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 Civic Duty
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6h
revised Continuity of distance function
added 12 characters in body
1d
comment Prove that the ball $b[f;0.125)$ is convex.
The idea is correct, but you could try improving the way you write the solution.
Jul
24
revised Continuity of distance function
added 255 characters in body
Jul
24
answered Continuity of distance function
Jul
23
comment Can I always write a bounded operator $T$ as $T=R^{*}S$
If $H$ is separable, then every linear combination of such products $R^*S$ has separable image, and this is also true taking limits. Thus if $K$ is non-separable, the identity $id_K$ does not belong to the closure of the linear span of such products.
Jul
23
answered Prove the Supremum is attained.
Jul
23
answered Solving this n x n matrix equation with special structure
Jul
20
answered In a C*-algebra $A$, $x$ is self-adjoint iff $\lim_{t\to 0}(1/t)(\Vert 1-itx\Vert-1)=0$.
Jul
20
revised In a C*-algebra $A$, $x$ is self-adjoint iff $\lim_{t\to 0}(1/t)(\Vert 1-itx\Vert-1)=0$.
added 105 characters in body
Jul
16
accepted Relation between tracial states on von Neumann algebras and their GNS representations
Jul
16
answered Show that a subspace is closed in a Hilbert space $H$
Jul
16
asked Relation between tracial states on von Neumann algebras and their GNS representations
Jul
5
answered If a C*-algebra $A=\overline{\bigcup S}$, where $S$ is a class of prime C*-subalgebras, then $A$ is prime.
Jul
3
awarded  Civic Duty
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.