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Give an example of a perfect set in $\mathbb R^n$ that does not contain any of the rationals.

(Or prove that it does not exist).

  • 0
    Hint: It exists, and you can even construct such a set of specified Lebesgue measure.2010-07-28
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    What is a perfect set? Also, this looks like a homework problem.2010-07-28
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    Why are you asking if you apparently know the answer?2010-07-28
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    @Kevin: Line Bundle is not asking homework questions - no-one would have homework on so many different areas at once2010-07-28
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    @Kevin, a perfect set is one which is equal to its derived set, as in http://en.wikipedia.org/wiki/Perfect_set2010-07-28
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    Mariano has started a discussion at meta: http://meta.math.stackexchange.com/questions/313/asking-questions-whose-answers-are-clearly-known-to-the-op2010-07-28
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    @Casebash: It may not be a homework question assigned to the asker, but I agree that it looks like one.2010-07-28
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    See also: http://math.stackexchange.com/q/381690/462 If $C$ is the standard Cantor set, then for comeager many $x$, we have that $C+x$ consists only of irrationals.2013-05-05

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