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Dec
9
awarded  Caucus
Dec
1
comment Solve this equation
What means $\langle E \rangle$ and $\langle N \rangle $ ?
Dec
1
answered Prove there exists a real number x such that
Dec
1
answered Question about the convergence of $\sum a_n$
Dec
1
answered $X$ compact metric implies $X$ separable
Nov
24
answered Limit of an Indicator function
Nov
24
awarded  Popular Question
Nov
23
comment What is the limit of this trig function?
Which limit are you looking for ?
Nov
8
comment Positive elements of a $C^*$-algbera form a poset
Nope, but I guess my question is yet a bit too advanced for my level of knowledge. I first have to dive into it.
Nov
8
comment Positive elements of a $C^*$-algbera form a poset
I think we can take the relation on all elements (despite my formulation.)
Nov
8
comment Positive elements of a $C^*$-algbera form a poset
I think this are the polynomials in $a-b$ which clearly form a commutative sub-algebra.
Nov
8
comment Positive elements of a $C^*$-algbera form a poset
I was more wondering about the fact why this relation is anti-symmetric.
Nov
8
asked Positive elements of a $C^*$-algbera form a poset
Nov
8
comment trying to prove: If $f$ is continious and is lebesgue-almost-everywhere constant, then it is constant
Ah, I understand what you mean. Clearly it does not take a single value on the complement.
Nov
8
comment trying to prove: If $f$ is continious and is lebesgue-almost-everywhere constant, then it is constant
Yes, but why should the function required to be continuous on the complement ?
Nov
8
comment trying to prove: If $f$ is continious and is lebesgue-almost-everywhere constant, then it is constant
Maybe I sould say more precisely that the staircase is constant on each open component of the complement of the cantor set.
Nov
8
answered trying to prove: If $f$ is continious and is lebesgue-almost-everywhere constant, then it is constant
Nov
8
comment Proving that in a Group the inverse of the inverse of an element is the element itself
Left and right cancellation follows from the uniqueness. To prove the latter assume $b^{-1} = a = a'$. Then $a = a*e = a*(b*a') = (a*b)*a'=e*a'=a'$.
Nov
8
comment Proving that in a Group the inverse of the inverse of an element is the element itself
You have to know that inverses are unique (which is an easy task to show) so that if $a*b=b*a= e$ then $a^{-1}=b$ and $b^{-1} = a$.
Nov
8
comment Proving that in a Group the inverse of the inverse of an element is the element itself
Yes, eventhough it is not written very readable.