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Jun
6
revised Showing that the closure of a totally bounded set is totally bounded
deleted 2 characters in body
Jun
4
revised pointwise convergence and monotony implies uniformly convergence
deleted 21 characters in body
Jun
4
answered pointwise convergence and monotony implies uniformly convergence
Jun
3
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 $\tau_1=\tau_2$.
May
28
revised Does every homeomorphism of a compact metric space lift to the Cantor set?
Simplified a lot an argument
May
28
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.
May
28
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?
May
28
comment Harmonic functions locally null on connected open set
This follows from the "identity principle". See Theorem 5 in these notes
May
28
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?"
May
28
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.
May
28
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!
May
28
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 $Q:C\to K$, $Q((x_n)_n)=x_0$).
May
28
revised Does every homeomorphism of a compact metric space lift to the Cantor set?
Added a result about minimal systems
May
27
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.
May
27
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
May
27
revised Does every homeomorphism of a compact metric space lift to the Cantor set?
added 7 characters in body
May
27
comment Radon-Nikodem Derivative of a purely nonatomic Borel Measure
So $X$ is a subspace of $\mathbb{R}^n$?
May
27
answered Density of a subset of a Hilbert space
May
27
revised Density of a subset of a Hilbert space
Improved formatting
May
27
comment Radon-Nikodem Derivative of a purely nonatomic Borel Measure
Could you give the definition of the density function of $\mu$?