10-327/Solution to Almost Disjoint Subsets - Revision history

Drorbn at 10:48, 20 October 2010

2010-10-20T10:48:13Z

← Older revision		Revision as of 06:48, 20 October 2010
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	*I don't have time to write out the whole proof, and haven't gone over it completely yet but it seems to work. Showing the the function is injective gives uncountablity. And proving that if they have an infinite intersection they have the same preimage, which is just a single point by injective, they are the same set. - John		*I don't have time to write out the whole proof, and haven't gone over it completely yet but it seems to work. Showing the the function is injective gives uncountablity. And proving that if they have an infinite intersection they have the same preimage, which is just a single point by injective, they are the same set. - John
	** Your solution works, though there are simpler solutions. [[User:Drorbn\|Drorbn]] 20:57, 19 October 2010 (EDT)		** Your solution works, though there are simpler solutions. [[User:Drorbn\|Drorbn]] 20:57, 19 October 2010 (EDT)

			Back to [[10-327/Classnotes for Thursday October 14]].

Drorbn at 00:57, 20 October 2010

2010-10-20T00:57:07Z

← Older revision		Revision as of 20:57, 19 October 2010
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	*I don't have time to write out the whole proof, and haven't gone over it completely yet but it seems to work. Showing the the function is injective gives uncountablity. And proving that if they have an infinite intersection they have the same preimage, which is just a single point by injective, they are the same set. - John		*I don't have time to write out the whole proof, and haven't gone over it completely yet but it seems to work. Showing the the function is injective gives uncountablity. And proving that if they have an infinite intersection they have the same preimage, which is just a single point by injective, they are the same set. - John
			** Your solution works, though there are simpler solutions. [[User:Drorbn\|Drorbn]] 20:57, 19 October 2010 (EDT)

Johnfleming at 00:06, 20 October 2010

2010-10-20T00:06:05Z

New page

I think that this collection satisfies the properties.

Let <math>A</math> be the set of all infinite sequences of 0's and 1's.
Let <math>x \in A</math> and <math> x=x_1,x_2,... </math> with <math> x_i \in \{0,1\}</math>
Let <math> p_i </math> be the ith prime number i.e. <math> p_1 = 2, p_2=3, p_3=5, </math> etc.

Let <math>f:A \to 2^{\mathbb{N}}</math>

Such that <math> f(x)=\bigcup_{n=0}^{\infty}\{\prod_{i=0}^n p_i^{x_i}\}</math>

ie <math>10101010101010.... \to \{2,2*5,2*5*11,2*5*11*17,...\}</math>

Then <math> f(A) </math> is a collection of sets with the desired properties (I think).

*I don't have time to write out the whole proof, and haven't gone over it completely yet but it seems to work. Showing the the function is injective gives uncountablity. And proving that if they have an infinite intersection they have the same preimage, which is just a single point by injective, they are the same set. - John