Let $A$ be the Cantor set and $B = \pi\, (-A \cup A)$.
Is B a perfect set with no rational numbers?
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Oops, sorry! I misread the question. – Asaf Karagila May 29 '14 at 16:27
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4Is $B=\pi(-A\cup A)$? You know $A$ contains $0$, right? – Andrés E. Caicedo May 29 '14 at 16:29
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I edited the question to use LaTeX. Please let me know if I have misrepresented your question inadvertently. – apnorton May 29 '14 at 16:30
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1If you just wanted to construct some such set, see http://math.stackexchange.com/q/1064/ – May 30 '14 at 01:49
1 Answers
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As Andres Caicedo pointed out, $B$ contains $0$.
It is reasonable to ask whether $B$ contains a nonzero rational number. This is equivalent to asking whether $A$ contains a rational multiple of $\pi$ (other than $0$).
If the answer is affirmative, then the base $3$ expansion of $\pi$ has a (complicated) infinite pattern in its digits. It is generally believed that the digits of $\pi$, in any number base, do not exhibit any infinite pattern. However, for all we know at present, the base 3 expansion of $\pi$ could have finitely many $1$s. (See Distribution of the digits of Pi and links there.) So, the answer to your question (with excluded $0$) is: nobody knows.
