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| ``Re: A nonempty perfect subset of R''
by cgibbard on 2003-11-16 04:36:59 |
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| Oops :)
Sorry about that - I was using a different definition than is here (essentially a point in the closure). Yes, the condition that it should actually be in the closure of the set with the point removed makes things more difficult.
It might be possible to translate a Cantor set by some irrational x to get rid of the rational points, but yes, you'd have to be careful about not allowing x added to any irrational point in the Cantor set to be rational. That sort of thing might get tricky to work with. I'll have to look at it a bit more.
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