10-327/Classnotes for Thursday November 4 - Revision history

Xwbdsb at 12:43, 19 December 2010

2010-12-19T12:43:28Z

← Older revision		Revision as of 08:43, 19 December 2010
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	Without loss of generality <math> \epsilon < \Delta (x)</math> otherwise the condition holds vacuously.		Without loss of generality <math> \epsilon < \Delta (x)</math> otherwise the condition holds vacuously.
	Suppose we construct a ball of radius r such that <math> r > \epsilon </math> around x so that it is a subset of some U in the cover, then we can construct a ball of radius <math>r - \epsilon</math> around y such that this ball is also in U. So <math> \Delta (y) \geq r-\epsilon </math> for all r such that a ball of radius r exists around x as in the proof. Which implies <math> \Delta (y) \geq sup(r) - \epsilon </math> implies <math> \Delta (y) \geq \Delta (x) - \epsilon </math>. Hopefully this makes sense and works - [[Johnfleming\|John]]		Suppose we construct a ball of radius r such that <math> r > \epsilon </math> around x so that it is a subset of some U in the cover, then we can construct a ball of radius <math>r - \epsilon</math> around y such that this ball is also in U. So <math> \Delta (y) \geq r-\epsilon </math> for all r such that a ball of radius r exists around x as in the proof. Which implies <math> \Delta (y) \geq sup(r) - \epsilon </math> implies <math> \Delta (y) \geq \Delta (x) - \epsilon </math>. Hopefully this makes sense and works - [[Johnfleming\|John]]
			-Yes. Makes sense. Nice and concise thanks! -Kai

Johnfleming at 02:35, 19 December 2010

2010-12-19T02:35:46Z

← Older revision		Revision as of 22:35, 18 December 2010
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	*I think you are correct, (if you follow the proof on the blackboard shots to the letter) as an example of what you are talking about take the open cover of [0,1] using sets of the form [0,x) for x<3/4 and (y,1] for y>1/4. Then <math>\Delta (x)\geq 1/4</math> and at 1/2 <math>\Delta (1/2)=1/4</math> but no ball of radius 1/4 around 1/2 fits inside any of the sets. So we will need to take a smaller value than <math>inf(\Delta (x))</math> as our value of <math>\delta_0</math> for Lebesgue's Lemma. Dividing by two as you suggest should work fine to fix this problem... Maybe it's just me, but in proof's like this I always feel the urge to divide the final answer by 2, just in case I mixed up some some strict inequality, with a non-strict one somewhere - [[Johnfleming\|John]]		*I think you are correct, (if you follow the proof on the blackboard shots to the letter) as an example of what you are talking about take the open cover of [0,1] using sets of the form [0,x) for x<3/4 and (y,1] for y>1/4. Then <math>\Delta (x)\geq 1/4</math> and at 1/2 <math>\Delta (1/2)=1/4</math> but no ball of radius 1/4 around 1/2 fits inside any of the sets. So we will need to take a smaller value than <math>inf(\Delta (x))</math> as our value of <math>\delta_0</math> for Lebesgue's Lemma. Dividing by two as you suggest should work fine to fix this problem... Maybe it's just me, but in proof's like this I always feel the urge to divide the final answer by 2, just in case I mixed up some some strict inequality, with a non-strict one somewhere - [[Johnfleming\|John]]
	**Thanks John very nice example. Can you also help me with this question? "delta(x) isn't a radius that we can fit a ball inside one of the U's. It is the supremum of all possible radius. Wouldn't that give us a problem in showing delta(y)>=delta(x)-epsilon when d(x,y)<epsilon?"-Kai		**Thanks John very nice example. Can you also help me with this question? "delta(x) isn't a radius that we can fit a ball inside one of the U's. It is the supremum of all possible radius. Wouldn't that give us a problem in showing delta(y)>=delta(x)-epsilon when d(x,y)<epsilon?"-Kai
			***I don't think there is any problem with this step.
			Without loss of generality <math> \epsilon < \Delta (x)</math> otherwise the condition holds vacuously.
			Suppose we construct a ball of radius r such that <math> r > \epsilon </math> around x so that it is a subset of some U in the cover, then we can construct a ball of radius <math>r - \epsilon</math> around y such that this ball is also in U. So <math> \Delta (y) \geq r-\epsilon </math> for all r such that a ball of radius r exists around x as in the proof. Which implies <math> \Delta (y) \geq sup(r) - \epsilon </math> implies <math> \Delta (y) \geq \Delta (x) - \epsilon </math>. Hopefully this makes sense and works - [[Johnfleming\|John]]

Xwbdsb at 23:13, 18 December 2010

2010-12-18T23:13:31Z

← Older revision		Revision as of 19:13, 18 December 2010
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	And once we found this delta(x_0) we should divide by 2 so that delta(x)>delta(x_0) for all x in X? Because it is not necessarily true that we can find a ball of radius delta(x_0) around x_0 to fit into ball?-Kai		And once we found this delta(x_0) we should divide by 2 so that delta(x)>delta(x_0) for all x in X? Because it is not necessarily true that we can find a ball of radius delta(x_0) around x_0 to fit into ball?-Kai
	*I think you are correct, (if you follow the proof on the blackboard shots to the letter) as an example of what you are talking about take the open cover of [0,1] using sets of the form [0,x) for x<3/4 and (y,1] for y>1/4. Then <math>\Delta (x)\geq 1/4</math> and at 1/2 <math>\Delta (1/2)=1/4</math> but no ball of radius 1/4 around 1/2 fits inside any of the sets. So we will need to take a smaller value than <math>inf(\Delta (x))</math> as our value of <math>\delta_0</math> for Lebesgue's Lemma. Dividing by two as you suggest should work fine to fix this problem... Maybe it's just me, but in proof's like this I always feel the urge to divide the final answer by 2, just in case I mixed up some some strict inequality, with a non-strict one somewhere - [[Johnfleming\|John]]		*I think you are correct, (if you follow the proof on the blackboard shots to the letter) as an example of what you are talking about take the open cover of [0,1] using sets of the form [0,x) for x<3/4 and (y,1] for y>1/4. Then <math>\Delta (x)\geq 1/4</math> and at 1/2 <math>\Delta (1/2)=1/4</math> but no ball of radius 1/4 around 1/2 fits inside any of the sets. So we will need to take a smaller value than <math>inf(\Delta (x))</math> as our value of <math>\delta_0</math> for Lebesgue's Lemma. Dividing by two as you suggest should work fine to fix this problem... Maybe it's just me, but in proof's like this I always feel the urge to divide the final answer by 2, just in case I mixed up some some strict inequality, with a non-strict one somewhere - [[Johnfleming\|John]]
			**Thanks John very nice example. Can you also help me with this question? "delta(x) isn't a radius that we can fit a ball inside one of the U's. It is the supremum of all possible radius. Wouldn't that give us a problem in showing delta(y)>=delta(x)-epsilon when d(x,y)<epsilon?"-Kai

Johnfleming at 21:22, 18 December 2010

2010-12-18T21:22:15Z

← Older revision		Revision as of 17:22, 18 December 2010
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	***I have some doubts with Lebesgue number lemma too.. this delta(x) isn't a radius that we can fit a ball inside one of the U's. It is the supremum of all possible radius. Wouldn't that give us a problem? Don't we need to subtract some small positive value and then find a valid radius? I am not sure if there should be some technicality involved here.-Kai		***I have some doubts with Lebesgue number lemma too.. this delta(x) isn't a radius that we can fit a ball inside one of the U's. It is the supremum of all possible radius. Wouldn't that give us a problem? Don't we need to subtract some small positive value and then find a valid radius? I am not sure if there should be some technicality involved here.-Kai
	And once we found this delta(x_0) we should divide by 2 so that delta(x)>delta(x_0) for all x in X? Because it is not necessarily true that we can find a ball of radius delta(x_0) around x_0 to fit into ball?-Kai		And once we found this delta(x_0) we should divide by 2 so that delta(x)>delta(x_0) for all x in X? Because it is not necessarily true that we can find a ball of radius delta(x_0) around x_0 to fit into ball?-Kai
			*I think you are correct, (if you follow the proof on the blackboard shots to the letter) as an example of what you are talking about take the open cover of [0,1] using sets of the form [0,x) for x<3/4 and (y,1] for y>1/4. Then <math>\Delta (x)\geq 1/4</math> and at 1/2 <math>\Delta (1/2)=1/4</math> but no ball of radius 1/4 around 1/2 fits inside any of the sets. So we will need to take a smaller value than <math>inf(\Delta (x))</math> as our value of <math>\delta_0</math> for Lebesgue's Lemma. Dividing by two as you suggest should work fine to fix this problem... Maybe it's just me, but in proof's like this I always feel the urge to divide the final answer by 2, just in case I mixed up some some strict inequality, with a non-strict one somewhere - [[Johnfleming\|John]]

Xwbdsb at 05:15, 18 December 2010

2010-12-18T05:15:38Z

← Older revision		Revision as of 01:15, 18 December 2010
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	** If you could find a ball of radius 7 around <math>x</math> which fits inside some set <math>U</math>, and you move <math>x</math> just a 1 unit away to <math>y</math>, then by the triangle inequality the ball of radius 6 around <math>y</math> is entirely contained inside the ball of radius 7 around <math>x</math> so it is entirely contained in <math>U</math>. [[User:Drorbn\|Drorbn]] 18:37, 6 November 2010 (EDT)		** If you could find a ball of radius 7 around <math>x</math> which fits inside some set <math>U</math>, and you move <math>x</math> just a 1 unit away to <math>y</math>, then by the triangle inequality the ball of radius 6 around <math>y</math> is entirely contained inside the ball of radius 7 around <math>x</math> so it is entirely contained in <math>U</math>. [[User:Drorbn\|Drorbn]] 18:37, 6 November 2010 (EDT)
	***I have some doubts with Lebesgue number lemma too.. this delta(x) isn't a radius that we can fit a ball inside one of the U's. It is the supremum of all possible radius. Wouldn't that give us a problem? Don't we need to subtract some small positive value and then find a valid radius? I am not sure if there should be some technicality involved here.-Kai		***I have some doubts with Lebesgue number lemma too.. this delta(x) isn't a radius that we can fit a ball inside one of the U's. It is the supremum of all possible radius. Wouldn't that give us a problem? Don't we need to subtract some small positive value and then find a valid radius? I am not sure if there should be some technicality involved here.-Kai
			And once we found this delta(x_0) we should divide by 2 so that delta(x)>delta(x_0) for all x in X? Because it is not necessarily true that we can find a ball of radius delta(x_0) around x_0 to fit into ball?-Kai

Xwbdsb at 04:51, 18 December 2010

2010-12-18T04:51:35Z

← Older revision		Revision as of 00:51, 18 December 2010
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	* Question: Regarding the proof of the Lebesque Number Lemma, let T(x) be the function we were working with in the proof. I am confused with how we reached the conclusion that if d(x,y)<E, then T(y)>= T(x) - E. I know that it was said that this is just an application of the triangle inequality, but I am having a bit of trouble seeing that. Hopefully someone can make this point a bit clearer for me. Thanks! Jason.		* Question: Regarding the proof of the Lebesque Number Lemma, let T(x) be the function we were working with in the proof. I am confused with how we reached the conclusion that if d(x,y)<E, then T(y)>= T(x) - E. I know that it was said that this is just an application of the triangle inequality, but I am having a bit of trouble seeing that. Hopefully someone can make this point a bit clearer for me. Thanks! Jason.
	** If you could find a ball of radius 7 around <math>x</math> which fits inside some set <math>U</math>, and you move <math>x</math> just a 1 unit away to <math>y</math>, then by the triangle inequality the ball of radius 6 around <math>y</math> is entirely contained inside the ball of radius 7 around <math>x</math> so it is entirely contained in <math>U</math>. [[User:Drorbn\|Drorbn]] 18:37, 6 November 2010 (EDT)		** If you could find a ball of radius 7 around <math>x</math> which fits inside some set <math>U</math>, and you move <math>x</math> just a 1 unit away to <math>y</math>, then by the triangle inequality the ball of radius 6 around <math>y</math> is entirely contained inside the ball of radius 7 around <math>x</math> so it is entirely contained in <math>U</math>. [[User:Drorbn\|Drorbn]] 18:37, 6 November 2010 (EDT)
			***I have some doubts with Lebesgue number lemma too.. this delta(x) isn't a radius that we can fit a ball inside one of the U's. It is the supremum of all possible radius. Wouldn't that give us a problem? Don't we need to subtract some small positive value and then find a valid radius? I am not sure if there should be some technicality involved here.-Kai

Drorbn at 22:37, 6 November 2010

2010-11-06T22:37:43Z

← Older revision		Revision as of 18:37, 6 November 2010
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	{{Template:10-327:Dror/Students Divider}}		{{Template:10-327:Dror/Students Divider}}

	Question: Regarding the proof of the Lebesque Number Lemma, let T(x) be the function we were working with in the proof. I am confused with how we reached the conclusion that if d(x,y)<E, then T(y)>= T(x) - E. I know that it was said that this is just an application of the triangle inequality, but I am having a bit of trouble seeing that. Hopefully someone can make this point a bit clearer for me. Thanks! Jason.		* Question: Regarding the proof of the Lebesque Number Lemma, let T(x) be the function we were working with in the proof. I am confused with how we reached the conclusion that if d(x,y)<E, then T(y)>= T(x) - E. I know that it was said that this is just an application of the triangle inequality, but I am having a bit of trouble seeing that. Hopefully someone can make this point a bit clearer for me. Thanks! Jason.
			** If you could find a ball of radius 7 around <math>x</math> which fits inside some set <math>U</math>, and you move <math>x</math> just a 1 unit away to <math>y</math>, then by the triangle inequality the ball of radius 6 around <math>y</math> is entirely contained inside the ball of radius 7 around <math>x</math> so it is entirely contained in <math>U</math>. [[User:Drorbn\|Drorbn]] 18:37, 6 November 2010 (EDT)

Jdw at 17:37, 6 November 2010

2010-11-06T17:37:06Z

← Older revision		Revision as of 13:37, 6 November 2010
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	See some blackboard shots at {{BBS Link\|10_327-101104-142342.jpg}}.		See some blackboard shots at {{BBS Link\|10_327-101104-142342.jpg}}.

			{{Template:10-327:Dror/Students Divider}}

			Question: Regarding the proof of the Lebesque Number Lemma, let T(x) be the function we were working with in the proof. I am confused with how we reached the conclusion that if d(x,y)<E, then T(y)>= T(x) - E. I know that it was said that this is just an application of the triangle inequality, but I am having a bit of trouble seeing that. Hopefully someone can make this point a bit clearer for me. Thanks! Jason.

Drorbn at 20:53, 4 November 2010

2010-11-04T20:53:15Z

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