21 reputation
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bio website
location Bhopal, India
age 22
visits member for 1 year, 9 months
seen Dec 17 '13 at 14:56

Undergrad at IISER Bhopal, pursuing integrated BS-MS in Mathematics.


Sep
24
awarded  Autobiographer
May
22
awarded  Supporter
Dec
15
comment In a metric space, if a set is compact, then it is closed: improving proof
I guess you need to read Hausdorff Separation property to understand what you are missing. en.wikipedia.org/wiki/Hausdorff_space
Dec
15
comment Bolzano-Weierstrass proof correction
oh, so sorry. e_i is. and sorry for the bad typesetting.
Dec
15
comment Bolzano-Weierstrass proof correction
You are taking natural numbers m_i such that m_i+1>m_i, and then taking e_i=1/m_i. Clearly m_i is decreasing. Consider the ball around x0 with radius e_i. x0 being accumulation point, this ball has at least one element of the sequence. Call this the ith term of your subsequence.
Dec
15
answered Bolzano-Weierstrass proof correction
Dec
15
answered Show that the set given is closed
Dec
15
comment Prove by Induction that $e^{nx}=(e^x)^n$
E(1x)=E(x)=E(x)^1.
Dec
14
answered Prove by Induction that $e^{nx}=(e^x)^n$
Dec
12
awarded  Teacher
Dec
12
answered Why must we distinguish between rational and irrational numbers?