4. Note that if the action of $G$ on $X$ is proper (eg, if $G$ is compact), then your map factors through a homeomorphism $G/G_{x_0}\to X$. Then the question is just whether the orbit map $G\to G/G_{x_0}$ is a fibration. I have a feeling this is not always ...


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I will give a class of metric examples. Let $G_1$ be an abelian group with a metric $d_1$. Let $G_2$ be a discrete nonabelian group, with metric $d_2(x,y)=1$ for all $x\not=y$. Let $G:= G_1\times G_2$, with $d( (x_1, x_2), (y_1, y_2)) = d_1(x_1,y_1) + d_2(x_2, y_2)$. Let $H$ be any subgroup of $G$. Let $U$ be a neighborhood of $1$. Wlog the ...


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It is very common for a topological space to be a complete uniform space in some uniformity, but it is less common for a topological space to satisfy the Baire category theorem since the proof of the Baire category theorem makes an essential use of countability of the metric rather than the uniformity.. Every paracompact space has a compatible complete ...


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This is basically immediate from the definition and Pontryagin duality. Pontryagin duality gives a contravariant involution on the category $LCAb$ of locally compact abelian groups which sends the subcategory $CAb$ of compact groups to the subcategory $Ab$ of discrete groups and conversely. The Bohr compactification $b:LCAb\to CAb$ is defined as the left ...


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Functional analysis abounds with such examples. There are many complete locally convex spaces which are countable unions of closed subspaces with empty interior. One such is the space of smooth functions on the line which have compact support, the test functions of L. Schwartz. An even simpler one is the space of finite sequences with the locally convex ...


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(In response to suggestion of Johannes Hahn): As indicated in my comment, and developed further in j.p.'s comment, a fairly intuitive "explanation" is provided by the fact that Lagrange's Theorem tells us that an element $x$ of a finite group $G$ is already central in $G$ once it commutes with more than half the elements of $G$ (since $C_{G}(x)$ is a ...


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[I realize I had misread the question, as I understood that the real subgroup required to be compact is the normal one. Since asking the question with the normal subgroup ($X$ in Dave's post) required to be compact seems much less trivial than the original question (answered by Dave), I'll include the full answer to the modified question.] To avoid ...


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If I understand the question, then the answer is no. Let $X = \{(x,1) : x \in \mathbb{R}\}$. Then the conjugation action of $K$ has only three orbits on $X$. All three orbits must be closed (since $K$ is compact). So the orbits are also open (in $X$), since there are only finitely many of them. One of these orbits is the singleton $\{(0,1)\}$, so the ...



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