10-327/Classnotes for Monday December 6: Difference between revisions

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*'''Question.''' The fact that the metric space of real-valued functions on the unit interval with uniform metric is complete uses the fact that [0,1] is compact right? If the function space is defined on a non-compact topological space is that necessarily complete?... -Kai [[User:Xwbdsb|Xwbdsb]] 00:01, 20 December 2010 (EST)
*'''Question.''' The fact that the metric space of real-valued functions on the unit interval with uniform metric is complete uses the fact that [0,1] is compact right? If the function space is defined on a non-compact topological space is that necessarily complete?... -Kai [[User:Xwbdsb|Xwbdsb]] 00:01, 20 December 2010 (EST)
** No, the compactness of <math>[0,1]</math> is not used. As we said in class, if <math>(f_n)</math> is Cauchy in the uniform metric, then for any <math>x</math>, the sequence <math>(f_n(x))</math> is Cauchy in <math>{\mathbb R}</math>, so it has a limit. Call that limit <math>f(x)</math>; it is not hard to show that <math>f</math> is continuous and that <math>f_n\to f</math>. Theorem 43.6 in Munkres is a slight generalization of this. [[User:Drorbn|Drorbn]] 07:12, 20 December 2010 (EST)
** No, the compactness of <math>[0,1]</math> is not used. As we said in class, if <math>(f_n)</math> is Cauchy in the uniform metric, then for any <math>x</math>, the sequence <math>(f_n(x))</math> is Cauchy in <math>{\mathbb R}</math>, so it has a limit. Call that limit <math>f(x)</math>; it is not hard to show that <math>f</math> is continuous and that <math>f_n\to f</math>. Theorem 43.6 in Munkres is a slight generalization of this. [[User:Drorbn|Drorbn]] 07:12, 20 December 2010 (EST)
Thanks Dror.

Everybody good luck on the exam!-Kai

Revision as of 11:45, 20 December 2010

See some blackboard shots at BBS/10_327-101206-142909.jpg.

Video: dbnvp Topology-101206

Dror's notes above / Student's notes below

Lecture Notes

  • Question. The fact that the metric space of real-valued functions on the unit interval with uniform metric is complete uses the fact that [0,1] is compact right? If the function space is defined on a non-compact topological space is that necessarily complete?... -Kai Xwbdsb 00:01, 20 December 2010 (EST)
    • No, the compactness of is not used. As we said in class, if is Cauchy in the uniform metric, then for any , the sequence is Cauchy in , so it has a limit. Call that limit ; it is not hard to show that is continuous and that . Theorem 43.6 in Munkres is a slight generalization of this. Drorbn 07:12, 20 December 2010 (EST)

Thanks Dror.

Everybody good luck on the exam!-Kai