0708-1300/Class notes for Thursday, October 11 - Revision history

Trefor: /* Typed Class Notes */

2007-11-07T20:36:04Z

Typed Class Notes

← Older revision		Revision as of 16:36, 7 November 2007
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	''Proof''		''Proof''

	Let A be ~~coverd~~ by <math>A_i</math>, and B by <math>B_i</math> (which,respectively, have volumes that sum less than <math>\epsilon</math> )		Let A be covered by <math>A_i</math>, and B by <math>B_i</math> (which,respectively, have volumes that sum less than <math>\epsilon</math> )

	But then <math>A\times B</math> is covered by <math>A_i\times B_i</math> which has <math>\sum_{i,j} vol(A_i\times B_j)= \sum_{i}vol(A_i)\times\sum_{j}vol(B_{j})</math> which can be made arbitrarily small by assumption.		But then <math>A\times B</math> is covered by <math>A_i\times B_i</math> which has <math>\sum_{i,j} vol(A_i\times B_j)= \sum_{i}vol(A_i)\times\sum_{j}vol(B_{j})</math> which can be made arbitrarily small by assumption.
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	''Proof''		''Proof''

	Let each <math>A_i</math> in the countable union be covered by <math>B_{j}^{i}</math> so that <math>\sum_{j}vol(B_{j}^{i})<\frac{\epsilon}{s^i}</math>		Let each <math>A_i</math> in the countable union be covered by <math>B_{j}^{i}</math> so that <math>\sum_{j}vol(B_{j}^{i})<\frac{\epsilon}{2^i}</math>

	Hence, <math>\mathcal{B} = \{B_{j}^{i}\}</math> covers the union and <math>\sum_{B^{i}_{j}\in\mathcal{B}}vol(B^{i}_{j}<\sum_{j}\frac{\epsilon}{s^i} = \epsilon'</math>		Hence, <math>\mathcal{B} = \{B_{j}^{i}\}</math> covers the union and <math>\sum_{B^{i}_{j}\in\mathcal{B}}vol(B^{i}_{j})<\sum_{j}\frac{\epsilon}{2^i} = \epsilon'</math>

	''Q.E.D''		''Q.E.D''

Trefor: /* Typed Class Notes */

2007-10-16T17:50:20Z

Typed Class Notes

← Older revision		Revision as of 13:50, 16 October 2007
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	==Typed Class Notes==		==Typed Class Notes==

			<span style="color: red;">The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.</span>

	''Note'': Assignment 1 was handed back		''Note'': Assignment 1 was handed back

Trefor at 20:18, 11 October 2007

2007-10-11T20:18:49Z

← Older revision		Revision as of 16:18, 11 October 2007
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	Strange as it may seem, the faded and barely visible Cantor aerogel in the square at the bottom right of the image above is still thick enough to block all diagonal light rays.		Strange as it may seem, the faded and barely visible Cantor aerogel in the square at the bottom right of the image above is still thick enough to block all diagonal light rays.

	~~'''~~Proof~~:'''~~		===Proof===

	----
	Take an arbitrary diagonal line intersecting the figure. Observe that in the first step figure the condition holds (the diagonal line intersects the black squares). Observe now that when moving to the second figure the given line will intersect a black square too. This means that if a diagonal line intersects the black set then it will intersect the set of the next step.		Take an arbitrary diagonal line intersecting the figure. Observe that in the first step figure the condition holds (the diagonal line intersects the black squares). Observe now that when moving to the second figure the given line will intersect a black square too. This means that if a diagonal line intersects the black set then it will intersect the set of the next step.

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	==Class Notes==		==Typed Class Notes==

	''Note'': Assignment 1 was handed back		''Note'': Assignment 1 was handed back

Trefor at 20:16, 11 October 2007

2007-10-11T20:16:26Z

Show changes

Drorbn at 14:02, 11 October 2007

2007-10-11T14:02:05Z

← Older revision		Revision as of 10:02, 11 October 2007
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	{{0708-1300/Navigation}}		{{0708-1300/Navigation}}

			==Today's Agenda: Intro to Sard==
			* The Cantor set <math>C</math>.
			* Sets of measure zero.
			* <math>C</math> is of measure zero.
			* But <math>C+C</math> is <math>[0,2]</math> (in two ways).
			* Thick Cantor sets.
			* "Measure zero" is a smooth invariant.
			* Measure 0 makes sense on manifolds.
			* Baby Sard for <math>C^1</math> functions <math>f:{\mathbb R}\to{\mathbb R}</math>:
			** Such an <math>f</math> with a Cantor set of singular points and values.
			** And yet a proof.
			* A counterexample for Sard for <math>C^1</math> functions <math>f:{\mathbb R}^2\to{\mathbb R}</math>.

	==The Cantor Aerogel==		==The Cantor Aerogel==

Franklin at 13:40, 11 October 2007

2007-10-11T13:40:16Z

← Older revision		Revision as of 09:40, 11 October 2007
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	----		----
	Observe that in the first step figure the condition holds (~~every~~ diagonal line intersects the black squares).		Take an arbitrary diagonal line intersecting the figure. Observe that in the first step figure the condition holds (the diagonal line intersects the black squares). Observe now that when moving to the second figure the given line will intersect a black square too. This means that if a diagonal line intersects the black set then it will intersect the set of the next step.
	Observe now that when moving to the second figure the given line will intersect a square too. This means that if a diagonal line intersects the black set then it will intersect the set of the next step.

	Since in each step the set is self similar i.e. in each quarter of the square you have the black set of the previous step, then the intersection point of the diagonal line is contained in the same quarter of the previous intersection. Since this ~~quarter~~ are nested and the diagonal of the ~~quarter~~ tents to zero, the sequence of intersection form a convergent sequence which converges inside the square (which is compact) and inside the segment of the line inside the square (which is compact too). Since the Cantor set is closed the limit point is in the Cantor set.		Since in each step the set is self similar i.e. in each quarter of the square you have the black set of the previous step, then the intersection point of the diagonal line is contained in the same quarter of the previous intersection. Since this quarters are nested and the diagonal of the quarters tents to zero, the sequence of intersection form a convergent sequence which converges inside the square (which is compact) and inside the segment of the line inside the square (which is compact too). Since the Cantor set is closed the limit point is in the Cantor set. It is inside the line segment as well.

	'''QED'''		'''QED'''

Franklin at 13:36, 11 October 2007

2007-10-11T13:36:58Z

← Older revision		Revision as of 09:36, 11 October 2007
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	Strange as it may seem, the faded and barely visible Cantor aerogel in the square at the bottom right of the image above is still thick enough to block all diagonal light rays.		Strange as it may seem, the faded and barely visible Cantor aerogel in the square at the bottom right of the image above is still thick enough to block all diagonal light rays.

			'''Proof:'''

			----
			Observe that in the first step figure the condition holds (every diagonal line intersects the black squares).
			Observe now that when moving to the second figure the given line will intersect a square too. This means that if a diagonal line intersects the black set then it will intersect the set of the next step.

			Since in each step the set is self similar i.e. in each quarter of the square you have the black set of the previous step, then the intersection point of the diagonal line is contained in the same quarter of the previous intersection. Since this quarter are nested and the diagonal of the quarter tents to zero, the sequence of intersection form a convergent sequence which converges inside the square (which is compact) and inside the segment of the line inside the square (which is compact too). Since the Cantor set is closed the limit point is in the Cantor set.

			'''QED'''

Drorbn at 01:13, 11 October 2007

2007-10-11T01:13:59Z

New page

{{0708-1300/Navigation}}

==The Cantor Aerogel==
[[Image:0708-1300-The Cantor Aerogel.png|520px]]

Strange as it may seem, the faded and barely visible Cantor aerogel in the square at the bottom right of the image above is still thick enough to block all diagonal light rays.