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 Jun6 revised Showing that the closure of a totally bounded set is totally bounded deleted 2 characters in body Jun4 revised pointwise convergence and monotony implies uniformly convergence deleted 21 characters in body Jun4 answered pointwise convergence and monotony implies uniformly convergence Jun3 comment A locally convex space is metrizable if and only if its topology is determined by a countable set of seminorms. If $V$ is a vector space and $\tau_1$ and $\tau_2$ are two vector space topologies on $V$ such that convergence to $0$ in $\tau_1$ is equivalent to convergence in $\tau_2$, then $\tau_1=\tau_2$. Indeed, this means that the identity $(V,\tau_1)\leftrightarrow(V,\tau_2)$ is continuous in both directions at $0$, and being linear, it is continuous in both directions, hence a homeomorphism, which means that $\tau_1=\tau_2$. May28 revised Does every homeomorphism of a compact metric space lift to the Cantor set? Simplified a lot an argument May28 comment Is the left translation $T_a(x) =ax$ a homomorphism? When dealing with groups of permutations, we (essentially) always assume the operation is composition. Note, however, that when $G$ is a group and $X$ is any set, we can form the set $G^X$, consisting of all functions $f:G\to X$ from $G$ to $X$, and this has a group structure given in the following way: For $f,g:X\to G$, we define a new function $f*g:X\to G$ by $(f*g)(x)=f(x)*g(x)$ for all $x\in X$. This defines a group structure on $G^X$. However, this group structure is completely different from the group structure on the permutations of $G$, as described above. May28 comment Is the left translation $T_a(x) =ax$ a homomorphism? If $X$ is any set, a permutation of $X$ is a bijection $f$ from $X$ to itself. We form the set $\mathfrak{S}_X$ of all permutations of $X$. This has a natural operation, namely composition: Given two permutations $f,g:X\to X$, we denote $f\circ g$, or simply $fg$, the map $fg:X\to X$ given by $fg(x)=f(g(x))$ for all $x\in X$. This gives a group structure on $\mathfrak{S}_X$ (indeed, the identity on $X$ is the unit of this group, and every bijection has an inverse). Does that solve your problem? May28 comment Harmonic functions locally null on connected open set This follows from the "identity principle". See Theorem 5 in these notes May28 comment X - Y in a finite set The Wikipedia page on relations has a nice introduction (I recommend you read until Section 1.1). Then you can look at the page on Partial Orders to see the definition. $X-Y$ denotes difference of sets: $X-Y=\left\{x\in X:x\not\in Y\right\}$. This question can be reformulated (more formally) as: "Let $A$ be a finite set and $B$ the power set of $A$. For $X,Y\in B$, write $X\sim Y$ is $X-Y$ is nonempty. Is $\sim$ a partial, or strict, order, and if so is it total?" May28 comment Is the left translation $T_a(x) =ax$ a homomorphism? He's not saying that each map $T_a$ is a homomorphism, but instead the map $T:G\to\mathfrak{S}_G$, where $\mathfrak{S}_G$ denotes the set of permutations of $G$ (which is a group by composition), given by $T(a)=T_a$, is a homomorphism. See that he wrote "so $T_{ab}=T_aT_b$", which means that $T$ is a homomorphism. May28 comment A inquality in matrix norm Welcome to math.SE! Please consider taking the time to read the faq to familiarise yourself with some of our common practices. In addition, this page should give you a start at learning how to typeset mathematics here so that your posts say what you want them to, and also look good. As this question appears to be homework, please consider reading this page for information about asking effective homework-related questions. Cheers! May28 comment Does every homeomorphism of a compact metric space lift to the Cantor set? @JimBelk I added a consequence, that every minimal infinite topological system is a factor of a minimal Cantor system. The finite case should be true as well, but I still don't know how to prove it in general. (For $\mathbb{Z}$ it is fairly easy: a minimal finite (compact) system consists of a finite set $K=\left\{1,2,\ldots,n\right\}$ and a permutation $\psi=(1\ 2\cdots n)$. Take the odometer $\phi$ on $C=K^\mathbb{N}$, which is minimal, and the quotient $Q:C\to K$, $Q((x_n)_n)=x_0$). May28 revised Does every homeomorphism of a compact metric space lift to the Cantor set? Added a result about minimal systems May27 comment Does every homeomorphism of a compact metric space lift to the Cantor set? @JimBelk Yes. Actually my proof that $D$ is perfect was not correct, since the metric I considered did not actually make $\lambda_g$ an isometry. I tried to apply your suggestion and apparently it works. May27 revised Does every homeomorphism of a compact metric space lift to the Cantor set? Corrected a false statement, generalized the result for non-free actions using the suggestion of the OP May27 revised Does every homeomorphism of a compact metric space lift to the Cantor set? added 7 characters in body May27 comment Radon-Nikodem Derivative of a purely nonatomic Borel Measure So $X$ is a subspace of $\mathbb{R}^n$? May27 answered Density of a subset of a Hilbert space May27 revised Density of a subset of a Hilbert space Improved formatting May27 comment Radon-Nikodem Derivative of a purely nonatomic Borel Measure Could you give the definition of the density function of $\mu$?