1d
comment Topological conditions on compact Hausdorff $X$ under which the unital C* algebra C(X) is separable
$C(X)$ separable iff $X$ is metriazable. Theorem 4.1.3 in Topics in Banach space theory. F. Albiac, N. Kalton
Jul
28
comment How many ideals are there in $B(H)^{**}$?
You could find something in Matthew Daws Ph.D. thesis
Jul
28
awarded Enlightened
Jul
28
awarded Nice Answer
Jul
27
comment The restriction of an open bounded linear operator
No take any quotient map $q$ and set $X_0=\ker q$
Jul
27
comment A question about conic sets of functionals.
here is +15 rep.
Jul
24
awarded Civic Duty
Jul
19
revised Completely continuous implies compact under separability assumption
added 76 characters in body
Jul
19
asked On $C_0(\Omega)$-module maps from $L_\infty(\Omega,\mu)$ to $L_q(\Omega,\nu)$
Jul
17
reviewed Close $i^{-1} F$ a sheaf if and only if $\varinjlim_{ U \subseteq X \text{ open}, ~ x,y \in U } F(U) \to F_x \times F_y$ is an isomorphism
Jul
17
reviewed Close Nonhomogeneous equations
Jul
17
reviewed Close A question of real analysis.
Jul
17
reviewed Close Finding first few terms in power series expansion of general solution
Jul
17
reviewed Close Advantage of Bootstrapping Confidence Intervals over Standard Error
Jul
17
reviewed Leave Open Compute $\int_1^e \frac{dx}{x(x+(\ln x)^2)}$
Jul
15
comment Distance to a closed subspace. Prove $d(x, Ker(T))=max\{|\phi(x)|: \phi \in Ker(T)^\perp, ||\phi||=1\}$.
possible duplicate of Distance between point and linear Space
Jul
14
revised Proof of $\lim_{s\to 1^+}\frac{\sum_p p^{-s}}{\ln(s-1)}=-1$
added 53 characters in body
Jul
14
comment Clever downvoting
@O.L. The problem is solved. Downvoters made to much downvotes and the system finally classified them as serial downvotes.
Jul
14
answered Show that $\sum_{\{a_1, a_2, \dots, a_k\}\subseteq\{1, 2, \dots, n\}}\frac{1}{a_1*a_2*\dots*a_k} = n$
Jul
14
comment Show that $\sum_{\{a_1, a_2, \dots, a_k\}\subseteq\{1, 2, \dots, n\}}\frac{1}{a_1*a_2*\dots*a_k} = n$
Another approach. Denote this sum by $S$. Consider $p(x)=\prod_{k=1}^n\left(x+\frac{1}{k}\right)$. If you expand $p(x)$ and substitute $x=1$ you'll get after carefull gazing $S+1$. So the desired sum is $S=p(1)-1=\prod_{k=1}^n\frac{k+1}{k}-1$. By telescoping that product equals to $n+1$, hence $S=n$
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