10-327/Classnotes for Monday December 6: Difference between revisions
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[http://katlas.math.toronto.edu/drorbn/index.php?title=Image:327-notes_for_120610.pdf Lecture Notes] |
[http://katlas.math.toronto.edu/drorbn/index.php?title=Image:327-notes_for_120610.pdf Lecture Notes] |
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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) |
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*Question. |
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** 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) |
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-Kai[[User:Xwbdsb|Xwbdsb]] 00:01, 20 December 2010 (EST) |
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Revision as of 07:12, 20 December 2010
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See some blackboard shots at BBS/10_327-101206-142909.jpg.
Video: 
Topology-101206
| Dror's notes above / Student's notes below |
- 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 [math]\displaystyle{ [0,1] }[/math] is not used. As we said in class, if [math]\displaystyle{ (f_n) }[/math] is Cauchy in the uniform metric, then for any [math]\displaystyle{ x }[/math], the sequence [math]\displaystyle{ (f_n(x)) }[/math] is Cauchy in [math]\displaystyle{ {\mathbb R} }[/math], so it has a limit. Call that limit [math]\displaystyle{ f(x) }[/math]; it is not hard to show that [math]\displaystyle{ f }[/math] is continuous and that [math]\displaystyle{ f_n\to f }[/math]. Theorem 43.6 in Munkres is a slight generalization of this. Drorbn 07:12, 20 December 2010 (EST)
