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seen Feb 14 '13 at 22:09

Oct
5
awarded  Popular Question
Sep
10
awarded  Popular Question
Jul
2
awarded  Curious
Jan
28
accepted Convergence of $\sum_{n=2}^\infty \frac{1}{n^\alpha \ln^\beta (n)} $
Jan
28
comment Convergence of $\sum_{n=2}^\infty \frac{1}{n^\alpha \ln^\beta (n)} $
Sorry for commenting in an old post but I think that by the ratio test if $\alpha > 1$ the series converges.
Dec
31
accepted Every finite set contains its supremum: proof improvement.
Dec
31
accepted Every sequence is composed of isolated points?
Dec
31
accepted In a metric space, compactness implies completness
Dec
30
asked Convergence of $\sum_{n=2}^\infty \frac{1}{n^\alpha \ln^\beta (n)} $
Dec
16
comment how to mathematically formulate the sign of a value?
So $\mathrm{sgn}(a) :=\left\{\begin{array}{cc} \frac{|a|}{a} & \text{if } a \neq 0 \\ 0 & \text{if } a = 0\end{array}\right. $
Dec
16
comment how to mathematically formulate the sign of a value?
What about $a= 0$? Is there convention for $\mathrm{sgn}(0)$?
Dec
16
comment Every sequence is composed of isolated points?
Corrected, thanks
Dec
16
asked Every sequence is composed of isolated points?
Dec
16
comment Every finite set contains its supremum: proof improvement.
It is supposed that I can not conclude that $\max A = \sup A$ immediately, so I need to use supremum properties... Should I order elements of $A$ insted of using $\max A$?
Dec
16
awarded  Custodian
Dec
16
reviewed Approve Every finite set contains its supremum: proof improvement.
Dec
16
asked Every finite set contains its supremum: proof improvement.
Dec
15
accepted In a metric space, if a set is compact, then it is closed: improving proof
Dec
15
revised In a metric space, if a set is compact, then it is closed: improving proof
deleted 74 characters in body
Dec
15
comment In a metric space, if a set is compact, then it is closed: improving proof
Thanks!, but I'm trying to understand this proof and unfortunately I've to rewrite it.