10-327/Solution to Almost Disjoint Subsets: Difference between revisions
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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 |
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** 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) |
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Back to [[10-327/Classnotes for Thursday October 14]]. |
Latest revision as of 05:48, 20 October 2010
I think that this collection satisfies the properties.
Let be the set of all infinite sequences of 0's and 1's. Let and with Let be the ith prime number i.e. etc.
Let
Such that
ie
Then 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
- Your solution works, though there are simpler solutions. Drorbn 20:57, 19 October 2010 (EDT)