Sudhanshu Srivastava
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 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?