10-327/Classnotes for Monday December 6 - Revision history

Xwbdsb at 00:14, 21 December 2010

2010-12-21T00:14:04Z

← Older revision		Revision as of 20:14, 20 December 2010
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	Everybody good luck on the exam!-Kai		Everybody good luck on the exam!-Kai

			Great course! Thank you very much for all your help Dror and all the classmates in this class.
			-Kai

Xwbdsb at 15:45, 20 December 2010

2010-12-20T15:45:45Z

← Older revision		Revision as of 11:45, 20 December 2010
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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

Drorbn at 12:12, 20 December 2010

2010-12-20T12:12:40Z

← Older revision		Revision as of 08:12, 20 December 2010
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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]

		⚫	*'''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.
			** 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)
⚫	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)

Xwbdsb at 05:01, 20 December 2010

2010-12-20T05:01:31Z

← Older revision		Revision as of 01:01, 20 December 2010
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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]

			*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)

Anne.d at 05:38, 11 December 2010

2010-12-11T05:38:45Z

← Older revision		Revision as of 01:38, 11 December 2010
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			[http://katlas.math.toronto.edu/drorbn/index.php?title=Image:327-notes_for_120610.pdf Lecture Notes]

Drorbn at 20:42, 6 December 2010

2010-12-06T20:42:15Z

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