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Feb 21, 2012 at 9:30 vote accept Julie
Feb 21, 2012 at 7:38 answer added Thierry de la Rue timeline score: 1
Feb 21, 2012 at 7:28 comment added Dan Petersen This question still doesn't make sense. You can't have uncountably many disjoint open balls in euclidean space.
Feb 21, 2012 at 6:28 history edited Julie CC BY-SA 3.0
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Feb 21, 2012 at 6:21 history edited Neil Strickland CC BY-SA 3.0
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Feb 21, 2012 at 6:19 answer added Neil Strickland timeline score: 1
Feb 21, 2012 at 6:12 history edited Julie CC BY-SA 3.0
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Feb 21, 2012 at 6:08 comment added Julie @ David and Yoav. Ah, thanks. I have updated my additional info.
Feb 21, 2012 at 4:46 comment added David Feldman BTW, I don't understand your "Additional info." An open set doesn't contain its boundary?!? Right?
Feb 21, 2012 at 4:45 comment added David Feldman $R^{mxn}$ is $\sigma$−compact so an uncountable open cover has a countable subcover. So you're good, right?
Feb 21, 2012 at 4:38 comment added Yoav Kallus Your additional info imply that $A_j$ is an open set which contains its boundary, making it closed. Did you mean to write that $A_j$ is the closure of an open set?
Feb 21, 2012 at 3:39 history edited Julie CC BY-SA 3.0
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Feb 21, 2012 at 3:11 comment added Julie Thanks for all the comments. I have added some more information to my question. Best, Julie
Feb 21, 2012 at 1:51 comment added user9072 If you allow uncountably many sets just to get what you claim you could make the Ah's singletons. So, by itself this cannot possibly tell much. I am afraid without further details there is not much specufuc to answer.
Feb 21, 2012 at 1:48 answer added Gerald Edgar timeline score: 0
Feb 21, 2012 at 1:42 history asked Julie CC BY-SA 3.0