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Sat, 14 Jun 2014 05:21:50 +0000hourly1http://wordpress.com/Comment on https www quora com What is it about… by Any inspiring biographies here http en wikipedia org… | riceissa
https://riceissa.wordpress.com/2014/06/14/https-www-quora-com-what-is-it-about/comment-page-1/#comment-27
Sat, 14 Jun 2014 05:21:50 +0000http://riceissa.wordpress.com/2014/06/14/https-www-quora-com-what-is-it-about/#comment-27[…] EDIT: see also https://riceissa.wordpress.com/2014/06/14/https-www-quora-com-what-is-it-about/ […]
]]>Comment on Two definitions for the least upper bound by This post http riceissa wordpress com 2014 03… | riceissa
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Wed, 16 Apr 2014 07:28:12 +0000http://riceissa.wordpress.com/?p=199#comment-9[…] post https://riceissa.wordpress.com/2014/03/18/two-definitions-for-the-least-upper-bound/ has been ported into Markdown and is hosted on my GitHub website here […]
]]>Comment on An exercise from Halmos’s set theory book by Luís Sequeira
https://riceissa.wordpress.com/2012/07/08/an-exercise-from-halmoss-set-theory-book/comment-page-1/#comment-7
Wed, 12 Dec 2012 14:34:53 +0000http://riceissa.wordpress.com/?p=4#comment-7For the second implication, it is probably easier to argue by contrapositive: assume C is not a subset of A; let x be an element in C that does not belong to A; then it is clear that x is in one of the sets and not the other.
]]>Comment on An exercise from Halmos’s set theory book by gowers
https://riceissa.wordpress.com/2012/07/08/an-exercise-from-halmoss-set-theory-book/comment-page-1/#comment-6
Fri, 10 Aug 2012 18:05:36 +0000http://riceissa.wordpress.com/?p=4#comment-6I agree that the question is interesting. However, I’m not sure that the proof you give is what Halmos is hinting at when he says, “Observe that the condition has nothing to do with .” It’s notable that the result is true when and also when is the universal set of which all three sets are presumably subsets. How can we exploit this to show the result for an arbitrary ? One idea is to use the fact that two sets are equal if their intersections with are equal and their intersections with are equal. These two cases amount to treating as the universal set and then treating it as the empty set. So we’re done. (I need to think a bit more about how I’d present that argument rigorously.)
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